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3 years ago

Subtract 10x - 7 from 2x – 6.

Mathematics

2 answers:

hichkok12 [17]3 years ago

8 0

(2x - 6) - (10x - 7) # Starting expression

-8x - 1 # Combine like terms

Final answer:

-8x - 1

Hope this helps!

Artyom0805 [142]3 years ago

7 0

The formula is
2x-6-10x-7
We group the like and unlike terms
2x-10x-6-7
-8x-14

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The equation C = 30x+120 represents the cost of producing gaming chairs. The equation R = 50x represents the revenue generated b

ANEK [815]

Answer:

6 chairs

Step-by-step explanation:

The break even Point on gaming chairs can be expressed as :

Revenue = Cost

Revenue, R = 50x

Cost, C = 30x + 120

R = C

50x = 30x + 120

50 - 30x = 120

20x = 120

Divide both sides by 20

20x/ 20 = 120 / 20

x = 6

Hence, 6 chairs must be sold for the company to break even.

4 0

3 years ago

A tennis player keeps track of the number of successful first serves he makes. During the first 8 service points of a game, only

Andru [333]

Answer:

He needs 7 more consecutive successful first serves to raise his first serve percentage to 60%.

Step-by-step explanation:

After n consecutive serves, his total number of serves is going to be n+8, since he has already served 8 times. In the best case, his number of successful first serves is n+2.

His percentage of succesful first serves is the division of the number of succesful first serves divided by the total number of serves. So

$P = \frac{n+2}{n+8}$

We want $P = 0.60$ . So

$0.6 = \frac{n+2}{n+8}$

$n+2 = 0.6*(n+8)$

$n + 2 = 0.6n + 4.8$

$n - 0.6n = 4.8 - 2$

$0.4n = 2.8$

$n = \frac{2.8}{0.4}$

$n = 7$

He needs 7 more consecutive successful first serves to raise his first serve percentage to 60%.

5 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$