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Alborosie
3 years ago
8

Solve the following equation

Mathematics
2 answers:
Gnoma [55]3 years ago
5 0
40x+8x^2=0 can be solved for x (there are two solutions):

Divide all 3 terms by the greatest common multiple (which is 8x):

40x+8x^2=0
------------- -----
       8x       8x

5 + x = 0             produces the root x = - 5.   

Setting 8x = 0 and solving for x produces the root x = 0.

Be certain to check these results.  substitute x = -5 into 40x+8x^2=0.  Is the resulting equation true or false?  Next, subs. x=-5 into 40x+8x^2=0.  Is the resulting equation true or false?
dexar [7]3 years ago
3 0
40x + 8x²= 0

Move the 8x² over the the left side of 40x, so that the equation is in correct order.

So, now we have :

8x² + 40x = 0

Factorize the left side.

What goes into both 8x² and 40x? 8x do! :)

8x(x + 5) = 0

Set the factors to equal zero.

8x = 0 AND x + 5 = 0

8x = 0

Divide both sides b 8.

x = 0

x + 5 = 0

Subtract 5 from both sides.

x = -5

So, x = 0 AND x = -5

(0, -5)

~Hope I helped!~


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If a yeti cooler is reRegularly priced for $349 but is being discounted by 15% how much money will you say if you buy it on sale
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349*.15=52.35
you save 52.35
6 0
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elaina wrote two correct variable expressions to represent the phrase “one-fourth of a number.” which variable expressions could
antiseptic1488 [7]
One-fourth of a number

1/4n and n/4
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Find the distance between points A and C round to the nearest tenth.
Taya2010 [7]

Well the first thing that came to my mind was to do rise/run and doing that I got 6/5 which I added together to get 11 but since that's obviously not correct I assumed it was a little under so the answer I got was 7.8

Really hope this helps!

6 0
3 years ago
A ramp has a horizontal distance of 10 feet and a vertical rise of 3 feet. Find the length of the ramp. Round to the nearest ten
Tatiana [17]

Answer:

Step-by-step explanation:

Tye length of the ramp can be gotten using Pythagoras theorem.

Horizontal distance = 10ft.

Vertical distance = 3ft

Length of the ramp² = 10²+3²

Length of the ramp² = 100+9

Length of the ramp² = 109

L² = 109

L =√109

L = 10.44ft

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4 0
3 years ago
1948 Olympics
mash [69]

Answer:

For 1948 Men's :Interquartile range is 1.5, Median is 58.3

For 2012 Men's: Interquartile range is 0.315, Median is 47.86

You can infer that in 2012 the swimmers were better and it was more competitive as the interquartile range was lower and the median was also lower as well

Step-by-step explanation:

1948 Men's 100m

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6,1:00.5

               ⬆                ⬆                 ⬆

         Low IQR      Median       High IQR

Low IQR is average of 57.8 and 58.1 = 57.95

High IQR  is average of 59.3 and 59.6 =  59.45

Median is average of 58.3 and 58.3 = 58.3

Interquartile range is High IQR- Low IQR

59.45 - 57.95=1.5

2012 Men's 100m

47.52, 47.53, 47.8, 47.84, 47.88, 47.92, 48.04, 48.44

                    ⬆                  ⬆                      ⬆

             Low IQR         Median          High IQR

Low IQR is average of 47.53 and 47.8 = 47.665

High IQR is average of 48.04 and 47.92 = 47.98

Median is average of 47.88 and 47.84 = 47.86

Interquartile range is High IQR-Low IQR

47.98-47.665=0.315

Read more on Brainly.com - brainly.com/question/14915771#readmore

8 0
4 years ago
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