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3 years ago

Pleaseee help!

Mathematics

1 answer:

user100 [1]3 years ago

7 0

Step-by-step explanation:

f(3)=?

f(x)=2x+5

put x=3,

f(3)=2(3)+5=6+5=11

------------

g(2)=?

g(x)=x^2-3

put x=2,

g(2)=2^2-3=4-3=1

-----------------

g(f(-1))=?

g(x)=x^2-3

and f(-1)=2(-1)+5= -2+5=3

so g(f(-1))=3^2-3=9-3=6

----------------

f(g(-1))=?

f(x)=2x+5

g(x)=g(-1)=(-1)^2-3=1-3= -2

f(g(-1))=2(-2)+5= -4+5=1

----------------

g(f(x))=?

g(x)= x^2-3

put x=f(x),

g(f(x))=f(x)^2-3=(2x+5)^2-3=4x+25+20x-3=24x+25

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Select all input values for which g(x)=8

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The difference between the roots of the quadratic equation x^2−14x+q=0 is 6. Find q.

pychu [463]

Answer :

The value of q for, the given quadratic equation is 40

Step-by-step explanation :

Given quadratic equation as :

x² - 14 x + q = 0

And , Difference between the roots of equation is 6

Let A , B be the roots of the equation

So, A - B = 6

The roots of the quadratic equation ax² + bx + c = 0 as can be find as :

x = $\frac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}$

x = $\frac{14\pm \sqrt{(-14)^{2}-4\times 1\times q}}{2\times 1}$

or, x = $\frac{-14\pm \sqrt{196-4 q}}{2}$

Or, x = $\frac{-14\pm \sqrt{196-4 q}}{2}$

So , The roots are

A = $-7 + \frac{\sqrt{196-4q}}{2}$

And B = $-7 - \frac{\sqrt{196-4q}}{2}$

∵ The difference between the roots is 6

So, A - B = 6

Or, ( $-7 + \frac{\sqrt{196-4q}}{2}$ ) - ( $-7 - \frac{\sqrt{196-4q}}{2}$ ) = 6

Or, ( - 7 + 7 ) + 2 ( $\sqrt{196-4q}$ = 6

Or, 0 + 2 ( $\sqrt{196-4q}$ = 6

∴ 196 - 4 q = 36

or, 4 q = 196 - 36

or 4 q = 160

∴ q = $\frac{160}{4}$

I.e q = 40

S0, The value of q = 40

Hence The value of q for, the given quadratic equation is 40 . Answer

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3 years ago