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3 years ago

Every hour of training raises the salary $

Mathematics

1 answer:

Nadya [2.5K]3 years ago

8 0

Answer:

20 dollars

Step-by-step explanation:

Difference in salary between any two hours with a 10 hour gap is 200 dollars

so for on hour it is 20$

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Find the area of the sector . Round to the nearest tenth

Eva8 [605]

Answer:

maybe its 30688399329932080?

Step-by-step explanation:

6 0

3 years ago

The radius or diameter is given. Find the missing measures to the nearest hundredth.

Fynjy0 [20]

Answer:

For any circle of radius r

The diameter of the circle = two times of the radius = 2r

And The Circumference= C = 2πr

14. If r = 8

So, d = 2r = 2 * 8 = 16 cm

C = 2πr = 2 * π * 8 = 50.27 cm (to the nearest hundredth.)

15. If d = 9 m

So, d = 2r

∴ r = d/2 = 9/2 = 4.5 m

And C = 2πr = 2 * π * 4.5 = 28.27 m (to the nearest hundredth.)

3 0

3 years ago

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

I will try to give as many details as possible.

First of all, I just would like to say:

$\text{Use } \LaTeX !$

Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! （＊＾Ｕ＾）人（≧Ｖ≦＊）/

$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

3 years ago

Natalie was given a 30 question test. For each correct question she answered, she earned 2 points. For every incorrect answer sh

RideAnS [48]

The number of questions Natalie answered incorrectly.

8 0

3 years ago

Jack said, "five times my age in 2 years is 100." How old is jack now?

denpristay [2]

10 ???? im Guessing.. Hope Its Right (: pretty sure it is ..

6 0

3 years ago

Read 2 more answers