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8090 [49]
3 years ago
9

What is the value of x?

Mathematics
1 answer:
Andrei [34K]3 years ago
6 0

Answer:

x=2

Step-by-step explanation:

We will use the secant-secant Formula:

(whole secant) x (external part) =         (whole secant) x (external part)

(1+x+4) * (x+4) = (11+x+1) * (x+1)

Combine like terms

(x+5) (x+4) = (x+12) (x+1)

FOIL

x^2 + 4x+5x + 20 = x^2 +  x+ 12x + 12

Combine like terms

x^2 + 9 x + 20 = x^2 + 13 x + 12

Bring everything to the left

x^2 + 9 x + 20- (x^2 + 13 x + 12) = x^2 + 13 x + 12-( x^2 + 13 x + 12)

x^2 + 9 x + 20- (x^2 + 13 x + 12) =0

Distribute the minus sign

x^2 + 9 x + 20- x^2 - 13 x - 12 =0

Combine like terms

-4x +8 = 0

Subtract 8 from each side

-4x+8-8=0-8

-4x=-8

Divide each side by -4

-4x/-4 = -8/-4

x=2

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4 years ago
Write the equation of the line f.
vekshin1

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y = -4

Step-by-step explanation:

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7 0
3 years ago
which function has real zeros at x = −10 and x = −6? f(x) = x2 16x 60 f(x) = x2 − 16x 60 f(x) = x2 4x 60 f(x) = x2 − 4x 60
LiRa [457]

Answer:

Option A is correct

The function x^2+16x+60 has real zeroes at x =-10 and x =-6

Explanation:

Given: The real zeroes or roots are x = -10, and x = -6

To find the quadratic function of degree 2.

x^2- (\alpha+\beta)x + \alpha\beta =0 where α,β are real roots.   ....[1]

Here, α= -10  and β= -6

Sum of the roots:

α+β =  -10+(-6) = -10-6 = -16

Product of the roots:

αβ = (-10)(-6)= 60

Substitute these value in equation [1] we have;

x^2-(-16)x+60 = x^2+16x+60

Therefore, the quadratic function for the real roots at x =-10 and x =-6 ;

x^2+16x+60

8 0
3 years ago
Read 2 more answers
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