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3 years ago

The vertices of a triangle are A (0, 2), B(-2,-5), and C(0,5). Find the new vertices.

Mathematics

1 answer:

Umnica [9.8K]3 years ago

3 0

Answer:

see explanation

Step-by-step explanation:

Given the translation rule

(x, y ) → (x + 3, y - 2 )

This means add 3 to the original x- coordinate and subtract 2 from the original y- coordinate, thus

A(0, 2 ) → A'(0 + 3, 2 - 2 ) → A'(3, 0 )

B(- 2, - 5 ) → B'(- 2 + 3, - 5 - 2 ) → B'(1, - 7 )

C(0, 5 ) → C'(0 + 3, 5 - 2 ) → C'(3, 3 )

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2.4x-9 < 1.8x+6 solve the inequality

nasty-shy [4]

Answer:

x<25

Step-by-step explanation:

Let's solve your inequality step-by-step.

2.4x−9<1.8x+6

Step 1: Subtract 1.8x from both sides.

2.4x−9−1.8x<1.8x+6−1.8x

0.6x−9<6

Step 2: Add 9 to both sides.

0.6x−9+9<6+9

0.6x<15

Step 3: Divide both sides by 0.6

0.6x/0.6 < 15/0.6

x<25

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2 years ago

Identify the missing c value in the following equation.

GalinKa [24]

Answer:

A.+4

Step-by-step explanation:

Graph pass (1 , 0) and (4 , 0)

y = (x - 1) (x - 4) = x² - 5x + 4

c = 4

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2 years ago

Suppose your friend's parents invest $15,000 in an account paying 7% compounded annually. What will the balance be after 6 years

kykrilka [37]

Answer:

$22,510.96

Step-by-step explanation:

Final investment value

$22,510.96

Total interest earned

$7,510.96

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Total monthly deposits

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6 0

3 years ago

At 1 P.M., the total snowfall is 5 centimeters. At 4 P.M., the total snowfall is 16 centimeters. What is the mean hourly snowfal

Yuri [45]

Answer:

Mr. Snowbully

Step-by-step explanation:

I think the mean hourly snowfall is named Mr. Snowbully

It's a funny name!

Well, the snow would grow at a rate of 4 cm every hour so...

5 0

2 years ago

A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for

tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of t are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

H₀: There is no difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ = 0.

Hₐ: There is a difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

$\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29$

$s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64$

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

$\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11$

$s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08$

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a t-test for two means.

Compute the test statistic value as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}$

$=\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}$

$=-4.123$

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

$m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}$

$=\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}$

$=26.55\\\approx 27$

Compute the critical value for α = 0.05 as follows:

$t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052$

*Use a t-table for the values.

Thus, the critical values of t are ± 2.052.

3 0

2 years ago