Someone please help find a ratio that is equivalent to 4/11
actually help and don't just say random stuff​

Step-by-step explanation:

nzkaoavdvsoasb xzakoakaidbzbsva

Answer:

See attached

Step-by-step explanation:

Solution is below

Answer:

New moon, Waxing Crescent, First Quarter, Waxing Gibbous, Full moon, Waning Gibbous, Last Quarter, and finally Waning Crescent.

Step-by-step explanation:

I HOPE THIS HELPS.

THANK U!

The slope of the points in this table is equal to -3.

In Mathematics and Geometry, the slope of any straight line can be determined by using this mathematical equation;

Slope (m) = (Change in y-axis, Δy)/(Change in x-axis, Δx)

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

Based on the information provided in the table, we can logically deduce the following data points on the line:

By substituting the given points into the slope formula, we have the following;

Slope, m = (4 - 7)/(1 - 0)

Slope, m = -3/1

Slope, m = -3.

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Answer:

D

Step-by-step explanation:

Answer:f(x)=|+1

Step-by-step explanation:

The reflection of the parent absolute value function over the x-axis is:

f(x) = -|x|

First, for a point (x, y), a reflection over the x-axis just changes the sign of the y-component, then the reflected point will be (x, -y).

For the case of a function, we have:

y = f(x).

Then if we reflect the function f(x) over the x-axis, we will get:

g(x) = -f(x).

In this case, we want to reflect the parent function |x| over the x-axis, so we will get:

f(x) = -|x|

So the correct option is the first one.

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Answer:

YES,

AS ,

1/2^-4

=2^4=16.

16>1

HOPE IT HELPS

The orthogonal projection of vector v onto vector w in R^3 is (-54/14, -159/70, -159/35).

To find the orthogonal projection of v onto w, we need to calculate the scalar projection of v onto w and multiply it by the unit vector of w. The scalar projection of v onto w is given by the formula:

proj_w(v) = (v⋅w) / (w⋅w) * w

where ⋅ denotes the dot product.

Calculating the dot product of v and w:

v⋅w = (-7)(-5) + (6)(-3) + (-6)(-6) = 35 + (-18) + 36 = 53

Calculating the dot product of w with itself:

w⋅w = (-5)(-5) + (-3)(-3) + (-6)(-6) = 25 + 9 + 36 = 70

Now, substituting these values into the formula, we have:

proj_w(v) = (53/70) * (-5,-3,-6) = (-54/14, -159/70, -159/35)

Therefore, the orthogonal projection of v onto w is (-54/14, -159/70, -159/35).

In simpler terms, the orthogonal projection of v onto w can be thought of as the vector that represents the shadow of v when it is cast onto the line defined by w. It is calculated by finding the component of v that aligns with w and multiplying it by the direction of w. The resulting vector (-54/14, -159/70, -159/35) lies on the line defined by w and represents the closest point to v along that line.

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The percent of change from 400 to 600 is 33.34%.

What is Percentage Increase?

It is given that the percent of change is from 400 to 600.

There is an increase from 400 to 600, so the percent of change is an increase.

\(\% \text Increase = \frac{\text FV - IV}{FV} \times 100\)

Here, the final value (FV) = 600 and initial value (IV) = 400.

Substituting the given values in the above formula, we get

\(\% \text Increase = \frac{600-400}{600} \times 100\\\implies \% \text Increase = \frac{200}{600} \times 100\\\implies \% \text Increase = 33.34 \%\)

Therefore, the percent of change from 400 to 600 is 33.34%.

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Answer:

2.4 makes it true!

I love you

yea yea yea yea yea

Step-by-step explanation:

Answer:

you forgot the pic

Step-by-step explanation:

The three Pythagorean trigonometric identities, which I’m sure one can find in any Algebra-Trigonometry textbook, are as follows:

sin² θ + cos² θ = 1

tan² θ + 1 = sec² θ

1 + cot² θ = csc² θ

where angle θ is any angle in standard position in the xy-plane.

Consistent with the definition of an identity, the above identities are true for all values of the variable, in this case angle θ, for which the functions involved are defined.

The Pythagorean Identities are so named because they are ultimately derived from a utilization of the Pythagorean Theorem, i.e., c² = a² + b², where c is the length of the hypotenuse of a right triangle and a and b are the lengths of the other two sides.

This derivation can be easily seen when considering the special case of the unit circle (r = 1). For any angle θ in standard position in the xy-plane and whose terminal side intersects the unit circle at the point (x, y), that is a distance r = 1 from the origin, we can construct a right triangle with hypotenuse c = r, with height a = y and with base b = x so that:

c² = a² + b² becomes:

r² = y² + x² = 1²

y² + x² = 1

We also know from our study of the unit circle that x = r(cos θ) = (1)(cos θ) = cos θ and y = r(sin θ) = (1)(sin θ) = sin θ; therefore, substituting, we get:

(sin θ)² + (cos θ)² = 1

1.) sin² θ + cos² θ = 1 which is the first Pythagorean Identity.

Now, if we divide through equation 1.) by cos² θ, we get the second Pythagorean Identity as follows:

(sin² θ + cos² θ)/cos² θ = 1/cos² θ

(sin² θ/cos² θ) + (cos² θ/cos² θ) = 1/cos² θ

(sin θ/cos θ)² + 1 = (1/cos θ)²

(tan θ)² + 1 = (sec θ)²

2.) tan² θ + 1 = sec² θ

Now, if we divide through equation 1.) by sin² θ, we get the third Pythagorean Identity as follows:

(sin² θ + cos² θ)/sin² θ = 1/sin² θ

(sin² θ/sin² θ) + (cos² θ/sin² θ) = 1/sin² θ

1 + (cos θ/sin θ)² = (1/sin θ)²

1 + (cot θ)² = (csc θ)²

3.) 1 + cot² θ = csc² θ

Answer: the distance between the points (-2, 5) and (-4, -5) is approximately 10.198 units.

Step-by-step explanation:

(-2, 5) and (-4,-5) are two points in the coordinate plane.

The first point (-2, 5) has an x-coordinate of -2 and a y-coordinate of 5. This point is 2 units to the left of the y-axis and 5 units above the x-axis.

The second point (-4, -5) has an x-coordinate of -4 and a y-coordinate of -5. This point is 4 units to the left of the y-axis and 5 units below the x-axis.

To find the distance between these two points, we can use the distance formula:

distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

Plugging in the coordinates of the two points, we get:

distance = sqrt[(-4 - (-2))^2 + (-5 - 5)^2] = sqrt[(-2)^2 + (-10)^2] = sqrt[104]

So the distance between the points (-2, 5) and (-4, -5) is approximately 10.198 units.

Answer:

839

Step-by-step explanation:

Just divide the number and then subtract

Answer: Perimeter = 36 units

Explanation:

The perimeter of the polygon is the sum o the length of each side.

We would find the length of each side by drawing right triangles as shown below

There are 4 right triangles labelled A, B, C and D

The required lengths are the hypotenuses of each triangle. We would apply the pythagorean theorem which is expressed as

hypotenuse^2 = one leg^2 + other leg^2

For triangle A

one leg = 5

other leg = 12

AD^2 = 5^2 + 12^2 = 25 + 144 = 169

AD = √169 = 13

AD = 13

Triangle D has the same dimensions as A. Thus,

CD = 13

For triangle B

one leg = 3

other leg = 4

Thus,

AB = 3^2 + 4^2 = 9 + 16 = 25

AB = √25 = 5

AB = 5

Triangle B has the same dimensions as C. Thus,

Thus,

BC = 5

Perimeter = AD + CD + AB + BC = 13 + 13 + 5 + 5

Perimeter = 36 units

Answer:

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Step-by-step explanation:

USE LINK!!!!

Answer:

A

Step-by-step explanation:

13/8, 1 3/4, 9/5

The top left graph represents the weight of the radioactive isotope over time.

An exponential function has the definition presented according to the equation as follows:

\(y = ab^x\)

In which the parameters are given as follows:

The parameter values for the function in this problem are given as follows:

a = 96, b = 0.5.

Hence the function is given as follows:

\(y = 96(0.5)^x\)

Two points on the graph of the function are given as follows:

(1,48) and (2, 24).

Hence the top left graph represents the weight of the radioactive isotope over time.

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Answer:

Graph W

Step-by-step explanation:

The given information describes a radioactive decay process, where the weight of the isotope decreases by half at regular intervals. This type of decay is characteristic of exponential decay.

Based on the description, the graph that represents the weight of the radioactive isotope over time would be a decreasing exponential curve, where the y-axis represents the weight of the isotope (in grams), and the x-axis represents time (in hours).

The initial weight of the isotope is 96 grams, and after each subsequent hour, the weight becomes half of what it was in the previous hour. Therefore, the correct graph would start at 96 grams (the initial weight when x = 0) and then decrease by half every hour. It would be a curve that gets closer and closer to zero but never quite reaches it.

Initial weight: 96 grams

After 1 hour: 96 / 2 = 48 grams

After 2 hours: 48 / 2 = 24 grams

After 3 hours: 24 / 2 = 12 grams

After 4 hours: 12 / 2 = 6 grams

After 5 hours: 6 / 2 = 3 grams

So, the points on the graph would be:

Therefore, the graph that represents the weight of the radioactive isotope over time is Graph W.

Answer:

m = -8

Step-by-step explanation:

I hope this helps!

Answer:

1,000,000,000

Step-by-step explanation:

Answer:

1000000000

Step-by-step explanation:

Literally just 1 with 9 0's behind.

Answer:

q = 27/10

Step-by-step explanation:

10/9 = 3/q

To find q, we must seperate it from the other numbers,

First we divide both sides by 3,

(10/9)/3 = (3/q)/3

10/27 = 1/q

Now we find the reciprocal of both sides to get the answer,

27/10 = q

q = 27/10

Answer:

\(150cm^2\)

Step-by-step explanation:

\(area =\) \(\frac{1}{2} * 12 * (y + x + y + z)\)

\(area =\)  \(6 * (x + 2y + z)\)

\((y + z)^2 + 12^2 = 20^2\) \(and\) \((x + y)^2 + 12^2 = 15^2\)

\((y + z) =\) \(\sqrt{20^2-12^2}\)

−−−−−−−− \(= 16cm\)

\((x + y) =\) \(\sqrt{15^2-12^2}\)

−−−−−−−− \(= 9cm\)

\(area =\)  \(\frac{1}{2} * 12 * (\sqrt{20^2-12^2} + \sqrt{15^2-12^2})\)

\(area = \\\)  \(6 * (16+9)\)

\(area = 150cm^2\)

Pythagorean Theorem

We can apply this by breaking down the trapezoid in two triangles.

\(base =\) \(\sqrt{(20^2 - 12^2}) + \sqrt{(15^2 - 12^2)\)

\(base =\) \(25\)

\(area =\) \(\frac{1}{2}bh\)

\(area =\) \(\frac{1}{2} * 25 * 12\)

\(area =\) \(150cm^2\)

Answer: The complete question is "A circle has a radius of \blue{3}3start color #6495ed, 3, end color #6495ed. An arc in this circle has a central angle of 20^\circ20 ∘ 20, degrees. What is the length of the arc?"

The length of the arc is 1.06667 units.

Step-by-step explanation:

According to the question the radius of the circle \(R=3 \, units\) and central angle of arc is \(\Theta =20^{o}\)

As we know that the length of the arc is given as: \(L=R\Theta\)

Where R is radius of the circle, L is the length of the arc and  \(\Theta\) is central angle in radian.

Now, \(\Theta =20^{o}\times \frac{\Pi }{180}=\frac{\Pi }{9} \, rad\)

Therefore, length of the arc is

\(L=3\times \frac{\Pi }{9}=\frac{\Pi }{3} =\frac{3.14}{3}=1.0466667 \, units\)

The domain of the rational function h(x) = (x - 3)/(x² - 4) is equal to (-∞, -2)  ∪ (-2, 2) ∪ (2, ∞).

The possible points that would lie on h(x) = x² - x + 1 are (3, 7).

True: (1, -1) is a point of the graph of f(x) = -2(x + 1)² + 7.

In Mathematics, a domain is the set of all real numbers for which a particular function is defined.

From the graph of h(x) = (x - 3)/(x² - 4), the domain in interval notation and set builder notation is as follows;

Domain = (-∞, -2)  ∪ (-2, 2) ∪ (2, ∞)

Domain = {x|x ≠ 2, -2}.

Next, we would determine possible x-value for the given quadratic function as follows;

h(x) = x² - x + 1

7 = x² - x + 1

7 - 1 = x² - x

6 = x² - x

x² - x - 6 = 0.

x² - 3x + 2x - 6 = 0.

x(x - 3) + 2(x - 3) = 0

(x + 2)(x - 3) = 0

x = 3 or x = -2.

For second quadratic function f(x) = -2(x + 1)² + 7, we have:

f(x) = -2(x + 1)² + 7

-1 = -2(1 + 1)² + 7

-1 = -2(2)² + 7

-1 = -8 + 7

-1 = -1 (True).

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Answer:

890 1/2 or  890.5

Step-by-step explanation:

Answer:

890.5 ft²

Step-by-step explanation:

Formula to find rectangular area

    Area = A = L × W

So put values

R 1        A 1 = 24.5 ×  18 = 441

R 2      A 2 = 14.5  × 18.5 = 268.25

R 3      A 3 = 12.5 ×  14.5 = 181.25

Add them up

441 + 268.25 + 181.25 = 890.5

Answer:

Step-by-step explanation:

In the Intermediate Phase of mathematics education, one topic that demonstrates a hierarchical and logical progression in patterns, functions, and algebra is the concept of "Linear Equations."

The topic of Linear Equations in the Intermediate Phase builds upon the foundation laid in earlier grades and serves as a stepping stone towards more advanced algebraic concepts. Here is an overview of the topic sequence and content area spread for Linear Equations:

Introduction to Variables and Expressions:

Students are introduced to the concept of variables and expressions, learning to represent unknown quantities using letters or symbols. They understand the difference between constants and variables and learn to evaluate expressions.

Solving One-Step Equations:

Students learn how to solve simple one-step equations involving addition, subtraction, multiplication, and division. They develop the skills to isolate the variable and find its value.

Solving Two-Step Equations:

Building upon the previous knowledge, students progress to solving two-step equations. They learn to perform multiple operations to isolate the variable and find its value.

Writing and Graphing Linear Equations:

Students explore the relationship between variables and learn to write linear equations in slope-intercept form (y = mx + b). They understand the meaning of slope and y-intercept and how they relate to the graph of a line.

Systems of Linear Equations:

Students are introduced to the concept of systems of linear equations, where multiple equations are solved simultaneously. They learn various methods such as substitution, elimination, and graphing to find the solution to the system.

Word Problems and Applications:

Students apply their understanding of linear equations to solve real-life word problems and situations. They learn to translate verbal descriptions into algebraic equations and solve them to find the unknown quantities.

The content area spread for Linear Equations includes concepts such as variables, expressions, equations, operations, graphing, slope, y-intercept, systems, and real-world applications. The progression from simple one-step equations to more complex systems of equations reflects a logical sequence that builds upon prior knowledge and skills.

By following this hierarchical progression, students develop a solid foundation in algebraic thinking and problem-solving skills. They learn to apply mathematical concepts and procedures in a systematic and logical manner, paving the way for further exploration of patterns, functions, and advanced algebraic topics in later phases of mathematics education.

Answer:

3190 is the answer may i have brainly?

Step-by-step explanation: