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3 years ago

15

The coordinates of the vertices of ABC are A(-5, 2) B(-2, 4), and C(-4, 7). What are the coordinates of the image of A after the

translation (x, y) = (x+6, y-5)?

1 answer:

Veseljchak [2.6K]3 years ago

8 0

Translation will reduce x coordinate by 2: (0,4), (5,6) and (3,2)

Rotation 90 CCW will change (x,y) to (-y,x): (-4,0),(-6,5),(-2,3)

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Hey can yall please help me its 4'1 minuse 1'10

neonofarm [45]

4’1= 4 feet and 1 inch
12 inches= 1 ft.
12(4)= 48 ; 48+ 1inch = 49 inches
1’10= 1 foot and 10 inches
1(12)= 12 + 10 = 22
49-22= 27 inches !

7 0

3 years ago

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F+6=18;11,12,13,14 <br><br> A.14<br> B.13<br> C.12<br> D.11

zaharov [31]

(c) abshjwjwkwjwjwjwjwmwkkwkwkskss

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3 years ago

4462 mm² in m²

Varvara68 [4.7K]

Answer:

0.004462m²

612000m²

0.0482m²

Step-by-step explanation:

Using the conversion

1mm² = 10^-6m²

4462mm² = x

Cross multiply

1 * x = 4462 * 10^-6

x = 4462 * 10^-6

x = 4.462 * 10^3 * 10^-6

x = 4.462*10^{3-6}

x = 4.462*10^-3

x = 0.004462m²

Hence 4462 mm² in m² is 0.004462m²

0.612 km² in m²

1km² = 10^6m²

0.612km² = y

Cross multiply

1 * y = 0.612 * 10^6

y = 0.612*1,000,000

y = 612000m²

Hence 0.612 km² in m² is 612000m²

482 cm² in m²

1cm² = 0.0001m²

482 cm² = z

z = 482 * 0.0001

z = 0.0482m²

Hence 482 cm² in m² is 0.0482m²

6 0

2 years ago

A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. A previous study indicates that the propo

never [62]

Answer:

A sample of 997 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A previous study indicates that the proportion of left-handed golfers is 8%.

This means that $\pi = 0.08$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a p-value of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

This is n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.327\sqrt{\frac{0.08*0.92}{n}}$

$0.02\sqrt{n} = 2.327\sqrt{0.08*0.92}$

$\sqrt{n} = \frac{2.327\sqrt{0.08*0.92}}{0.02}$

$(\sqrt{n})^2 = (\frac{2.327\sqrt{0.08*0.92}}{0.02})^2$

$n = 996.3$

Rounding up:

A sample of 997 is needed.

3 0

3 years ago

Plz need help this is hw

KengaRu [80]

Answer: A= 1,-5 b=5,-3

Step-by-step explanation: i dont have one

4 0

2 years ago

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