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3 years ago

4/5 x 25 ????????????

Mathematics

2 answers:

Lady_Fox [76]3 years ago

8 0

20 should be the answer

coldgirl [10]3 years ago

7 0

It is 20 i think because 4/5x25 equal 20

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lawyer [7]

Yes because if you graph it, it will pass the vertical line test

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3 years ago

There are 2 questions pls answer fast

Furkat [3]

Answer:

Q1: B. UHWXUQV

Q2: None of these answers equal 12 so I'll just tell you the answers for all of them in case you mistyped.

a. 66

b. 82

c. 120

d. 109

Im assuming you meant 120. If so, your answer is c.

5 0

3 years ago

Use the linear combination method to solve the system of equations. â€"2. 1x 4y = 5. 3 2. 1x â€" 5. 5y = â€"0. 2 What is the sol

RoseWind [281]

The solution to the systems of equations is (7, 3)

Given the systems of equations expressed as:

-x + 4y = 5 ....................1

x - 5y = -2 ...................... 2

Add both equations to have:

-x+x + 4y - 5y = 5 - 2

4y-5y = -3

-y = -3

y = 3

Substitute y = 3 into equation 1:

-x + 4(3) = 5

-x + 12 = 5

-x = 5 - 12

-x = -7

x = 7

Hence the solution to the systems of equations is (7, 3)

Learn more on simultaneous equations here: brainly.com/question/148035

4 0

2 years ago

Aberdine makes a $2,900 purchase on her credit card that has a 21% APR. If she only makes the minimum monthly payment of $70, wh

masya89 [10]

Answer:

Step-by-step explanation:

i took the test

4 0

3 years ago

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

5 0

3 years ago