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2 years ago

PLZ HELP ME!! if F(x) = x - 7 and G(x) = x, what is G(FX))?

Mathematics

2 answers:

Aleonysh [2.5K]2 years ago

3 0

Answer:

alright so this doesn't make any sense so

Step-by-step explanation:

i can't really help but 12 is the answer

kap26 [50]2 years ago

3 0

Answer:

G(F(x)) = x-7

Step-by-step explanation:

G(F(x)) is composition of functions that means to plug in F(X) into G function.The new variable will be F(X) instead of X.

G(x) = x, F(x) = x - 7

G(F(x)) = x-7

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Explain how you got the answer

Gnoma [55]

Answer:

$\huge\boxed{Center = (3,3),Radius = 2\sqrt{15} }$

Step-by-step explanation:

<u><em>Given equation is</em></u>

$x^2 + y^2 -6x-6y -42 = 0$

Adding 42 to both sides

$x^2 + y^2 -6x-6y = 42\\$

Completing squares

$x^2 -6x+y^2 -6x = 42\\(x)^2 - 2(x)(3) +y^2 - 2(y)(3) = 42$

Adding (3)² => 9 and (3)² => 9 to both sides

$(x-3)^2+(y-3)^2 = 42+9+9\\(x-3)^2 + (y-3)^2 = 60\\(x-3)^2 (y-3)^2 = (2{\sqrt{15})^2}$

Comparing it with $(x-h)^2+(y-k)^2 = r^2$ where Center = (h,k) and Radius = r

We get:

Center = (3,3)

Radius = $2\sqrt{15}$

7 0

3 years ago

The radius of circle C is 6 units and the measure of central angle ACB is StartFraction pi Over 2 EndFraction radians. What is t

monitta

Answer:

Given: The radius of circle C is 6 units and the measure of central angle ACB is StartFraction pi Over 2 EndFraction radians.

What is the approximate area of the entire circle?

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What is the approximate area of the entire sector created by central angle ACB?

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What is the approximate area of the shaded region only?

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6 0

3 years ago

Read 2 more answers

What are the answer to these questions:

Tomtit [17]

1. 40/60 + 15/60 + 24/60 + 10/60 = 1.48

2.30/60 + 20/60 + 15/60 + 12/60 + 10/60 = 1.45

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2 years ago

What is the volume of a sphere with a radius of 6 mm?

DanielleElmas [232]

Maybe 6 but I’m not sure.

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2 years ago

Solve the following quadratic equation using the quadratic formula. Separate multiple answers with a comma if necessary.

Valentin [98]

Answer:

$y^2 -4y +6=0$

$y =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where $a = 1, b= -4 ,c =6$

And replacing we got:

$y = \frac{-(-4) \pm \sqrt{4^2 -4(1)(6)}}{2*1}$

And solving we got:

$y = \frac{4 \pm \sqrt{-8}}{2} =2 \pm 2\sqrt{2} i$

Where $i =\sqrt{-1}$

And the possible solutions are:

$y_1=2 + 2\sqrt{2} i , y_2 = 2 - 2\sqrt{2} i$

Step-by-step explanation:

For this case we use the equation given by the image and we have:

$-y^2 +4y -6=0$

We can rewrite the last expression like this if we multiply both sides of the equation by -1.

$y^2 -4y +6=0$

Now we can use the quadratic formula given by:

$y =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where $a = 1, b= -4 ,c =6$

And replacing we got:

$y = \frac{-(-4) \pm \sqrt{4^2 -4(1)(6)}}{2*1}$

And solving we got:

$y = \frac{4 \pm \sqrt{-8}}{2} =2 \pm 2\sqrt{2} i$

Where $i =\sqrt{-1}$

And the possible solutions are:

$y_1=2 + 2\sqrt{2} i , y_2 = 2 - 2\sqrt{2} i$

4 0

2 years ago