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3 years ago

Find the midpoint of the line segment with the given endpoints (-4,4) (5,-1)

Mathematics

2 answers:

malfutka [58]3 years ago

4 0

<h3>Answer :</h3>

A : (x₁ , y₁) = (-4 , 4)

B : (x₂ , y₂) = (5 , -1)

Formula of $\tt{midpoint}$ is given by

◈ (x , y) = (x₂ + x₁)/2 , (y₂ + y₁)/2

_________________________________

⇒ (x , y) = (x₂ + x₁)/2 , (y₂ + y₁)/2

⇒ (x , y) = [5+(-4)]/2 , [-1+4]/2

⇒ (x , y) = 1/2 , 3/2

seraphim [82]3 years ago

3 0

Answer:

(1/2), (3/2)

Step-by-step explanation:

midpoint fomula

(x1+x2)/ 2, (y1+y2)/2

(5-4)= 1 1/2 = (1/2)

(-1+4)=3 3/2=(3/2)

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The Perfect square numbers between 100 and 200 are ____ Pls help.

svetlana [45]

Step-by-step explanation:

The perfect square numbers between 100 and 200 are :

10^2 =100

11^2 =121

12^2 =144

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14^2 =196

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3 years ago

Find this quotient. -36.86 ______ -3.8 The quotient is:

VMariaS [17]

Answer:

9.7

Step-by-step explanation:

First convert the numbers into whole numbers by multiplying both numbers by 100. Then divide the numbers by using long division.

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3 years ago

Rectangle ABCD was dilated to create rectangle A'B'C'D. Triangle A B C D is dilated to form triangle A prime B prime C prime D.

Julli [10]

Answer:

$AB = 6$

Step-by-step explanation:

Given

$BC = 3.8$

$A'B' = 15$

$B'C' = 9.5$

Required

Determine the length of AB

Because A'B'C'D is a dilation of ABCD, then the following relationship must exist:

$A'B' : AB = B'C' : BC$

Substitute values for A'B', B'C' and BC

$15 : AB = 9.5 : 3.8$

Express the ratio as fractions:

$\frac{AB}{15} = \frac{3.8}{9.5}$

Multiply through by 14

$15 * \frac{AB}{15} = \frac{3.8}{9.5} * 15$

$AB = \frac{3.8}{9.5} * 15$

$AB = \frac{3.8* 15}{9.5}$

$AB = \frac{57}{9.5}$

$AB = 6$

8 0

3 years ago

Read 2 more answers

Find the measure of the missing angles! Pls help thank you

Bond [772]

Answer:

Step-by-step explanation:

d = 52°

e = 77°

f = 180° - ( 52° + 77° ) = 51°

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3 years ago

I’m trying to find the answer for these 2 questions. Please!

Elanso [62]

1) For the second semester, we recognize that the number of actual students (75) is 15 times the ratio number (5) used to represent it. We also recognize that the total number of ratio units (5+4) for the second semester is the same as the total number (2+7) for the first semester.

If each ratio unit reprsents 15 students, in the first semester there were

... 2×15 = 30 students in art

... 7×15 = 105 students in gym

2) We can use the same sort of logic for the second problem as for the first. 6 ratio units represents $936 before money is moved, so each represents $156.

After the move, the account balances are

... savings: 4×$156 = $624

... checking: 3×$156 = $468

_____

Since the numbers of "ratio units" add to the same value in each case for both problems, we can solve this the way we have done—by determining the value a "ratio unit" stands for. If the problem had been posed differently, we might have been required to find the total number of students (dollars), then reapportion them according to the new ratios.

Essentially, when you have

... part1 : part2

you can find the total from

... part1 : total = part1 : (part1 + part2)

... (part1 + part2)/(part 1) × (actual value of part 1) = (actual value of total)

4 0

3 years ago