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Andrej [43]
3 years ago
12

18,000 divided by 25%

Mathematics
2 answers:
DanielleElmas [232]3 years ago
7 0
Correct answer - 72000.
Mice21 [21]3 years ago
6 0

Answer:

4500

Step-by-step explanation:

18000 / 100 =180

180*25 = 4500

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Describe how the graph of y | x| -4 is like the graph of y= |x| and how it is different.
zhannawk [14.2K]
The graph y=|x|-4 is obtained from the graph y=|x| dy <span>moving down 4 units the graph y=|x|  along the y-axis (see, if x=0, then for y=|x|, y=0 and for y=|x|-4, y=-4).
</span>
These two graphs have the same form.
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4 0
3 years ago
Read 2 more answers
The locations:
GREYUIT [131]

The quadrilateral is a trapezoid and the area of the quadrilateral is 85.04 square units

<h3>How to determine the quadrilateral?</h3>

The vertices are given as:

A:(-2, 3) B:(4, -6) C:(10, 2) D:(6, 8)

Next, we plot the vertices (see attachment)

From the attached graph, we can see that the quadrilateral is a trapezoid

<h3>How to determine the area?</h3>

From the plot, we have the following features:

Height: AD

Parallel sides: CD and AB

Calculate the lengths using:

d = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2}

So, we have:

AD = \sqrt{(-2 -6)^2 + (3 -8)^2}

AD = \sqrt{89}

CD = \sqrt{(10 -6)^2 + (2 -8)^2}

CD = \sqrt{52}

AB = \sqrt{(-2 -4)^2 + (3 +6)^2}

AB = \sqrt{117}

The area is then calculated as:

Area = 0.5 * (CD + AB) * AD

This gives

Area = 0.5 * (√52 + √117) * √89

Evaluate

Area = 85.04

Hence, the area of the quadrilateral is 85.04 square units

Read more about areas at:

brainly.com/question/24487155

#SPJ1

7 0
2 years ago
I need to transpose the following equation to find the term (g)<br><br> V^2=2gh
nataly862011 [7]
v^2=2gh\\\\2gh=v^2\ \ \ \ |divide\ both\ sides\ by\ 2h\\\\\boxed{g=\frac{v^2}{2h}}
7 0
3 years ago
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Simplify the cube root in simplest radical form.<br> 9x374<br> Please help
algol [13]

Answer:

D. xy\sqrt[3]{9y}

Step-by-step explanation:

\sqrt[3]{9x^3y^4}

\sqrt[3]{9}\sqrt[3]{x^3}\sqrt[3]{y^4}

The \sqrt[3]{x^3} cancels out to become x:

\sqrt[3]{9}x\sqrt[3]{y^4}

Split the y^4=y*y^3

\sqrt[3]{9}x\sqrt[3]{y^3}\sqrt[3]{y^1}

\sqrt[3]{y^3} =y

xy\sqrt[3]{9} \sqrt[3]{y}

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xy\sqrt[3]{9y}

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3 years ago
Someone give me the answers to these please
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Answer:

ITs blury

Step-by-step explanation:

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3 years ago
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