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bixtya [17]
3 years ago
15

How Do I solve -7(2v-2)+4v=4(v+1)

Mathematics
1 answer:
diamong [38]3 years ago
4 0
V = 5/7 in fraction from and in decimal form 0. 714285 . Hope this helped . Please mark me brainliest
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20 POINTS , ALSO GIVING BRAINLIEST , PLEASE HURRY !!
liq [111]

Answer:

.5236

Step-by-step explanation:

Formula for sphere volume:

V = \frac{4}{3} \pi r^{3}

substiture 0.5 for r:

\frac{4}{3} \pi 0.5^{3} = .5236

5 0
2 years ago
Mandy has 8 ounces of fish food to feed her fish . she feeds the fish 0.4 of fish food at each feeding how many times can Mandy
sergejj [24]
20 times

This is just simple division. Do 8 divided by 0.4 and you should get 20

Hope this helps!
-Coconut;)
8 0
2 years ago
#7) Solve 8c - 3c + 3 = 13.
nordsb [41]

Hello!

\large\boxed{c = 2}

8c - 3c + 3 = 13

Begin by combining like terms:

5c + 3 = 13

Subtract 3 from both sides:

5c + 3 - 3 = 13 - 3

5c = 10

Divide both sides by 5:

5c/5 = 10/5

c = 2.

6 0
3 years ago
Factor these expressions into an equivalent form.
Marrrta [24]
(c+8)(c-8)

3(2y+5)(2y-5)

There are no like terms so it's still ab^2-b
7 0
2 years ago
Read 2 more answers
At a college, 69% of the courses have final exams and 42% of courses require research papers. Suppose that 29% of courses have a
laila [671]

Answer:

a) 0.82

b) 0.18

Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

P(F and R)=0.29.

a)

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

b)

P( NEITHER of two requirements)=P(F' and R')=?

According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.

3 0
2 years ago
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