Ask question

Ask question

All categories

3 years ago

15

The length of EF⎯⎯⎯⎯⎯ is 40 millimeters. What is the length of E′F′⎯⎯⎯⎯⎯⎯⎯⎯ after a dilation with a scale factor of 2/5 ? Enter

your answer in the box. mm

1 answer:

valentina_108 [34]3 years ago

3 0

Answer:

16 mm

Step-by-step explanation:

Dilation multiplies every length by the dilation factor.

E'F' = (2/5)EF = (2/5)(40 mm) = 16 mm

You might be interested in

A<br> ОЈ<br> 6(3+2d) = 54<br> I need that answer

disa [49]

Answer:

d = 3

Step-by-step explanation:

6*3=18
2d*6=12d
54-18 = 12d
36/12 = 12d/12
d = 3

8 0

2 years ago

Read 2 more answers

The measure of angle A is 30 and the measure of angle B is 124 . What is the measure of angle C in degrees?

professor190 [17]

Answer: 26

124 + 30 = 154

180 - 154 = 26

7 0

3 years ago

Help me please find the equation of line.

Aleksandr [31]

Answer:

The correct answer is A

Step-by-step explanation

6 0

2 years ago

Read 2 more answers

What is the equation of the line that passes through the point (-8,0)and has a slope of -1/4

Dahasolnce [82]

Answer:

y = -1/4x - 2

Step-by-step explanation:

I'm going to write this equation in slope intercept form, which looks like this: y = mx + b. First, write the equation using the slope given. It should look like this:

y = -1/4x + b

Next, you must find the y intercept by plugging in the point to the equation. This is what it looks like: 0 = -1/4(-8) + b. Multiply -1/4 by -8 to get 2. This is what your equation should look like now: 0 = 2 + b. Subtract 2 from both sides of the equation to get -2 = b. Go back to your original equation and plug -2 in for b. This is your final equation:

y = -1/4x - 2

Hope this helped!

8 0

3 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

$r = 5$

Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

<u><em>Solved</em></u>

6 0

2 years ago