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photoshop1234 [79]

3 years ago

8

Identify the transformation that maps the regular pentagon with a center (0, -2) onto itself.

2 answers:

Gnesinka [82]3 years ago

8 0

<span>Rotate 144° about the point (0, -2)</span>

puteri [66]3 years ago

7 0

1) you could rotate 360 degrees about the center (0,-2)
2) you could reflect over the x axis, then over the y axis, then over the x axis, and then over the y axis, and you're back where you started those are just some choices

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Point B is the image of point A when point A is rotated about the origin. What is known about point A and B?

IRINA_888 [86]

Answer:

Point B is the result of a 90° counterclockwise rotation.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What is the solution to the system of equations? Use the substitution method.

kap26 [50]

Hello,
answer A

4x+2y=11
x-2=-2y==>2y=-x+2

4x-x+2=11==>3x=9==>x=3

2y=-3+2==>y=-1/2

5 0

3 years ago

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$ $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

4 0

3 years ago

The public library has an approximate length of 1 x 10^5 millimeters and width of 8 x 10^4 millimeters. what is the area, in squ

Wewaii [24]

Area = length*width
.. = (1*10^5 mm)*(8*10^4 mm)
.. = (1*8)*10^(5+4) mm^2
.. = 8*10^9 mm^2

Selection A is appropriate.

3 0

3 years ago

Change the expression to a single square root, or its opposite:

aleksley [76]

Answer:

a) $2\sqrt{2}=\sqrt{8}$

b) $-7\sqrt{3} =-\sqrt{147}$

c) $\frac{1}{3} \sqrt{18b} =\sqrt{2.b}$

d) $5\sqrt{y} =\sqrt{25y}$

e) $-6\sqrt{2a} =-\sqrt{72a}$

f) $-0.1\sqrt{200c}=- \sqrt{2c$

Step-by-step explanation:

a) $2\sqrt{2}=( \sqrt{2} )^2.\sqrt{2}=(\sqrt{2})^3 = \sqrt{2^3} =\sqrt{8}$

b) $-7\sqrt{3} =-(\sqrt{7} )^2\sqrt{3} =-\sqrt{7^2.3} =-\sqrt{147}$

c) $\frac{1}{3} \sqrt{18b} =\frac{1}{3} \sqrt{9.2.b} =\frac{1}{3} \sqrt{3^2.2.b} =\frac{1}{3} \times 3\sqrt{2.b} =\sqrt{2.b}$

d) $5\sqrt{y} =\sqrt{5^2} \sqrt{y}=\sqrt{25y}$

e) $-6\sqrt{2a} =-\sqrt{6^2}\sqrt{2a} = -\sqrt{36.2a} =-\sqrt{72a}$

f) $-0.1\sqrt{200c}=-\frac{1}{10} \sqrt{10^2.2c} =-\frac{1}{10}\times10 \sqrt{2c}=- \sqrt{2c$

4 0

4 years ago