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Archy [21]
3 years ago
13

Graph the image of this figure after a dilation with a scale factor of 1/3

Mathematics
1 answer:
aliya0001 [1]3 years ago
8 0

Answer:

(-1,1), (1,3), (2,2)

Step-by-step explanation:

To dilate an object, we need to multiply the x and y values by the given scale factor.

In this case the scale factor is 1/3 --> 1/3(x, y)

Before-> After dilation

1/3(-3,3) = (-1,1)

1/3(3,9) = (1,3)

1/3(6,6) = (2,2)

Please leave a 'thanks' if this helps!

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Which of the following equations represents the linear relationship shown in the table below?
Sphinxa [80]

Answer:

B

Step-by-step explanation:

The equation is:

y = 2x + 3

Put x as 2.

y = 2(2) + 3

y = 4 + 3

y = 7

Put x as 3.

y = 2(3) + 3

y = 6 + 3

y = 9

Put x as 4.

y = 2(4) + 3

y = 8 + 3

y = 11

Put x as 5.

y = 2(5) + 3

y = 10 + 3

y = 13

6 0
3 years ago
Gina puts $ 4500 into an account earning 7.5% interest compounded continuously. How long will it take for the amount in the acco
Elza [17]

~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5150\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years \end{cases}

5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}

4 0
1 year ago
A study of peach trees found that the average number of peaches per tree was 725. The standard deviation of the population is 70
TEA [102]

Answer:

She needs to sample 189 trees.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

The standard deviation of the population is 70 peaches per tree.

This means that \sigma = 70

How many trees does she need to sample to obtain an average accurate to within 10 peaches per tree?

She needs to sample n trees.

n is found when M = 10. So

M = z\frac{\sigma}{\sqrt{n}}

10 = 1.96\frac{70}{\sqrt{n}}

10\sqrt{n} = 1.96*70

Dividing both sides by 10:

\sqrt{n} = 1.96*7

(\sqrt{n})^2 = (1.96*7)^2

n = 188.2

Rounding up:

She needs to sample 189 trees.

3 0
3 years ago
Please help me, thank you
Pavel [41]

Answer:

the answer is -3. Your welcome

4 0
3 years ago
Can someone help with this
mixas84 [53]

Step-by-step explanation:

-10-2x = 3x+15

5x = -25

x= -5

5 0
2 years ago
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