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adoni [48]
3 years ago
9

The ratio of the measures of the sides of a triangle is 4:7:5. If the perimeter of the triangle is 128 yards, find the length of

the longest side.
Mathematics
1 answer:
andrew11 [14]3 years ago
3 0

Answer:

56

Step-by-step explanation:

4+7+5=16

7/16*128=56

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What is x given ABC~DBE.<br> Show your work
alexandr1967 [171]

x = 37.5 (or) \frac{75}{2}

Solution:

Given \triangle A B C \sim \triangle D B E

AC = 50, DE = 30, EC = 25, BE = x, BC = 25 + x

To find the value of x:

Property of similar triangles:

If two triangles are similar then the corresponding angles are congruent and the corresponding sides are in proportion.

$\frac{BE}{BC} =\frac{DE}{AC}

$\frac{x}{25+x} =\frac{30}{50}

Do cross multiplication, we get

50x=30(25+x)

50x=750+30x

Subtract 30x from both sides of the equation.

20x=750

Divide by 20 on both sides of the equation, we get

x = 37.5 (or) \frac{75}{2}

Hence the value of x is 37.5 or \frac{75}{2}.

5 0
3 years ago
8=4y-8+4 solve the equation
Yuliya22 [10]

Answer:

y=3

Step-by-step explanation:

4y-8+4=8

4y-4=8

4y-4+4=8+4

4y=12

4y/4=12/4

y=3

3 0
3 years ago
Read 2 more answers
Flora made 4 withdrawals of $80 each from her bank account. What was the overall change
Len [333]

Answer:

$320 if your saying how much in all its basically 8 times 4 add a zero to the end

Step-by-step explanation:

4 0
2 years ago
1. Jose and Carlos received P 450 as prize for
Ipatiy [6.2K]

Answer:

1.Jose received 5/9×P450=P250

2.A dozen =12

12/3=4×165=660

3.40/1002/5

4. 100/60×9=15

5. 60/100×40=24

8 0
2 years ago
A consumer organization estimates that over a​ 1-year period 20​% of cars will need to be repaired​ once, 8​% will need repairs​
Sladkaya [172]

Answer:

Probability that a car need to be repaired​ once = 20% = 0.20

Probability that a car need to be repaired​ twice = 8% = 0.08

Probability that a car need to be repaired​ three or more = 2% = 0.02

a) If you own two​ cars what is the probability that  neither will need​ repair?

Probability that a car need to be repaired​ once , twice and thrice or more= 0.20+0.08+0.02=0.3

Probability that car need no repair = 1-0.3=0.7

Neither car will need repair=0.7 \times 0.7=0.49

​b) both will need​ repair?

Probability both will need​ repair = 0.3 \times 0.3=0.09

c)at least one car will need​ repair

Neither car will need repair=0.7 \times 0.7=0.49

Probability that at least one car will need​ repair= 1-0.49 = 0.51

6 0
2 years ago
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