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3 years ago

8

If you vertically stretch the absolute value parent function, f(x) = |x|, by a factor of 5, which is the equation of the new fun

ction?

1 answer:

faltersainse [42]3 years ago

3 0

I believe it is f(x)=5|x|
I suggest you get a second opinion, though

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Find the value of x in the isosceles triangle shown below

egoroff_w [7]

Answer:

x = 10

Step-by-step explanation:

if there 2 same triangle then they are right triangle

By using Pythagorean theorem for 1 of right triangle, we have

$\sqrt{74} ^{2}$ = $7^{2}$ + $(\frac{x}{2}) ^{2}$

74 = 49 + $\frac{x^{2}}{4}$

74 - 49 = $\frac{x^{2}}{4}$

25 = $\frac{x^{2}}{4}$

therefore, x = $\sqrt{25*4}$ = $\sqrt{100}$ = 10

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2 years ago

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In ΔMNO, m = 780 inches, o = 760 inches and ∠O=164°. Find all possible values of ∠M, to the nearest degree.

Mademuasel [1]

Answer:

So the answer is both

180 and 328 (but it may or may not show you "not possible, if so, then you got it right)

Step-by-step explanation:

SinA/a= SinB/b

1. SinM/780 = Sin164/760

2. 780sin164/760 = 0.2828909704

3. M = sin^-1 (0.2828909704)= 16.4328229 or 16

Check for possibility

180-16= 164

164 + 16= 180 (not possible)

164 + 164= 328 (not possible)

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3 years ago

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Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

What solid figure has five faces and five vertices? rectangular prism cone triangular prism square pyramid

Marizza181 [45]

Answer:

Square pyramid

Step-by-step explanation:

The flat square on the bottom is 1 face and there are 4 triangle faces.

The flat square on the bottom has 4 vertices and there is the pointy 1 at the top that holds the 4 triangles.

Hope dis helps,

Have a great day.

8 0

3 years ago

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Use the graph to write a linear function that relates y to x.

allsm [11]

Answer:

um we need the photo sry

Step-by-step explanation:

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3 years ago