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dolphi86 [110]
3 years ago
11

Find two consecutive ODD integers such that: The square of the larger is 120 more than the square of the smaller

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
6 0

Answer: See explanation

Step-by-step explanation:

Le the number be x and x+2

The information given can be formed into and equation as:

(x + 2)² = x² + 120

(x + 2)(x + 2) = x² + 120

x² + 2x + 2x + 4 = x² + 120

Collect like terms

x² - x² + 2x + 2x = 120 - 4

4x = 116

x = 116/4

x = 29

The smallest number is 29

The larger number is x + 2 = 29 + 2 = 31

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3 years ago
Find the length of a diagonal of a square with sides 10in. long
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Hey!!!
Answer is....
diagonals of square bisect each other at right angles.
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{x}^{2}  +  {x}^{2}  =  {10}^{2}
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{x }^{2}  = 50
x =  \sqrt{50}
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3 0
3 years ago
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Answer:

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5 0
2 years ago
A rocket is launched from the top of a 99-foot cliff with an initial velocity of 122 ft/s.
Harman [31]
Remember that c is the initial height. Since we the rocket is in a 99-foot cliff, c=99. Also, we know that the velocity of the rocket is 122 ft/s; therefore v=122
Lets replace the values into the the vertical motion formula to get:
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Notice that the rocket hits the ground at the bottom of the cliff, which means that the final height is 99-foot bellow its original position; therefore, our final height will be h=-99
Lets replace this into our equation to get:
-99=-16 t^{2} +122t+99
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Now we can apply the quadratic formula t= \frac{-b+or- \sqrt{ b^{2} -4ac} }{2a} where a=-16, b=122, and c=198
t= \frac{-122+or- \sqrt{ 122^{2}-(4)(-16)(198) } }{(2)(-16)}
t= \frac{-122+ \sqrt{27556} }{-32} or t= \frac{-122- \sqrt{27556} }{-32}
t= \frac{-122+166}{-32} or t= \frac{-122-166}{-32}
t= \frac{-11}{8} or t=9

Since the time can't be negative, we can conclude that the rocket hits the ground after 9 seconds.
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3 years ago
Read 2 more answers
The formula for the area of a square is s2, where s is the side length of the square. What is the area of a square with a side l
levacccp [35]

36

area of square = s²

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area = 6² = 36



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3 years ago
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