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antoniya [11.8K]

2 years ago

13

According to the fundamental theorem of algebra which polynomial function has exactly 8 roots? A.F(x)=(3x^2-4x-5)(2x^6-5) B.f(x)

=(3x^4+2x)^4 C. (4x^2-7)^3 D. F(x)=(6x^8-4x^5-1)(3x^2-4)

1 answer:

choli [55]2 years ago

3 0

I think it’s a because I added from the therum

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If m∠1 =63° , find m∠2 and m∠3 by calculation

Neko [114]

Since ∠1 is 63 than ∠2 is 63 too because they are alternate interior. To find ∠3 you subtract 180 by the measure of ∠2 because both of these angles create a straight line.

180 - 63 = ∠3
117 = ∠3

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2 years ago

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2 years ago

In the function y=5x^2-2, what effect does the number 5 have on the graph, as compared to the graph of y=x^2?

allochka39001 [22]

General Idea:

The Rules for Transformations of Functions are given below:

If f(x) is the original function, a > 0 and c > 0; Then

$f(x)+k \; \; shift \; f(x) \; UPWARD \; k \; units\\\\f(x)-k \; \; shift \; f(x) \; DOWNWARD \; k \; units\\\\f(x+h) \; \; shift \; f(x) \; LEFT \; h \; units\\\\f(x-h) \; \; shift \; f(x) \; RIGHT \; h \; units\\\\-f(x) \; \; REFLECT \; f(x) \; in \; the \; x-axis\\\\f(-x) \; REFLECT \; f(x) \; in \; the \; y-axis\\\\a \cdot f(x), \; a>1 \; STRETCH \; f(x) \; vertically \; by \; a \; factor \; of \; a\\\\a \cdot f(x), \; 0$

$f(ax), \; a>1 \; SHRINK \; f(x) \; horizontally \; by \; a \; factor \; of \;\frac{1}{a} .\\\\f(ax), \; 0$

Applying the concept:

In the function y=5x^2-2, the effect that the number 5 have on the graph, as compared to the graph of y=x^2 is given below:

C.it stretches the graph vertically by a factor of 5

4 0

3 years ago

Read 2 more answers

What is the area of the composite figure

deff fn [24]

to find the area of the figure, you first need to find the area of the separate figures: a triangle and a square.

first, find the area of the triangle:

multiply 13 and 5 to get 65.

divide 65 by 2.

32.5.

then, find the area of the rectangle:

multiply 5 and 14 to get 70.

add those answers up, and you get:

102.5

3 0

3 years ago

4. Gloria the grasshopper is working on her hops.

aivan3 [116]

The path that Gloria follows when she jumped is a path of parabola.

The equation of the parabola that describes the path of her jump is $\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

The given parameters are:

$\mathbf{Height = 20}$

$\mathbf{Length = 28}$

<em>Assume she starts from the origin (0,0)</em>

The midpoint would be:

$\mathbf{Mid = \frac 12 \times Length}$

$\mathbf{Mid = \frac 12 \times 28}$

$\mathbf{Mid = 14}$

So, the vertex of the parabola is:

$\mathbf{Vertex = (Mid,Height)}$

Express properly as:

$\mathbf{(h,k) = (14,20)}$

A point on the graph would be:

$\mathbf{(x,y) = (28,0)}$

The equation of a parabola is calculated using:

$\mathbf{y = a(x - h)^2 + k}$

Substitute $\mathbf{(h,k) = (14,20)}$ in $\mathbf{y = a(x - h)^2 + k}$

$\mathbf{y = a(x - 14)^2 + 20}$

Substitute $\mathbf{(x,y) = (28,0)}$ in $\mathbf{y = a(x - 14)^2 + 20}$

$\mathbf{0 = a(28 - 14)^2 + 20}$

$\mathbf{0 = a(14)^2 + 20}$

Collect like terms

$\mathbf{a(14)^2 =- 20}$

Solve for a

$\mathbf{a =- \frac{20}{14^2}}$

$\mathbf{a =- \frac{20}{196}}$

Simplify

$\mathbf{a =- \frac{5}{49}}$

Substitute $\mathbf{a =- \frac{5}{49}}$ in $\mathbf{y = a(x - 14)^2 + 20}$

$\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

Hence, the equation of the parabola that describes the path of her jump is $\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

See attachment for the graph

Read more about equations of parabola at:

brainly.com/question/4074088

7 0

2 years ago