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OverLord2011 [107]

3 years ago

8

What are the coordinates for an enlarged parallelogram, A'B'C'D', if the scale factor for the dilation is?? 4 A 0 /( 7, 2); }' 7

, 2); C(0, 2); D' ( 14, 2) ON 7.7); B (7.7); C'(0, --7); D' (-14, -7) ON( 2.7); B (2,7); C'0,-7); D'(-4,-7) ON(-7,-2);B(7.-7); C'(0,7); D'(-14,7) 4

1 answer:

Sonbull [250]3 years ago

7 0

Answer:

B

Step-by-step explanation:

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3] What is - 2y + 10 + 2y - 8 simplified?

inna [77]

Answer:

d. 2

Step-by-step explanation:

-2y +10+2y-8

10-8. ( -2y+2y = 0)

2

7 0

2 years ago

Read 2 more answers

The math club needs to raise more than $552. $.50 for a trip to state competition. The club has raise 12% of the funds which is

ratelena [41]

Each of the members must raise at least $69.46.

To start they need to raise $552.50. However, they have already raised 12% of it. They only need to raise 88% of it.

552.50 x 0.88 = 486.2

Divide the remaining amount by 7 to determine how much each individual must raise.

486.2 / 7 = 69.46

4 0

3 years ago

Brent plays three sports: basketball, baseball, and soccer. He calculated the mean absolute deviation of the points he scored in

Flauer [41]

Answer:

basketball

I hope this is useful

8 0

2 years ago

Solve for x: |3x + 12| = 18

Alla [95]

Answer:

X = 2 or X = -10

Step-by-step explanation:

We know either3x+12=18or3x+12=−18

3x+12=18(Possibility 1)

3x+12−12=18−12(Subtract 12 from both sides)

3x=6

3x

3

=

6

3

(Divide both sides by 3)

x=2

3x+12=−18(Possibility 2)

3x+12−12=−18−12(Subtract 12 from both sides)

3x=−30

3x

3

=

−30

3

(Divide both sides by 3)

x=−10

Answer:

x=2 or x=−10

4 0

3 years ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

4 years ago