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Anit [1.1K]
3 years ago
14

Is Gross pay more or less than net pay?

Mathematics
1 answer:
GREYUIT [131]3 years ago
6 0
Be more than net pay
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Wendy says that soccer balls cost 2 1/2 times as much as baseball cost. do you agree? explain?​
max2010maxim [7]
It depends on the quality of the balls. It also depends on where you are buying them. Do you have any more details?
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3 years ago
Which of these number is irrational?<br><br> a) 9.6<br> b) √45 <br> c) 3/8<br> d) 3.4 x 10 ⁵
bearhunter [10]

Answer:

I think (D)

Step-by-step explanation:

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3 years ago
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Si se invierten $ 500 a una tasa de 5% anual. Hallar el valor futuro a 3 años si el interés es compuesto trimestralmente.
9966 [12]

Answer:

the future value is $5800.38

Step-by-step explanation:

Given that

The invested amount i.e present value is $500

The rate is 5 % per year so quarterly rate is 5% ÷ 4 = 1.25%

The time period is 3 per year so for quartely it is 3 × 4 = 12

We need to find out the future value

So as we know that

Future value = Present value × (1 + rate of interest)^time

= $500 × (1 + 0.0125)^12

= $580.38

hence, the future value is $5800.38

5 0
2 years ago
What is the length of each piece????PLEASE HELP!!!!
Alex787 [66]
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3 0
3 years ago
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I know you want to answer this question.
Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

\frac{1}{2} ^{x-4} - 3 = 4^{x-3} - 2

First, convert 4^{x-3} to base 2:

4^{x-3} = (2^{2})^{x-3}

\frac{1}{2} ^{x-4} - 3 = (2^{2})^{x-3} - 2

Next, convert \frac{1}{2} ^{x-4} to base 2:

\frac{1}{2} ^{x-4} = (2^{-1})^{x-4}

(2^{-1})^{x-4} - 3 =  (2^{2})^{x-3} - 2

Apply exponent rule: (a^{b})^{c} = a^{bc}:

(2^{-1})^{x-4} = 2^{-1*(x-4)}

2^{-1*(x-4)} - 3 = (2^{2})^{x-3} - 2

Apply exponent rule: (a^{b})^{c} = a^{bc}:

(2^{2})^{x-3} = 2^{2(x-3)}

2^{-1*(x-4)} - 3 = 2^{2(x-3)} - 2

Apply exponent rule: a^{b+c} = a^{b}a^{c}:

2^{-1(x-4)} = 2^{-1x} * 2^{4}, 2^{2(x-3)} = 2^{2x} * 2^{-6}

2^{-1 * x} * 2^{4} - 3 = 2^{2x} * 2^{-6} - 2

Apply exponent rule: (a^{b})^{c} = a^{bc}:

2^{-1x} = (2^{x})^{-1}, 2^{2x} = (2^{x})^{2}

(2^{x})^{-1} * 2^{4} - 3 = (2^{x})^{2} * 2^{-6} - 2

Rewrite the equation with 2^{x} = u:

(u)^{-1} * 2^{4} - 3 = (u)^{2} * 2^{-6} - 2

Solve u^{-1} * 2^{4} - 3 = u^{2} * 2^{-6} - 2:

u^{-1} * 2^{4} - 3 = u^{2} * 2^{-6} - 2

Refine:

\frac{16}{u} - 3 = \frac{1}{64}u^{2} - 2

Add 3 to both sides:

\frac{16}{u} - 3 + 3 = \frac{1}{64}u^{2} - 2 + 3

Simplify:

\frac{16}{u} = \frac{1}{64}u^{2} + 1

Multiply by the Least Common Multiplier (64u):

\frac{16}{u} * 64u = \frac{1}{64}u^{2} + 1 * 64u

Simplify:

\frac{16}{u} * 64u = \frac{1}{64}u^{2} + 1 * 64u

Simplify \frac{16}{u} * 64u:

1024

Simplify \frac{1}{64}u^{2} * 64u:

u^{3}

Substitute:

1024 = u^{3} + 64u

Solve for u:

u = 8

Substitute back u = 2^{x}:

8 = 2^{x}

Solve for x:

x = 3

4 0
3 years ago
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