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3 years ago

Which series of transformations maps △ABC to △A′B′C′?

Mathematics

2 answers:

denis-greek [22]3 years ago

5 0

Answer:

Step-by-step explanation:

If you translate left one unit, then reflect across the x axis, you get the ending figure

Alik [6]3 years ago

4 0

Answer:

Step-by-step explanation:

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Find the measure of each angle in the triangle.

Maru [420]

Answer:

25, 70, 85

Step-by-step explanation:

x+2x+20+3x+10=6x+30

6x+30=180

6x=150

x=25

The angles are 25, 70, and 85

6 0

3 years ago

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Use the following function rule to find f(r+4). simplify your answer .. f(n)=n+4

Fudgin [204]

N=54........................

7 0

3 years ago

Your team gets an ice sculpture in the shape of a basketball. It has a radius of 12 inches. What is the volume of the ice? Use 3

horsena [70]

The answer is 7234.56 inches cubed. The equation to find the volume of a sphere is 4/3*pi*r^3.

The equation would be 4/3 * 3.14 * 12^3.

4/3 * 3.14 * 1728

4/3 * 5425.92

7234.56

4 0

3 years ago

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In a solar system, two comets pass near the sun, which is located at the origin. Comet E is modeled by quantity y plus 16 end qu

Leya [2.2K]

Explanation:

The equation of comet E is given below as

$\frac{(y+16)^2}{400}+\frac{x^2}{144}=1$

The equation of comet H is modelled below as

$\frac{(y+13)^2}{144}-\frac{x^2}{25}=1$

Given, that the Sun is located at the origin.

This is an equation of an ellipse. So, the path traveled by Comet e is an elliptic path.

Comparing with the Standard equation of ellipse: below,we will have

$\frac{(y-k)^2}{b^2}+\frac{(x-h)^2}{a^2}$

Whise center is

$\begin{gathered} center=(h,k) \\ vertices=(h\pm a,0) \end{gathered}$

By comparing coefficient, we will have

$\begin{gathered} a^2=144 \\ a=12 \\ b^2=400 \\ b=20 \\ k=-16 \\ h=0 \\ (h,k)=(0,-16) \end{gathered}$

Hence,

The vertex of comet E will be

$\begin{gathered} (h\pm a,0) \\ (0+12,0),(0-12,0) \\ (12,0),(-12,0) \end{gathered}$

The vertex of comet E is

$(12,0),(-12,0)$

Part C:

It is the foci forboth comets

$\begin{gathered} cometE: \\ c=\sqrt{b^2-a^2} \\ c=\sqrt{400-144} \\ c=\sqrt{256} \\ c=16 \\ \\ For\text{ comet F:} \\ c=\sqrt{a^2-b^2} \\ c=\sqrt{144+25} \\ c=\sqrt{169} \\ c=13 \end{gathered}$

Hence,

The foci will be

$\begin{gathered} (16\pm16,0) \\ cometE \\ (0,0),(32,0) \\ cometF: \\ (13\pm13,0) \\ (0,0),(26,0) \end{gathered}$

5 0

1 year ago

100 is 10 times as much as...

Lorico [155]

Answer:

Step-by-step explanation:

3 0

3 years ago