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castortr0y [4]
4 years ago
10

Write an algebraic expression for "71 more than the product of 184 and d."

Mathematics
1 answer:
Lerok [7]4 years ago
5 0

Answer:

71 + 184d

Step-by-step explanation:

I'm sorry, I don't know how to explain how I got this. I hope it helps though.

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Find the slope of the line through each pair of points(-13,-18),(0.-17)
Jet001 [13]
The slope formula is

=2−1/2−1

For the first one it would be:

-17-(-18)/0-(-13)=1/13

m=1/13



Second one:

m=−16−−9/−17−−18= -7/1

m=-7

For the there one:

Slope =−12−−13/6−−1=1/7

m=1/7





I hope it was helpful. If you have any questions let me know.

3 0
3 years ago
HELP PLEASE} I have the question on the image.
MatroZZZ [7]
9+13+9+10+10+9+10+10+11+9=100
100/10=10
Hope That Helped
8 0
3 years ago
Read 2 more answers
Three wooden cube blocks each have a side length of 6cm. What is the combined volume of the three wooden blocks in cm?
aleksklad [387]

Answer:

c

Step-by-step explanation:

what you do is you find how many sides there are witch is 6 and multiply that by 6 witch would equal 36 now all you do is multiply 3x36 witch equals 108cm square

4 0
3 years ago
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A line includes the points (-3, -8) and (9, 4). What is its equation in slope-intercept form?
liubo4ka [24]

Answer:

Answer is in the photo :)

Step-by-step explanation:

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8 0
3 years ago
The street sign shown is a regular hexagon with side lengths of (7.6x+10.4) centimeters. The perimeter of the sign is 737.28 cen
lutik1710 [3]

Answer:

Equation:

6(7.6x+10.4)=737.28

Solution:

x=14.8\ centimeters

Step-by-step explanation:

<u>Perimeter of a hexagon </u>

We are given the length of the side of a regular hexagon

l=7.6x+10.4\ centimeters

And we also know the perimeter of the hexagon is

P=737.28\ centimeters

The perimeter of a regular hexagon is 6 times its side, so

P=6l=6(7.6x+10.4)

The equation that will give the relation to solve for x is

6(7.6x+10.4)=737.28

Reducing

7.6x+10.4=122.88

7.6x=122.88-10.4=112.48

\displaystyle x=\frac{112.48}{7.6}

x=14.8\ centimeters

5 0
3 years ago
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