Factor out the greatest common factor.
4x³+16x² +8x

(a) f(x) =  \(log_2(x)\) is a logarithmic function

(b) g(x) = ∜x is a root function

(c) h(x) = \(2x^3/(1 - x^2)\) is a rational function

(d) u(t) = 1 - 1.1t + \(2.54t^2\) is a polynomial of degree 2

(e) v(t) = \(5^t\)is an exponential function

(f) w(θ) = sin θ \(cos^{2}\theta\) is a trigonometric function.

(a) f(x) = \(log_2(x)\) is a logarithmic function. Logarithmic functions have the logarithm of the independent variable as the output. Here, the logarithm base is 2.

(b) g(x) = ∜x is a root function. Root functions have the square root or higher roots of the independent variable as the output. Here, the root is a cube root.

(c) h(x) = \(2x^3/(1 - x^2)\) is a rational function. Rational functions are functions that are expressed as the quotient of two polynomials. Here, the numerator is a cubic polynomial and the denominator is a quadratic polynomial.

(d) u(t) = 1 - 1.1t + \(2.54t^2\) is a polynomial of degree 2. Polynomials are functions that are expressed as a sum of powers of the independent variable, with coefficients. The degree of a polynomial is the highest power of the independent variable.

(e) v(t) = \(5^t\) is an exponential function. Exponential functions have the independent variable as the exponent. Here, the base is 5.

(f) w(θ) = sin θ \(cos^{2}\theta\) is an algebraic function and a trigonometric function. Algebraic functions are functions that can be expressed using arithmetic operations and algebraic expressions. Trigonometric functions are functions that involve the ratios of the sides of a right triangle. Here, the function is a combination of sine and cosine functions.

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The complete question is -

Classify each function as a power function, root function, polynomial (state its degree), rational function, algebraic function, trigonometric function, exponential function, or logarithmic function.

(a) f(x) =  \(log_2(x)\)    

(b) g(x) = ∜x    

(c) h(x) = \(2x^3/(1 - x^2)\)    

(d) u(t) = 1 - 1.1t + \(2.54t^2\)                    

(e) v(t) = \(5^t\)    

(f) w(θ) = sin θ \(cos^{2}\theta\)

The volume of the solid obtained by rotating the region bounded by the curve y = 1 - (x - 5)² in the first quadrant about the y-axis is 51π cubic units.

To find the volume of the solid obtained by rotating the region bounded by the curve y = 1 - (x - 5)² in the first quadrant about the y-axis, we can use the method of cylindrical shells.

The formula for the volume using cylindrical shells is:

V = 2π ∫ [a, b] x * h(x) dx

Where:

- V is the volume of the solid

- π represents the mathematical constant pi

- [a, b] is the interval over which we are integrating

- x is the variable representing the x-axis

- h(x) is the height of the cylindrical shell at a given x-value

In this case, we need to solve for x in terms of y to express the equation in terms of y.

Rearranging the given equation:

x = 5 ± √(1 - y)

Since we are only interested in the region in the first quadrant, we take the positive square root:

x = 5 + √(1 - y)

Now we can rewrite the volume formula with respect to y:

V = 2π ∫ [c, d] x * h(y) dy

Where:

- [c, d] is the interval of y-values that correspond to the region in the first quadrant

To determine the interval [c, d], we set the equation equal to zero and solve for y:

1 - (x - 5)² = 0

Expanding and rearranging the equation:

(x - 5)² = 1

x - 5 = ±√1

x = 5 ± 1

Since we are only interested in the region in the first quadrant, we take the value x = 6:

x = 6

Now we can evaluate the integral to find the volume:

V = 2π ∫ [0, 1] x * h(y) dy

Where h(y) represents the height of the cylindrical shell at a given y-value.

Integrating the expression:

V = 2π ∫ [0, 1] (5 + √(1 - y)) * h(y) dy

To find h(y), we need to determine the distance between the y-axis and the curve at a given y-value. Since the curve is symmetric, h(y) is simply the x-coordinate at that point:

h(y) = 5 + √(1 - y)

Substituting this expression back into the integral:

V = 2π ∫ [0, 1] (5 + √(1 - y)) * (5 + √(1 - y)) dy

Now, we can evaluate this integral to find the volume

V = 2π ∫ [0, 1] (5 + √(1 - y)) * (5 + √(1 - y)) dy

To simplify the integral, let's expand the expression:

V = 2π ∫ [0, 1] (25 + 10√(1 - y) + 1 - y) dy

V = 2π ∫ [0, 1] (26 + 10√(1 - y) - y) dy

Now, let's integrate term by term:

\(V = 2\pi [26y + 10/3 * (1 - y)^\frac{3}{2} - 1/2 * y^2]\)] evaluated from 0 to 1

V = \(2\pi [(26 + 10/3 * (1 - 1)^\frac{3}{2} - 1/2 * 1^2) - (26 * 0 + 10/3 * (1 - 0)^\frac{3}{2} - 1/2 * 0^2)]\)

V = 2π [(26 + 0 - 1/2) - (0 + 10/3 - 0)]

V = 2π (25.5)

V = 51π cubic units

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Answer:

\(\displaystyle \frac{\pi(5e^4+8e^2-1)}{2e^4}\approx9.526\)

Step-by-step explanation:

This can be solved with either the washer (easier) or the shell method (harder). For the disk/washer method, the slice is perpendicular to the axis of revolution, whereas, for the shell method, the slice is parallel to the axis of revolution.  I'll show how to do it with both:

Shell Method (Horizontal Axis)

\(\displaystyle V=2\pi\int^d_cr(y)h(y)\,dy\)

Radius: \(r(y)=2-y\) (distance from y=2 to x-axis)

Height: \(h(y)=2-(-\ln y)=2+\ln y\) (\(y=e^{-x}\) is the same as \(x=-\ln y\))

Bounds: \([c,d]=[e^{-2},1]\) (plugging x-bounds in gets you this)

Plugging in our integral, we get:

\(\displaystyle V=2\pi\int^1_{e^{-2}}(2-y)(2+\ln y)\,dy=\frac{\pi(5e^4+8e^2-1)}{2e^4}\approx9.526\)

Washer Method (Parallel to x-axis)

\(\displaystyle V=\pi\int^b_a\biggr(R(x)^2-r(x)^2\biggr)\,dx\)

Outer Radius: \(R(x)=2-e^{-x}\) (distance between \(y=2\) and \(y=e^{-x}\))

Inner Radius: \(r(x)=2-1=1\) (distance between \(y=2\) and \(y=1\))

Bounds: \([a,b]=[0,2]\)

Plugging in our integral, we get:

\(\displaystyle V=\pi\int^2_0\biggr((2-e^{-x})^2-1^2\biggr)\,dx\\\\V=\pi\int^2_0\biggr((4-4e^{-x}+e^{-2x})-1\biggr)\,dx\\\\V=\pi\int^2_0(3-4e^{-x}+e^{-2x})\,dx\\\\V=\pi\biggr(3x+4e^{-x}-\frac{1}{2}e^{-2x}\biggr)\biggr|^2_0\\\\V=\pi\biggr[\biggr(3(2)+4e^{-2}-\frac{1}{2}e^{-2(2)}\biggr)-\biggr(3(0)+4e^{-0}-\frac{1}{2}e^{-2(0)}\biggr)\biggr]\\\\V=\pi\biggr[\biggr(6+4e^{-2}-\frac{1}{2}e^{-4}\biggr)-\biggr(4-\frac{1}{2}\biggr)\biggr]\)

\(\displaystyle V=\pi\biggr[\biggr(6+4e^{-2}-\frac{1}{2}e^{-4}\biggr)-\frac{7}{2}\biggr]\\\\V=\pi\biggr(\frac{5}{2}+4e^{-2}-\frac{1}{2}e^{-4}\biggr)\\\\V=\pi\biggr(\frac{5}{2}+\frac{4}{e^2}-\frac{1}{2e^4}\biggr)\\\\V=\pi\biggr(\frac{5e^4}{2e^4}+\frac{8e^2}{2e^4}-\frac{1}{2e^4}\biggr)\\\\V=\pi\biggr(\frac{5e^4+8e^2-1}{2e^4}\biggr)\\\\V=\frac{\pi(5e^4+8e^2-1)}{2e^4}\approx9.526\)

Use your best judgment when deciding on what method you use when visualizing the solid, but I hope this helped!

Answer:

its yes

Step-by-step explanation:

Probability is defined as the likelihood or the certainty that an event is going to occur or happen.

The probability of randomly choosing a red and then a green marble is \(3/76\).

The total number of marbles = 20

The number of green marbles= 3

The number of blue marbles = 12

The number of red marbles = 5

The probability of choosing a red marble = Number of red marbles / Total number of marbles

= 5/20

In simplest fraction form = \(\frac{1}{4}\)

We are told in the question that you keep the red marble you choose, So this means the total number of marbles left reduces to 19

The probability of choosing a green marble is =  Number of green marbles / New total number of marbles

= 3/19

Therefore, the probability of randomly choosing a red and then a green marble is

P (Red) x P(Green)

= 1/4  x 3/19

= \(3/76\)

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Solution :

Given :

2y''' + 15y'' + 24y' + 11y= 0

Let x = independent variable

\((a_0D^n + a_1D^{n-1}+a_2D^{n-2} + ....+ a_n) y) = Q(x)\)  is a differential equation.

If \(Q(x) \neq 0\)

It is non homogeneous then,

The general solution  = complementary solution + particular integral

If Q(x) = 0

It is called the homogeneous then the general solution =  complementary solution.

2y''' + 15y'' + 24y' + 11y= 0

\($(2D^3+15D^2+24D+11)y=0$\)

Auxiliary equation,

\($2m^3+15m^2+24m +11 = 0$\)

-1  | 2    15    24     11

    | 0   -2    - 13    -11  

      2    13    11       0

∴ \(2m^2+13m+11=0\)

The roots are

\($=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$\)

\($=\frac{-13\pm \sqrt{13^2-4(11)(2)}}{2(2)}$\)

\($=\frac{-13\pm9}{4}$\)

\($=-5.5, -1$\)

So, \(m_1, m_2, m_3 = -1, -1, -5.5\)

Then the general solution is :

\($= (c_1+c_2 x)e^{-x} + c_3 \ e^{-5.5x}$\)

Answer:

3

−

9

=

−

3

9

3

−

6

=

-21

Step-by-step explanation:

3x​−9y=−39, 3x−6y=−21 : x=5, y=6

Steps

[ ​3x−9y=−39 3x−6y=−21 ]

Show Steps

Isolate x for 3x−9y=−39: x=−13+3y

Substitute x=−13+3y

[ 3(−13+3y)−6y=−21 ]

Show Steps

Simplify

[ −39+3y=−21 ]

Show Steps

Isolate y for −39+3y=−21: y=6

For x=−13+3y

Substitute y=6

x=−13+3· 6

Simplify

x=5

The solutions to the system of equations are:

Answer:

Approximately 36.5885

Approximately 4.2426

Step-by-step explanation:

Please reference the drawing I've provided (sorry it's kind of awful. Drawing on a computer is hard :(

Problem 1)

So, the trapezoid can be split into a right triangle and a rectangle. To find the perimeter, we just need to find all the side lengths and add them.

We already know the dimensions of the rectangle. So, we need to find the dimensions of the right triangle.

We know that the height is 9 since it's opposite to the rectangle. Importantly, we know that the angle opposite to 9 is 60°.

This means that the other angle is 30°. So, the right triangle is a "special right triangle." In the 30-60-90 triangle, the hypotenuse is 2x, the side opposite to 60° is x√3, and the side opposite to 30° is x.

So, we know that 9 is the side opposite to 60°. Substitute and solve for x:

\(9=x\sqrt{3}\\ x=\frac{9}{\sqrt{3} } \\ x=\frac{9\sqrt3}{3}=3\sqrt3\)

So, x is 3√3. This means that the side opposite of angle 30 or d is 3√3.

And since x is 3√3, this means that the hypotenuse is 2(3√3) or 6√3.

Therefore, the perimeter of the figure would be:

\(9+6+6+3\sqrt3+6\sqrt3\\=21+9\sqrt3\\\approx36.5885\)

Problem 2:

For the bookcase, we simply need to find the value of c.

The braces that will be built is simply the hypotenuse of the right triangles. Therefore:

\(a^2+b^2=c^2\\\)

Plug in 3 for a and b:

\(3^2+3^2=c^2\\c^2=9+9=18\\c=\sqrt{18}=\sqrt{9\cdot 2}=3\sqrt2\approx4.2426\)

Therefore, each of the braces will be approximately 4.2426 feet long.

Edit: Grammar

Answer:

\(x^6 - \frac{7x^5}{5} - 3x^3 + C\)

The lengths and breadths of the rectangular shaped figures indicates;

a) i) Rs 6,900

b) 1500 m

c) (i) 180 m, (ii) 180 m, (iii) 45 m

Creative section B) (i) 86 meters, (ii) 430 meters

b) (i) 288 meters

A rectangle is a quadrilateral that has congruent facing sides and four interior right angles.

The possible part of the question includes;

The length of the fence = 27 meters

The breadth = 19 meters

The perimeter of the fencing = 2 × (27 + 19) = 92 meters

The length of three round of fencing = 3 × 92 meters = 276 meters

(i) The cost per meter = Rs 25

The total cost of fencing = Rs 25/m × 276 m = Rs 6,900

b) The length of the square track = 75 m

The distance at covered after the 5th round = 5 × 4 × 75 m = 1500 m

c) (i) The length of wire required for one round = 540 m/3 = 180 m

(ii) The perimeter of the pond = 180 m

(iii) The length of the pond = 180 m/4 = 45 m

Creative Section;

(i) The perimeter = 2 × (25 + 18) = 86 meters

(ii) The distance the girl runs after five rounds = 5 × 86 meters = 430 meters

b) i) The perimeter required to fence Mrs. Kanchhi's the vegetable garden with 3 rounds = 3 × 2 × (30 m + 18 m) = 288 m

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Answer:

\(\displaystyle\)\(\displaystyle f'(1)=-\frac{9}{e^3}\)

Step-by-step explanation:

Use Quotient Rule to find f'(x)

\(\displaystyle f(x)=\frac{3x^2+2}{e^{3x}}\\\\f'(x)=\frac{e^{3x}(6x)-(3x^2+2)(3e^{3x})}{(e^{3x})^2}\\\\f'(x)=\frac{6xe^{3x}-(9x^2+6)(e^{3x})}{e^{6x}}\\\\f'(x)=\frac{6x-(9x^2+6)}{e^{3x}}\\\\f'(x)=\frac{-9x^2+6x-6}{e^{3x}}\)

Find f'(1) using f'(x)

\(\displaystyle f'(1)=\frac{-9(1)^2+6(1)-6}{e^{3(1)}}\\\\f'(1)=\frac{-9+6-6}{e^3}\\\\f'(1)=\frac{-9}{e^3}\)

Answer:

\(f'(1)=-\dfrac{9}{e^{3}}\)

Step-by-step explanation:

Given rational function:

\(f(x)=\dfrac{3x^2+2}{e^{3x}}\)

To find the value of f'(1), we first need to differentiate the rational function to find f'(x). To do this, we can use the quotient rule.

\(\boxed{\begin{minipage}{5.5 cm}\underline{Quotient Rule for Differentiation}\\\\If $f(x)=\dfrac{g(x)}{h(x)}$ then:\\\\\\$f'(x)=\dfrac{h(x) g'(x)-g(x)h'(x)}{(h(x))^2}$\\\end{minipage}}\)

\(\textsf{Let}\;g(x)=3x^2+2 \implies g'(x)=6x\)

\(\textsf{Let}\;h(x)=e^{3x} \implies h'(x)=3e^{3x}\)

Therefore:

\(f'(x)=\dfrac{e^{3x} \cdot 6x -(3x^2+2) \cdot 3e^{3x}}{\left(e^{3x}\right)^2}\)

\(f'(x)=\dfrac{6x -(3x^2+2) \cdot 3}{e^{3x}}\)

\(f'(x)=\dfrac{6x -9x^2-6}{e^{3x}}\)

To find f'(1), substitute x = 1 into f'(x):

\(f'(1)=\dfrac{6(1) -9(1)^2-6}{e^{3(1)}}\)

\(f'(1)=\dfrac{6 -9-6}{e^{3}}\)

\(f'(1)=-\dfrac{9}{e^{3}}\)

Therefore, **x < 7** is the inequality that may be utilized to find x

A mathematical statement known as an inequality compares two expressions using an inequality sign, such as (less than), > (greater than), or (less than or equal to).

For instance, the inequality x + 2 5 signifies that "x + 2 is less than 5".

Let x represent how many days the client paid for pet sitting.

$15 per day plus a $20 fixed service fee equals the total cost of pet sitting.

We are aware that the customer's purchase was under $125. Consequently, we can write:

20 + 15x < 125

Putting this disparity simply:

15x < 105

x < 7

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Step-by-step explanation:

the inside expression of an absolute value expression can be positive or negative, but the result is only the positive one.

therefore, for our example here the negative case would be

2.5x - 6.8 = -12.9

which gives us

2.5x = -6.1

x = -6.1/2.5 = -2.44

and the positive case would be

2.5x - 6.8 = 12.9

and that gives us

2.5x = 19.7

x = 19.7/2.5 = 7.88

Answer:

\(m_{perpendicular}\) = 6

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

given

x + 6y = - 24 ( subtract x from both sides )

6y = - x - 24 ( divide through by 6 )

y = - \(\frac{1}{6}\) x - 4 ← in slope- intercept form

with slope m = - \(\frac{1}{6}\)

given a line with slope m then the slope of a line perpendicular to it is

\(m_{perpendicular}\) = - \(\frac{1}{m}\) = - \(\frac{1}{-\frac{1}{6} }\) = 6

Answer: the first sone should be it

Step-by-step explanation:

Answer:

rate: 4                             rate: -2

starting: 3                       starting: 30

y= -4x + 3                        y= -2x + 30

rate: 2                            rate: 6

starting: -4                     starting: 6

y= 2x -4                          y= 6x + 6

Step-by-step explanation:

\(m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}\)

^equation to find slope

To also find b, the starting value,  you need to take any point given and substitute it back into the equation with the slope.

However, these problems give you the y-int already [it is the point like (0,y)]

7-3/1-0 = 4/1

rate: 4

starting: 3

y= 4x + 3

28-30/1-0= -2/1

rate: -2

starting: 30

y= -2x + 30

-2 - (-4)/1-0 = 2/1

rate: 2

starting: -4

y= 2x -4

6-0/1-0 = 6/1

rate: 6

starting: 6

y= 6x + 6

The answer : (x-5)^2 .....................................(x-5)square

For example, for f(x)= x square . For x=1 ,then, y=1 .

But consider the g function. the minimum value that g function has is x is 5. When x is 5, y is 0 . So (x-5)square is the equation of this function.

The complete equation of the blue graph is (x-5)^2.

Function is a type of relation, or rule, that maps one input to specific single output.

In mathematics, a function is an expression, rule, or law that describes the relationship between one variable (the independent variable) and another variable (the dependent variable) (the dependent variable). In mathematics and the physical sciences, functions are indispensable for formulating physical relationships.

Linear function is a function whose graph is a straight line

We are given that;

Red graph equation= X^2

Now,

In blue graph it shifted by 5 units to the right

For blue graph

x=5

Putting the value of x in red graph equation

=(x-5)^2

=x^2 + 25 - 10x

Therefore, by the given graph function will be (x-5)^2.

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The variable n represents time in seconds

The variable n represents age

The variable n represents amount of money in dollars.

The variable n represents height in inches.

In each expression:

L. The variable n represents Jerry's time in seconds for the 100-yard dash. Lisa's time is n-5 seconds, indicating that Lisa runs 5 seconds faster than Jerry.

M. The variable n represents Pedro's father's age in years. Pedro's age is n + 3 years old, meaning Pedro is 3 years younger than his father.

N. The variable n represents Lucia's sister's amount of money in dollars. Lucia has n-10 dollars, indicating that Lucia has $10.00 less than her sister.

O. The variable n represents Jack's younger brother's height in inches. Jack is 2 times taller than his younger brother, so Jack's height is 2 times n inches.

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Answer:

1. 9x-8y=9

2. -18x+16y= -18 ⇒ -9x+8y= -9 ⇒ 9x-8y=9

as we see both equations are same, it means the lines overlap and there is infinite number of solutions

Step-by-step explanation:

Since we have three sides and trying to find an angle, use Law of Cosines

\(c = \sqrt{ {a}^{2} + {b}^{2} - 2ab \times \cos(d) } \)

where c is

Solving for the angle d.

\( \cos {}^{ -1 } (( \frac{ {c}^{2} - {b}^{2} - {a}^{2} }{ - 2ab} ) = d\)

where d is in degrees

The angle d is the opposite of the side c.

It does not matter what a and b is(they can be either 4 or 5)

let

c=7

a=4

b=5

\( \cos {}^{ - 1} ( \frac{ - 1}{5} ) = d\)

This angle must be within 0 and 180

We get

\(d = 101.54\)

Answer:

6 to the 3rd power ÷ 4 + 2 ×9(256-17×4)

6 to the 3rd power ÷ 4 + 2 ×9(256-68)

6 to the 3rd power ÷ 4 + 2 ×9×188

216 ÷ 4 + 2 × 1692

216 ÷ 4 + 3384

54 + 3384

=3438

===========================================================

Reason:

This graph passes the vertical line test, so it is a function.

The vertical line test is where we try to pass a single vertical line through more than one point on the curve. If such a thing is possible, then we say the curve fails the vertical line test and it's not a function.

Graph D for instance can have a vertical line through x = 1 and intersect the curve at (1,4) and (1,-4) simultaneously. The input x = 1 leads to multiple outputs (y = 4 and y = -4). Graph D fails the vertical line test for this very reason, and it is not a function. Graphs A and B are similar stories. Any vertical line itself is automatically not a function.

To have a function, any x input in the domain must lead to exactly one and only one y output in the range.

---------

Notice with graph C it is impossible to have a single vertical line intersect more than one point on the curve. So this is why graph C passes the vertical line test. Each input leads to exactly one output.

9514 1404 393

Answer:

  $11.69 per pound

Step-by-step explanation:

To find price per pound, divide price by pounds:

  $46.76/(4 lb) = $11.69/lb

Answer:

Step-by-step explanation:

67

Alijah's taxable income puts him in the first tax bracket, over $8,350 but not surpassing $33,950. Hence, his tax due is a base of $835 plus 15% of the income that exceeds $8,350. This results in a total tax payable of $850. Option B is the correct answer.

Alijah's taxable income falls within the first bracket of the tax table given, being over $8,350 but not over $33,950.

This bracket requires him to pay a tax of $835.00 plus 15% of the amount over $8,350.

He has a surplus of $100 over the $8,350 threshold (i.e., $8,450 - $8,350), and 15% of $100 equals $15.

Therefore, the total tax he owes would be the fixed amount of $835.00 plus the $15.00 calculated, which sums up to $850.00.

Thus, looking at the provided options, option B ($850.00) would be the correct answer.

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Cutting squares with side length approximately 3.8 inches from each corner will result in a box with the maximum volume.

To determine the value of x that will maximize the volume of the box, we can start by visualizing the box and expressing its volume in terms of x.

When we cut out squares with side lengths x from each corner of the rectangular cardboard, the resulting shape, when folded, will form the box without a top. The height of the box will be equal to the side length of the squares, which is x. The length and width of the base of the box will be the dimensions of the cardboard after the squares are cut out, which will be (30 - 2x) and (10 - 2x) respectively.

The volume of the box can be calculated by multiplying the dimensions of the base and the height:

V(x) = (30 - 2x)(10 - 2x)x.

To find the value of x that maximizes the volume, we can take the derivative of V(x) with respect to x, set it equal to zero, and solve for x.

dV/dx = 0.

Let's calculate the derivative of V(x):

dV/dx = [(30 - 2x)(10 - 2x)]'x + (30 - 2x)(10 - 2x) * 1.

Simplifying and setting it equal to zero:

[(30 - 2x)(10 - 2x)]'x + (30 - 2x)(10 - 2x) = 0.

Solving for this;

x = 3.8 inches

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Answer:

C

Step-by-step explanation:

 2

+2

-----

 4

Answer:

51

Step-by-step explanation:

Answer:

4x+2y=4

3x-y=-7

first one by 3

second by 4

12x+6y=12

12x-4y=-28

subtract

10y=40

y=4

plug in

3x-4=-7

add 4

3x=-3

x=-1

(-1,4)

c

Hope This Helps!!!

Answer: third option

Step-by-step explanation:

There are 2 ways to go about systems of equations normally, and that's elimination and substitution. For this particular problem, I would recommend elimination.

we see the first equation 4x+2y=4

we want to be able to simplify it as much as possible (so it's easier to solve)

so we divide all the terms by 2, giving 2x+y=2.

the second equation can't be simplified, so we set up the elimination method.

   2x+y=2

  -3x-y=-7

multiplying both equations by 2 and 3, we subtract them and get that y equals 4, which is clearly the answer.