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Readme [11.4K]
3 years ago
5

What's the answer? (please help soon!)

Mathematics
1 answer:
nignag [31]3 years ago
3 0
Absolute value make it positive

15-|-5|=15-5=10

|-4|+6=4+6=10

-|7+3|=-|10|=-(10)=-10

|-10|=10

so it is -|7+3| that doesn't belong
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Sections of prefabricated fencing are each 4 1/3 feet long. how long are 6 1/2 sections placed end to end
zalisa [80]

we know that

4\frac{1}{3}= \frac{(4*3+1)}{3} =\frac{13}{3}

6\frac{1}{2}= \frac{(6*2+1)}{2}=\frac{13}{2}

To find how long are 6\frac{1}{2}  sections placed end to end

Multiply the number of sections by the length of one section

so

\frac{13}{2} *\frac{13}{3} =\frac{169}{6}\ ft

\frac{169}{6}\ ft =28\frac{1}{6}\ ft

therefore

<u>the answer is</u>

28\frac{1}{6} \ ft


6 0
3 years ago
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A man flies a kite with a 100 foot string the angle of elevation of the string is 52 how high off the ground is the kite
Slav-nsk [51]
The kite is approximately 78.80 feet off of the ground.


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4 0
3 years ago
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If R is the midpoint of TS, TR = 5x - 12, and RS = 3x + 8, find TS.
GenaCL600 [577]

Answer:

5x -12 = 3x +8 (set the two = each other because they are the same length)

2x- 12= 8 (subtract 3x from both sides)

2x = 20 (add 12 to both sides)

x=10 (what x= for both expressions)

5(10) -12 (plug it into the first one to see what the length is and to see they're =

50 - 12 ( I already multiplied, now subtract)

38 (what the length of TR is)

3(10) +8 (plug it in again but into the other expression)

30+8 (multiply and add)

38 (the two have the same answer, so the x-value is correct.)

38+38= 76 (add the lengths of RS and TR and you get the length of TS)

Step-by-step explanation:

I hope this helps :)

6 0
3 years ago
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9÷ 6/5 simplest form
jarptica [38.1K]
15/2 or 7 1/2 or 7.5
8 0
3 years ago
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Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y
mariarad [96]

Answer:

the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

\frac{\pi }{2}  [e^2 - 1 ]  or 10.036

Step-by-step explanation:

Given the data in the question;

y = y = e^{(x - 1 ), y = 0, x = 1, x = 2.

Now, using the integration capabilities of a graphing utility

y = y = e_2}^{(x - 1 )_, y = 0

Volume = \pi \int\limits^2_1 ( e^{x-1)^2} - (0)^2 dx

Volume = \pi \int\limits^2_1 ( e^{x-1)^2  dx

Volume = \pi \int\limits^2_1 e^{2x-2}dx

Volume = \frac{\pi }{e^2} \int\limits^2_1 e^{2x}dx

Volume = \frac{\pi }{e^2}  [\frac{e^{2x}}{2}]^2_1

Volume = \frac{\pi }{2e^2}  [e^4 - e^2 ]  

Volume = \frac{\pi }{2}  [e^2 - 1 ]  or 10.036

Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

\frac{\pi }{2}  [e^2 - 1 ]  or 10.036

   

3 0
3 years ago
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