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devlian [24]
3 years ago
10

the sum of two consecutive integers is -5.if 4 is subtracted from the smaller integer and 2 is added,what is the quotient of the

two resulting integers?
Mathematics
1 answer:
Lerok [7]3 years ago
4 0

Let the two consecutive integers be x and x+1. Given that

x+x+1=-5\\ 2x+1=-5\\ x=-3

The consecutive integers are -3,-2.

4 is subtracted from the smaller integer and 2 is added to the smaller integer the resulting integers are -3-4=-7 and -3=2=-1.

The quotient is \frac{-7}{-1} =7

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Y= -8x + 10/9 what is the y intercept
4vir4ik [10]

Answer:

10/9

Step-by-step explanation:

This equation is in slope-intercept form, this is referring to the y-intercept. Slope-intercept is y=mx+b, where m is the slope and b is the y-intercept. Therefore, 10/9 is the intercept while -8 is the slope.

4 0
3 years ago
Write an equation for each sentence and then solve.
Whitepunk [10]

Answer:

Step-by-step explanation:

Let the two consecutive numbers be 2n and 2n+2

The sum of these numbers in 2n + (2n+2).

Hence

2n + (2n+2) = 34

4n + 2 = 34

4n  = 32

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8 0
3 years ago
Solve the two-step equation. − 2 3 x – 21 4 = 27 4
svetoff [14.1K]

Answer:

Step-by-step explanation:

I'm going to assume that the problem is -23x -214 = 274

And in that case...

1. First isolate the x-value by adding 214 to both sides of the equation:

-23x = 488

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7 0
4 years ago
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Solve the system by the elimination method.
joja [24]

Answer: B

Step-by-step explanation:

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8 0
3 years ago
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What are the two missing sides of the triangle ??
igor_vitrenko [27]
Remark
It's a right triangle so the Pythagorean Theorem applies. All you have to do is put the right things in the right places of the formula.

Givens
a = x
b = x + 4
c = 20

Formula and Substitution.
a^2 + b^2 = c^2
x^2 + (x + 4)^2 = 20^2

Solution
x^2 + x^2 + 8x + 16 = 20 Collect the like terms on the left.
2x^2 + 8x + 16 = 20  Subtract 20 from both sides.
2x^2 + 8x + 16 - 20 = 0  
2x^2 + 8x - 4 = 0         Divide through by 2
x^2 + 4x - 2 = 0

Use the quadratic formula
a = 1
b = 4
c = - 2

\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} 
 
\text{x = }\dfrac{ -4 \pm \sqrt{4^{2} + 4(1)(2) } }{2} 


From which x = (-4 +/- sqrt(24) ) / 2
x1 = (- 4 +/- sqrt(4*6) ) / 2
x1 = (- 4 +/- 2 sqrt(6) ) / 2
x1 = -2 + sqrt(6)
x2 = -2 - sqrt(6)  This is an extraneous root. No line can be minus.

x1 = + 0.4495
x2 = x + 4 = 4.4495
6 0
3 years ago
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