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Paraphin [41]
3 years ago
14

the sum of three numbers is 79. The second number is 5 times the first, while the third is 16 more than the first. Find the numb

er
Mathematics
1 answer:
swat323 years ago
3 0
Let the 1st number be x; 2nd number be y; 3rd number be z.

x + y + z = 79

x = number we are looking for.
y = x * 5 ==> 5 times the first
z = x + 16 ==> 16 more than the first

Therefor,

x + (x * 5) + (x+16) = 79

1st step, multiply the 2nd number: x * 5 = 5x

x + 5x + x + 16 = 79

Add all like numbers:

7x + 16 = 79

To get x, transfer 16 to the other side and change its sign from positive to negative.

7x = 79 - 16
7x = 63

To get x, divide both sides by 7

7x/7 = 63/7
x = 9

To check. Substitute x by 9.

x + (x * 5) + (x+16) = 79

9 + (9 * 5) + (9 + 16) = 79
9 + 45 + 25 = 79
79 = 79 equal. value of x is correct.

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3 years ago
Read 2 more answers
-3x^{2}-21x-54 what are the zeros? (Solutions)
Irina-Kira [14]
The only way to solve if it is equal to something
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54

remember if we can do
xy=0 then assume x and y=0

so factor

0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is

if you have
ax^2+bx+c=0 then
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}

x^2+7x+18
a=1
b=7
c=18

x=\frac{-7+/- \sqrt{7^{2}-4(1)(18)} }{2(1)}
x=\frac{-7+/- \sqrt{49-72} }{2}
x=\frac{-7+/- \sqrt{-23} }{2}
i=√-1
x=\frac{-7+/- i\sqrt{23} }{2}



the zerose would be
x=\frac{-7+ i\sqrt{23} }{2} or \frac{-7- i\sqrt{23} }{2}




4 0
3 years ago
Show me a step by step work and solve the problem
AveGali [126]
What’s the question?
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3 years ago
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