Graph the line with the equation y = 1/3 x + 1

To find two different bases for the vector space V consisting of all quadratic polynomials, we need to find two sets of vectors that are linearly independent and span V.

One possible basis for V is \(\(\{1, x, x^2\}\)\). This set of vectors is linearly independent, meaning that no vector in the set can be written as a linear combination of the others. Any quadratic polynomial can be written as a linear combination of these vectors. For example, \(\(p(x) = ax^2 + bx + c\)\)can be written as \(\(a(1) + b(x) + c(x^2)\)\).

Another possible basis for V is \(\(\{x^2, x^2 + x, x^2 + 2x\}\)\). This set of vectors is also linearly independent and spans V. Any quadratic polynomial can be written as a linear combination of these vectors. For example, \(\(p(x) = ax^2 + bx + c\)\) can be written as \(\((-a+b)(x^2) + (b+2c)(x^2 + x) - c(x^2 + 2x)\).\)

In both cases, we have found two different bases for \(\(V\)\), and we have shown how to write a quadratic polynomial as a linear combination of each basis.

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The area of the sector is determined as 39.3 cm².

The area of sector ABC is calculated as follows;

The area of a sector can be calculated using the following formulas,

Area of a Sector of Circle = (θ/360º) × πr²

where;

The angle subtended by this sector = 0.785 radians = 45⁰

The radius of the sector = 10 cm

The area of the sector is calculated as follows;

A = ( 45 / 360) x π x ( 10²)

A = 39.3 cm²

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The answer is 6.85.

Consider, x-y axis, and side of equilateral triangle is ' a '.

Given,

\($$\begin{aligned}&A P=3 \text { units } \\&B P=2 \text { units } \\&C P=4 \text { units }\end{aligned}$$\)

Let x and y are point in x-y axis.

Using, distance formula here

\($A P^{2}=(x-0)^{2}+(y-0)^{2}$\)

\($3^{2}=x^{2}+y^{2}$\)

\($x^{2}+y^{2}=$\)__________(i)

\($p c^{2}=(x-a)^{2}+(y-0)^{2}$$4^{2}=x^{2}-2 a x+a^{2}+y^{2}$\)

from (i)

\($16=x^{2} 9-2 a x+a^{2}$\)

\($2 a x=a^{2}+9-16$\)

\($\left[x=\frac{\left(a^{2}-7\right)}{2 a}\right]$\)

\($B p^{2}=\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{\sqrt{3}}{2} a\right)^{2}$\)

\($2^{2}=x^{2}-a x+\frac{a^{2}}{4}+y^{2}-\sqrt{3} a b y+\frac{3}{4} a^{2}$\)

\($4=\left(x^{2}+y^{2}\right)+\frac{a^{2}+3 a^{2}}{4}-a\left(\frac{a^{2}-7}{2 a}\right)-\sqrt{3} a y$\)

from (i)

\($4=9+a^{2}-a^{2}+7-\sqrt{3} a y$\)

\($\sqrt{3} a y=16-4$\)

put x and y in equarion_______(i)

\($\left(\frac{12}{\sqrt{3} a}\right)^{2}+\left(\frac{a^{2}-7}{2 a}\right)^{2}=9$\)

\($\frac{144}{3 a^{2}}+\frac{\left(a^{2}-7\right)^{2}}{4 a^{2}}=9$\)

\($144 \times 4+3\left(a^{2}-7\right)^{2}=12 \times 9 a^{2}$\)

\($576+3\left(a^{2}-7\right)^{2}=108 a^{2}$\)

\(put $a^{2}=t$$$576+3(t-7)^{2}=108 t$$\)

\($\begin{aligned} 576+3\left(t^{2}-14 t+49\right) &=108 t \\ 576+3 t^{2}-42 t+147 &=108 t \end{aligned}$\)

\($3 t^{2}-150 t+429=0$$t^{2}-50 t+143=0$$t=50 \pm \sqrt{50^{2}-4 \times 143}$\)

\(t=\frac{50 \pm \sqrt{1928}}{2}t=\frac{50 \pm 43.91}{2}\)

\(t=\frac{50 \pm 43.91}{2}\\t=46.9545, \quad t=3.0455\)

\($\begin{array}{ll}a^{2}=46.9545, \\ a=\sqrt{46.9545}, & a^{2}=3.0455 \\ a, a=1.745\end{array}$\)

\($a=6.85$\)

for this point P is outside Therefore, it is not consider as solution (because in question inside point ls mention)

\(a=6.85 units\)

What is equilateral triangle?

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To prove that W is a subspace of R[x], we need to show that it satisfies three conditions: closure under addition, closure under scalar multiplication, and the presence of the zero vector.

Closure under addition:

Let f(x) and g(x) be polynomials in W. This means that f(x)(x^2 + 1) and g(x)(x^2 + 1) are both divisible by x^2 + 1.

We want to show that their sum, f(x)(x^2 + 1) + g(x)(x^2 + 1), is also divisible by x^2 + 1.

Expanding the sum, we have:

f(x)(x^2 + 1) + g(x)(x^2 + 1) = (f(x) + g(x))(x^2 + 1).

Since f(x) + g(x) is a polynomial and (x^2 + 1) is a factor of both f(x)(x^2 + 1) and g(x)(x^2 + 1), it follows that (f(x) + g(x))(x^2 + 1) is divisible by x^2 + 1. Therefore, W is closed under addition.

Closure under scalar multiplication:

Let f(x) be a polynomial in W, which means f(x)(x^2 + 1) is divisible by x^2 + 1.

We want to show that for any scalar c, the product cf(x)(x^2 + 1) is also divisible by x^2 + 1.

Expanding the product, we have:

cf(x)(x^2 + 1) = (cf(x))(x^2 + 1).

Since cf(x) is a polynomial and (x^2 + 1) is a factor of f(x)(x^2 + 1), it follows that (cf(x))(x^2 + 1) is divisible by x^2 + 1. Therefore, W is closed under scalar multiplication.

Presence of zero vector:

The zero vector in R[x] is the polynomial 0. We can see that 0(x^2 + 1) = 0, which is divisible by x^2 + 1. Therefore, the zero vector is in W.

Since W satisfies all three conditions (closure under addition, closure under scalar multiplication, and presence of zero vector), we can conclude that W is a subspace of R[x].

Now, let's consider R[x]/W, the quotient space of R[x] by W.

The elements of R[x]/W are the equivalence classes of polynomials in R[x], where two polynomials are considered equivalent if their difference is in W.

In other words, for polynomials f(x) and g(x) in R[x], f(x) and g(x) belong to the same equivalence class if and only if f(x) - g(x) is divisible by x^2 + 1.

The dimension of R[x]/W can be determined by finding a basis for the quotient space. A basis for R[x]/W consists of representatives of the equivalence classes.

In this case, the equivalence classes can be represented by polynomials of the form r(x)(x^2 + 1), where r(x) is any polynomial.

So, a basis for R[x]/W is {x^2 + 1}.

Since the basis has one element, the dimension of R[x]/W is 1.

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Benny’s arcade has video game machines. The average time between machine failures is x hours. Jimmy, the maintenance engineer, can repair a machine in y hours on average.

The machines have an exponential failure distribution, and Jimmy has an exponential service-time distribution.Exponential failure distribution can be used to model the time between machine failures, provided the failures are random. This exponential distribution function has a characteristic that the probability of a machine failing at any point in time is the same, regardless of how long the machine has been in use.

The probability that a machine is operating successfully at a particular point in time is called the reliability of the machine. If R(t) is the reliability of a machine at time t, then the exponential distribution function for failures is given by:R(t) = e−λt where λ is the failure rate per unit time, and t is the time that the machine has been operating since the last failure.The average time between machine failures is given by the inverse of the failure rate, i.e. x = 1/λ.If Jimmy has an exponential service-time distribution,

then the probability that he will take exactly y hours to repair a machine is given by:f(y) = λexp(−λy)For an exponential distribution, the expected value is equal to the inverse of the rate, i.e. E(Y) = 1/λ.In this case, the expected time for Jimmy to repair a machine is y = E(Y) = 1/λ.Since the expected time to repair is y, and the expected time between failures is x, then the expected time to failure is given by:x + y = 1/λ + 1/μwhere μ is the service rate per unit time.Hence, the expected time between failures and repairs.

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Answer:

w

Step-by-step explanation:

Answer:$6.50

Step-by-step explanation:

if you divide 91 by 14 you get 6.5 which makes one candle $6.50

Answer:

The company should choose the prism because it has approximately 14.1 in.³ less wasted space than the cylinder.

Step-by-step explanation:

To determine which package has the least amount of wasted space, we need to compare the volumes of the cylinder and the square prism. The volume of a cylinder is given by V = πr²h, where r is the radius and h is the height. In this case, the radius is 1.5 inches and the height is 6 inches, so the volume of the cylinder is V_cylinder = π(1.5)²(6) ≈ 42.4 in.³.

The volume of a square prism is given by V = l²h, where l is the length of the base and h is the height. In this case, the base lengths are 3 inches and the height is 6 inches, so the volume of the square prism is V_prism = (3)²(6) = 54 in.³.

To determine the wasted space, we subtract the volume of two balls (each with a volume of 4/3πr³) from the volume of each package. Since the radius of the balls is 1.5 inches, the volume of two balls is (4/3π(1.5)³) × 2 ≈ 14.1 in.³.

Comparing the wasted space, we find that the prism has approximately 14.1 in.³ less wasted space than the cylinder. Therefore, the company should choose the prism as it minimizes wasted space

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Answer:

what is this question...

Step-by-step explanation:

great than 2 so the answer could never be 2

Answer:

1.64

Step-by-step explanation:

8.20 / 5 = 1.64

Answer:

It D

Step-by-step explanation:

Answer:

(1,-1)

Step-by-step explanation:

-2x-1 = x+2

-3x = 3

x = -1

substitute

y = (-1) + 2

y = 1

point: (1,-1)

Answer:

third answer

Step-by-step explanation:

Answer:

area = 3.14 mi² (to the nearest hundredth)

Step-by-step explanation:

Area of a circle = \(\pi r^2\)

(where r is the radius)

Given:

Substitute r = 1 into the equation:

⇒ area = \(\pi\) × 1²

⇒ area = \(\pi\) mi²

⇒ area = 3.14 mi² (to the nearest hundredth)

Answer:

π

Step-by-step explanation:

→ State the formula for the area of a circle

π × r²

→ Substitute in the numbers

π × 1² = π

Answer:

The polynomial is positive.

Answer:

about 11 boys were on the track team last year.

The zeros of the polynomial P(x) = x^3 + 3x^2 - 4, where all the zeros are integers, can be found by factoring the polynomial or using synthetic division. The zeros of the polynomial are the values of x for which P(x) equals zero.

To find the zeros of the polynomial P(x) = x^3 + 3x^2 - 4, we set the polynomial equal to zero and solve for x. The equation becomes x^3 + 3x^2 - 4 = 0.

Unfortunately, there is no simple factoring pattern or rational root theorem that guarantees the presence of integer zeros. Therefore, we can use numerical methods or synthetic division to find the zeros.

Using numerical methods, we can evaluate the polynomial at different integer values of x until we find the zeros. By evaluating P(x) for various integer values, we find that P(-4) = 0 and P(1) = 0. Hence, the zeros of the polynomial are x = -4 and x = 1.

Therefore, the zeros of the polynomial P(x) = x^3 + 3x^2 - 4 are x = -4 and x = 1.

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Answer:

84 feet

Step-by-step explanation:

 he will need 84 ft to cover the entire garden

Answer:

38 ft of fencing

Step-by-step explanation:

The perimeter of a rectangle is 2L + 2W, where L and W are the length and width, respectively. The dimensions of the garden are given, 7 x 12 ft.

The amount of fencing needed is 2(7) + 2(12) = 14 + 24 = 38 ft of fencing

The required expression for the term of the given sequence is aₙ = 625 × (1/5)ⁿ⁻¹.

In order to check a geometric sequence, find the ratios of the consecutive terms of the given sequence. If they are equal, the given sequence is geometric.

The given sequence is as follows,

625, 125, 25, 5, 1

The first term is given as a₁ = 625

And, the ratio of the consecutive terms are,

125/625 = 1/5

25/125 = 1/5

Since the ratio of two consecutive terms are equal, the given sequence is a geometric sequence.

The nth term of a geometric sequence is given as aₙ = a₁rⁿ⁻¹.

Thus, the expression for the given sequence is aₙ = 625 × (1/5)ⁿ⁻¹.

Hence, the term for the given sequence can be written in general form as  aₙ = 625 × (1/5)ⁿ⁻¹.

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Step-by-step explanation:

sorry but i need points Thank you

Answer:

-3,2 is the answer

Step-by-step explanation:

hope you like it

Answer:

there is 3 terms I think

Step-by-step explanation:

The height of the tide in a small beach town is measured along a seawall. Water levels oscillate between 5 feet at low tide and 15 feet at high tide. On a particular day, low tide occurred at 6 am and high tide occurred at noon. Approximately every 12 hours, the cycle repeats.

To find an equation to model the water levels, we can use a sinusoidal equation. Let h(t) be the height of the water at time t (in hours). We know that h(6) = 5 and h(12) = 15. Using this information, we can find the equation:

h(t) = 10 sin (πt/6) + 10
To find an equation that models the water levels in a small beach town, given that the height of the tide is measured along a seawall, we need to use the following information:

Water levels oscillate between 5 feet at low tide and 15 feet at high tide. Low tide occurred at 6 am, and high tide occurred at noon. The cycle repeats approximately every 12 hours. Let the water level at low tide be represented by y = 5, and the water level at high tide be represented by y = 15. We can write these points as (0,5) and (12,15), respectively. Since the water levels oscillate every 12 hours, we can create a sine function that models this pattern. We can use the sine function y = a sin(bx + c) + d, where a is the amplitude (half the height of the wave), b is the frequency (number of waves per unit time), c is the phase shift (horizontal displacement of the wave), and d is the vertical displacement of the wave. Using the given information, we can determine the values of a, b, c, and d:a = (15 - 5)/2 = 5, since the amplitude is half the height of the wave. b = 2π/12 = π/6, since the wave repeats every 12 hours (or twice a day), and the period is 2π/b.c = -π/2, since the graph starts at high tide (the maximum point), not at the midpoint between low and high tide (the x-axis).d = (15 + 5)/2 = 10, since the midpoint between low and high tide is the vertical axis of the sine function. Therefore, the equation that models the water levels is: y = 5 sin(π/6 x - π/2) + 10.

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Answer:

4+3x

Step-by-step explanation:

4+3x

Answer: (5,5)

Step-by-step explanation:

The perpendicular to y=x has slope -1, and thus the equation of the perpendicular from (1,9) is \(y=-x+10\).

Finding the point where they intersect, since both equations are set equal to y, we know that -x+10=x, and thus x=5.

If x=5, then y=5 as well.

So, F has coordinates (5,5).

Answer:

95

Step-by-step explanation:

Let x be the number of children

10x+35 = money

15x+5 = money

Set  the equations equal

10x+35 = 15x+5

Subtract 10x from each side

35 = 5x+5

Subtract 5 from each side

30 =5x

Divide by 5

30/5 =5x/5

6 =x

There are 6 children

10x+35 = money

10(6) +35

60+35

95

The amount of money is 95

Answer:

114

Step-by-step explanation:

If this answer helped you then please mark it as brainliest so I can get to the next rank

Answer:

16

Step-by-step explanation:

x^2-8x+k is a quadratic expression of the form ax^2 + bx + c.  Here a = 1, b = -8 and c = k.  Focus on x^2-8x and complete the square as follows:  Take half of the coefficient of x (that is, take half of -8) and square the result:

(-4)^2 = 16; if we now write x^2-8x+ 16, we'll have the square of (x - 4):  (x -4)^2.

Thus, k = 16 turns x^2-8x+k into a perfect square.