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alex41 [277]
3 years ago
8

PLEASE PLEASE PLEASE HELP ILL GIVE BRAINEST PLEAESE​

Mathematics
2 answers:
VMariaS [17]3 years ago
6 0
The answer is D btw because there are two of the same x values
AveGali [126]3 years ago
5 0

Answer:

The last one

Step-by-step explanation:

There are two of the same x-values, meaning it is not a function.

Hopefully this helps!

Brainliest please?

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What value of x makes this equation true? x/3 −6=10
Setler [38]

Answer:

48

Step-by-step explanation:

x/3 - 6 = 10

      +6   +6

x/3=16

16*3=48

5 0
3 years ago
What is 7(x + 6) solve it
Elis [28]

Answer:

49

Step-by-step explanation:

First you must do x times 6, x is really just 1. After you do that, you times it by 7.

8 0
3 years ago
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Suppose you have a ​$8-off coupon at a fabric store. You buy fabric that costs ​$7.50 per yard. Write an equation that models th
zhannawk [14.2K]

Answer:

= 7.5x-8

Step-by-step explanation:

Total cost = cost per yard * yard of fabric - coupon

total cost = 7.5 x - 8

y = 7.5x-8

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3 years ago
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PLSSS HELP ASP PLSS HELPPPPPP
AleksandrR [38]

Answer:

9,5 i dont know the other one

Step-by-step explanation:

3 0
2 years ago
Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees a
Ivenika [448]

Answer:

<h2>70 possible groups</h2>

Step-by-step explanation:

Given the total number of employee to be equal to 12 temporary employee.

Number of women employee = 5 women

Number of men employee = 12 - 5 = 7 men

If 4 will be hired as permanent employees, the possible ways they can be grouped so that the 4 temporary employees consists of 3 women and 1 man can be done by applying the combination formula.

Combination has to do with selection. For example, if r objects are to be selected from n pool of similar objects, this can be done in <em>nCr </em>different ways.

nCr = \frac{n!}{(n-r)!r!}

According to question, we are to select 3 women from 5 women and 1 man from 7 men. Based on similarity in sex, this can be done in (5C3* 7C1) different ways .

5C3 * 7C1\\=  \frac{5!}{(5-3)!3!} * \frac{7!}{(7-1)!1!}\\\\= \frac{5!}{2!3!} * \frac{7!}{6!1!}\\\\=  \frac{5*4*3!}{3!*2} * \frac{7*6!}{6!*1}\\\\= \frac{20}{2} * \frac{7}{1}\\  \\= 10*7\\\\= 70\ possible\ groups

8 0
3 years ago
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