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tiny-mole [99]
3 years ago
13

Michael is a salesperson. Be sold an emerald for 637.75 and earns 8% commission. How much commission did Michael earn?

Mathematics
1 answer:
Zolol [24]3 years ago
8 0

Answer:

He earns $51.02

Step-by-step explanation:

If Michael earns an 8% commission, then he is paid 8% of what he sells it for. Therefore, he earns 8% of $637.75.

Your equation is y = 0.08 × 637.75

y = 51.02

He earns $51.02

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Solve the system by elimination​
mihalych1998 [28]

Answer:

(-1, -2)

Step-by-step explanation:

4x - 7y = 10

3x + 2y = -7

The common multiple of the x coefficient is 12.

Multiple 3 to 4 to get 12; 4 to 3 to 12:

3(4x - 7y = 10)

4(3x + 2y = -7)

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12x + 8y = -28

Subtract the two equations:

-29y = 58

y = -2

Substitute y = -2 to either 4x - 7y = 10 or 3x + 2y = -7. Typically, do the easier equation to solve:

3x + 2(-2) = -7

3x - 4 = -7

3x = -3

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The order pair solution is like a coordinate: (x, y)

You are welcome!

Kayden Kohl

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4 0
2 years ago
) f) 1 + cot²a = cosec²a​
notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}

Another identity is:

\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}

Therefore:

\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}

Another identity:

\displaystyle \large{\sin^2x+\cos^2x=1}

Therefore:

\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}

Hence proved, this is proof by using identity helping to find the specific identity.

6 0
3 years ago
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