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grigory [225]
3 years ago
12

A store pays $840 for an oil painting. The store marks up the price by 46%. What is the amount of the mark-up? please help!!!! I

XL is driving me crazy
Mathematics
1 answer:
Setler [38]3 years ago
3 0

Answer:

A store pays $840 for an oil painting. The store marks up the price by 46%. What is the amount of the mark-up?

=

$386.40

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784/ 93 = 8.43010752688
93 * <span>8.43010752688 = 784</span>

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Which set of numbers includes only integers?
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Answer:

{14,23,654,0,9}

Step-by-step explanation:

The first three contain negative numbers and decimal points therefor the last one {14,23,654,0,9} is correct.

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Hank’s pickup can travel 60 miles on 5 gallons of gas. How many gallons will Hank’s pickup need to travel 36 miles
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60 / 5 = 12 miles per gallon


36 miles = 12 miles / gal *  x gallons

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PLS HELP<br> THE SQUARE ROOT OF -17
jek_recluse [69]

Answer:

i√17 ≈ 4.123106i

Step-by-step explanation:

The square root of a negative number is "i" times the square root of its absolute value. It is an imaginary number.

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i is generally the symbol used for √(-1). In some fields, such as electrical engineering, where i has a different meaning, the symbol j may be used instead.

6 0
3 years ago
Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.
gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant  is  mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is  mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)

The partial derivatives of a function are f x and f y.

\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\

&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}

In conclusion,  the area is

A=3 √4 units ^2

Read more about the plane

brainly.com/question/1962726

#SPJ4

5 0
1 year ago
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