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IRISSAK [1]
3 years ago
9

Find the value of x in the triangle.

Mathematics
1 answer:
Studentka2010 [4]3 years ago
5 0

Answer:

164

Step-by-step explanation:

Remember, everytriangle is 180 all together. So add what you see and subtract it from 180.

9+7= 16

180-16= 164

Hope it helps

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Solve the following inequality 5x+12+3x<36
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Answer:

x<3

Step-by-step explanation:

5x+12+3x<36

First subtract 12 from both sides.

5x+3x<24

Next combine like terms on the left side (5x and 3x)

8x<24

Next divide 8 from both sides

x<3

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3 years ago
Identify the graph of (x+3)^2 - 4y for T(-7,2) and write and equation of the translated or rotated graph in general form.
Firlakuza [10]

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Step-by-step explanation:

C

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Which is the decimal expansion of the following 1/5
amid [387]

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A rectangle has a length two times the width. Describe how you would find the
IRINA_888 [86]

Answer:

perimeter = 6w

area = 2w^2

Step-by-step explanation:

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4 0
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Read 2 more answers
Please help with #12
ExtremeBDS [4]

Answer:

a. 1 1/8 b. 8/9

Step-by-step explanation:

You can set this up as a proportion to solve.  For part a. we know that 2/3 of the road is 3/4 mile long.  2/3 + 1/3 = the whole road, so we need how many miles of the road is 1/3 its length.  Set up the proportion like this:

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

Cross multiplying gives you:

\frac{2}{3}x=\frac{1}{3}*\frac{3}{4}

The 3's on the right cancel out nicely, leaving you with

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

To solve for x, multiply both sides by 3/2:

\frac{3}{2}*\frac{2}{3}x=\frac{1}{4}*\frac{3}{2} gives you

x=\frac{3}{8}

That means that the road is still missing 3/8 of a mile til it's finished.  The length of the road is found by adding the 3/4 to the 3/8:

\frac{3}{4}+\frac{3}{8}=\frac{6}{8}+\frac{3}{8}=\frac{9}{8}

So the road is a total of 1 1/8 miles long.

For b. we need to find out how much of 1 1/8 is 1 mile:

1 mile = x * 9/8 and

x = 8/9.  When 1 mile of the road is completed, that is 8/9 of the total length of the road completed.

8 0
3 years ago
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