Help with number 22. Image above.

Mathematics

1 answer:

stellarik [79]2 years ago

3 0

I think the answer to this question is letter A

You might be interested in

If employment is expected to increase by 14.8% between 01/2009 and 01/2016 and the employees number 19, 100 in 01/2009, how many

AlexFokin [52]

Based on the rate of increase, the number of employees that will be there in 2016 is<u> 21,927 people.</u>

<h3>Number of employees in 2016</h3>

This can be calculated by the formula:

= Employees in 2009 x ( 1 + rate of increase)

Solving gives

= 19,100 x (1 + 14.8%)

= 21,926.8

= 21,927 people

In conclusion, there will be 21,927 people in 2016.

Find out more on rates of increase at brainly.com/question/3040628.

3 0

2 years ago

At a local fitness center, members pay a $10 membership fee and $5 for each aerobics class. Nonmembers pay $6 for each aerob

sveta [45]

One way to solve it is writing two equations based on the info so

y = 5x + 10 for members
y = 6x for nonmembers

(the + 10 isn't included in the second equation because nonmembers won't pay a membership fee)

and to find their intersection point (where the values will be the same), you just set them equal to each other:

5x + 10 = 6x

subtract 5x

x = 10 classes

7 0

3 years ago

Which value of a makes the equation 5a+1 1/2=4 true?

ss7ja [257]

Answer:

5a+1 1/2=4

a = 1/2

Step-by-step explanation:

4 0

2 years ago

Lena’s brother loaned her $15. She now has at least $50 to purchase new boots. Write inequality to find m, the amount of money L

anyanavicka [17]

I believe it would be >50-15=m

8 0

3 years ago

Find the quotient. Write your answer in standard form. : 3-i/3+i

s344n2d4d5 [400]

The answer is <span>C. 4/5 - 3/5 i
</span>
$\frac{3-i}{3+i} = \frac{(3-i)(3-i)}{(3+i)(3-i)} \\ \\ 
(a+b)(a-b) = a^{2} -b^{2} \\ \\ 
\frac{(3-i)(3-i)}{(3+i)(3-i)}= \frac{ (3-i)^{2} }{3^{2}-i^{2} } \\ \\ 
(a-b)^{2} =a^{2}-2ab+b^{2} \\ \\ 
 \frac{ (3-i)^{2} }{3^{2}-i^{2} } = \frac{3^{2}-2*3*i+i^{2} }{3^{2}-i^{2} } = \frac{9-6i+i^{2}}{9-i^{2}} \\ \\ 
i = \sqrt{-1} \\ i^{2}=-1 \\ \\ 
\frac{9-6i+i^{2}}{9-i^{2}} = \frac{9-6i-1}{9+1} = \frac{8-6i}{10}= \frac{8}{10}- \frac{6i}{10}= \frac{4}{5}- \frac{3}{5}i$

8 0

2 years ago