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3 years ago

7

Please help 10 points 16/5√x

1 answer:

r-ruslan [8.4K]3 years ago

7 0

I’m not sure exactly what’s being asked but I think this is the correct answer

16 √x
———-
5

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kell works at an after-school program at an elementray school. on monday he work for 1.5 hours and recived 12.6 dollars, on tues

Rina8888 [55]

Answer:

Kell would make more money working 10 hours than Markio.

Step-by-step explanation:

To find how much Kell makes per hour, you have to divide how much he worked by how much he earned. So on Monday, Kell worked 1.5 hours and received 12.6 dollars. Divide 12.6/1.5 and you'll find you're answer.

Monday: 12.6/1.5=<u>8.4</u>

On Tuesday, the problem states that he worked 2.5 hours and received 21 dollars. Divide 21/2.5 and you'll find the answer.

Tuesday: 21/2.5= <u>8.4</u>

On Wednesday, the problem states that he worked 4 hours and received 33.60. Divide 33.60/4 and you'll get you're answer.

Wednesday: 33.60/4= <u>8.4</u>

So we can conclude that Kell makes $8.40 per hour.

Now the problem states, Markio gets paid 7 dollars per hour. To find the answer to the question, you have to multiply how much they make per hour times 10 hours.

Markio: 7x10=70

Kell: 8.4x10=84

So to answer the question, Kell would make more money working 10 hours.

Hope it helps!

3 0

2 years ago

What’s is the answer

kvasek [131]

When you are multiplying an exponent directly into a number/variable with an exponent, you multiply the exponents together.

For example:

$(x^{2} )^{3} = x^6$

$(x^{3} )^5=x^{15}$

When you are multiplying a variable with an exponent by another variable with an exponent, you add the exponents together.

For example:

$(x^{2} )(x^{3})=x^{5}$

$(x^{1} )(x^{2})=x^{3}$

$(\frac{(x^{-3})(y^{2})}{(x^{4})(y^{6})} )^{3}=\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}$

You multiply 3 into each exponent in the numerator and the denominator

$\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}= \frac{y^{6}}{(x^{9})(x^{12})(y^{18})}$

When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive.

$\frac{y^{6}}{(x^{21})(y^{18})} = \frac{1}{(x^{21})(y^{12})}$

When you have something like this:

$\frac{x^{2}}{x^5}$

You subtract the exponents together, so:

$\frac{x^2}{x^5} = x^{2-5} = x^{-3} = \frac{1}{x^3}$

Your answer is the second option

3 0

3 years ago