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3 years ago

Please HELP I WILL REWARD BRAINLIEST ANSWER IM VERY CONFUSED

Mathematics

1 answer:

ozzi3 years ago

4 0

Dilation is either : enlargement or shirking.

Simply you multiply the ordered pair times the dilation vector (Origin-Centered)

D.F = dilation factor;

d.f = 5

C' = d.fC

when you multiply a constant times an ordered pair , it multiples into each of the pair.

C' = 5 ( -5 , 1 ) = ( -25, 5 )

In case you were asked for A' , B' (test yourself before checking)

A' = 5 (-4,3) = ( -20 , 15)

B' = 5 ( 2, 3 ) = ( 10, 15)

Dividing the dilated by the original pair results in the dilation factor

A'/A = (-20/15)/(-4,3) = 5(-4,3)/(-4,3) = 5

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The sum of 2 consecutive odd integers is 52. If x represents the smaller integer, then the larger integer is

julsineya [31]

Answer:

Step-by-step explanation:

2x+2=52

2x=50

x=25

x+2=27

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3 years ago

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An urn contain 2 white marbles, 3 green marbles, and 5 red marbles. A marble is drawn and then replaced. Then a second marble is

11111nata11111 [884]

Answer:

1/5

Step-by-step explanation:

To find this you would need to multiply the probability of pulling a white marble to the probability of pulling out a green marble.

1)First you would need the probability of pulling out a white marble. There are 10 marbles in total and out of those 2 are white. So the probability of pulling out a white marble would be 2/10. If you simplify that you would get 1/5 for the probability of pulling out a white marble.

2)Next, you would find the probability of pulling out a green marble. Using the same process that we used to find the probability of pulling out a white marble, we would find the answer to be 3/10. All that we did here was green marbles/total marbles. By filling that in we got 3/10 for the probability of pulling out a green marble.

3)Now all that is left is doing probability of pulling a white marble × probability of pulling out a green marble. This would be 1/5 × 3/10. After solving the answer would be 3/15 which we would simplify down to 1/5 as our final answer.

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3 years ago

What is the product 4n/4n-4 n-1/n+1

olga_2 [115]

4n/4n-4 × n-1/n+1
= 4n(n-1)/(4n-4)(n+1)
= (4n^2 - 4n)/(4n^2 + 4n - 4n-4)
= (4n^2 - 4n)/(4n^2 - 4)
= 4(n^2 - n)/4(n^2 - 1)
=(n^2 - n)/(n^2 - 1)

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3 years ago

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Ratling [72]

Answer:

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Step-by-step explanation:

8 0

3 years ago

An instructor wants to write a test with 25 questions where each question is worth 3, 4, or 5 points based on difficulty. He wan

irakobra [83]

ANSWER

Find out the how many 3, 4, and 5 point questions could there be.

To proof

Let us assume that the 3 points based question be = x

Let us assume that the 4 points based question be = y

Let us assume that the 5 points based question be = z

As given

An instructor wants to write a test with 25 questions

than the equation become in the form

x + y + z = 25

he quiz to be worth a total of 100 points.

than the equation is becomes

3x + 4y + 5z =100

As given

He wants the number of 3-point questions to be 4 less than the number of 4-point questions

x = y -4

Than the three equation are

x + y + z = 25 ,3x + 4y + 5z =100 and x = y -4

put x = y -4 in the x + y + z = 25 ,3x + 4y + 5z =100

than

2y + z = 29, 7y +5z= 112

multiply 2y + z = 29 by 5 and subtracted 7y +5z= 112

10 y -7y + 5y -5y = 145 -112

3y = 33

$y = \frac{33}{3}$

y =11

put in the x = y -4

x = 11-4

x= 7

put the value of x ,y in the x + y + z = 25

7 + 11 +z =25

18 +z = 25

z = 25 -18

z=7

therefore

numbers of the 3 point question be = 7

numbers of the 4 point question be = 11

numbers of the 5 point question be = 7

Hence proved

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3 years ago