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DanielleElmas [232]
3 years ago
12

find each comission, given the sale and the comission rate 1. $2,500, 8% 2. $ 2,00, 7.5% 3. $600, 4.5%

Mathematics
2 answers:
Iteru [2.4K]3 years ago
8 0

Answer:

1. $200 2. $150 3.$27

Step-by-step explanation:

1. 2,500 x .08 = 200

2. 2,000 x .075 = 150

3. 600 x .045 = 27

likoan [24]3 years ago
6 0

Answer:

1. $200 2. $150 3.$27

Step-by-step explanation:


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2/7 +2/5 find the sum and difrence
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Answer:

24/35  and the difference between two fractional numbers with same or equal denominators 5/7 and 2/7.

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Six times a number increased by 5
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I'm not quite sure what you asking here but if your asking how to answer it in "math terms" than I'll be happy to explain :)

So "times" means multiply and "increase" means add so righting this in math terms would be 
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Find the equation of the line that passes through the pair of points. (-5,3) (-5,0)
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x = -5

Step-by-step explanation:

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What is 45 x 45 x 65B <br> what is 45 dived by 45 <br> what is acarology
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If by you mean kilobytes, then:

131.62500 kilobytes

45/45=1

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Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

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3 years ago
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