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3 years ago

12

Naja is paid p dollars per hours she works. for every hour she works over 40 hours, she is paid time and half which means she is

paid 1.5 times her normal hourly rate. She worked 50 hours last week. fill in the blanks!!!
40(___) + ___ (1.5p)

need answer ASAP!!

2 answers:

Andru [333]3 years ago

6 0

Answer:

40p + 10 (1.5p)

Step-by-step explanation:

= 40*p + (50-40) * (1.5p)

= 40p + 10 (1.5p)

Hope this help you :3

hram777 [196]3 years ago

6 0

Answer:

40p + 10 (1.5p)

Step-by-step explanation:

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Her salary 4 years ago would have been $30,000.

If 45,000 is 15,000 less than double her previous salary, we must first add 15,000 to 45,000 to find what double her salary was.

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I hope this helps!

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2 years ago

Two more than 5 times a number is greater than or equal to 3

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Answer:

n ≥ 1/5

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3 years ago

In a parking garage the number of suv's is 40% greater than the number of automobiles. there are 98 suv's. how many automobiles

chubhunter [2.5K]

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3 years ago

How do I solve this?

blagie [28]

Answer:

We'll solve that in parts.

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The bottom of that figure is made of 3 parts.

A) a 2 by 2 right triangle. area equals (2*2)/2 = 2

B) a 2 by 2 square whose area equals 2*2 = 4

C) a 6 by 2 right triangle whose area equals (6*2) / 2 = 6

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Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

Read 2 more answers