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3 years ago

Let f(x) = 3x - 6 and g(x) = x - 2. Find f(g(5)) - ie. (f o g)(5) and its domain

1 answer:

4 0

Do this by finding g(5) first. Replacing x with 5 in g(x) gives you that g(5) = 3. Now replace that 3 in the x in f(x) to get that f(3) = 3. As far as the domain goes,, they are both lines and the domain for lines is all real numbers.

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34kurt

Function B best models the researcher's data because it passes through most of the points.

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4 0

1 year ago

Y=2x+5 and y=3x+11 solve by substitution.

kondaur [170]

Answer:

Answer: x= -6 and y= -7

Step-by-step explanation:

Substitute 2x+5 for y in y=3x+11:

y=3x+11

2x+5=3x+11

2x+5+−3x=3x+11+−3x(Add -3x to both sides)

−x+5=11

−x+5+−5=11+−5(Add -5 to both sides)

−x=6

−x−1=6−1(Divide both sides by -1)

x=−6

Step: Substitute −6 for x in y=2x+5:

y=2x+5

y=(2)(−6)+5

y=−7(Simplify both sides of the equation)

-Source: MathPapa

6 0

3 years ago

HELP ME PLZ<br> ******************

lorasvet [3.4K]

Answer:

504 km

Step-by-step explanation:

divide 180 by 15 (because 3 hours is 180 minutes)

after you divide you get 12

then multiply 12 by the 42 kilometers and you get 504

7 0

3 years ago

Read 2 more answers

The formula d = rt can be used to calculate the distance (d) an object travels using its rate of speed (r) and the time it trave

riadik2000 [5.3K]

Start with d=rt. Now, solve for r by isolating it. since r is multiplied by t, divide both sides by t. d/t=r. So your answer would be the 2nd choice.

3 0

3 years ago

How do I complete these questions a bit confused . (extra credit work )

otez555 [7]

Answer:

(1)

$a = \frac{3\sqrt 3}{2}$

$b = \frac{3}{2}$

(2)

$a = \sqrt 6$

$b = \sqrt 2$

Step-by-step explanation:

Solving (1):

Considering

$\theta = 60^o$

We have:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

This gives:

$\sin(60^o) = \frac{a}{3}$

Solve for a

$a = 3 * \sin(60^o)$

$\sin(60^o) = \frac{\sqrt 3}{2}$

So:

$a = 3 * \frac{\sqrt 3}{2}$

$a = \frac{3\sqrt 3}{2}$

To solve for b, we make use of Pythagoras theorem

$3^2 = a^2 + b^2$

This gives

$3^2 = (\frac{3\sqrt 3}{2})^2 + b^2$

$9 = \frac{9*3}{4} + b^2$

$9 = \frac{27}{4} + b^2$

Collect like terms

$b^2 = 9 - \frac{27}{4}$

Take LCM and solve

$b^2 = \frac{36 - 27}{4}$

$b^2 = \frac{9}{4}$

Take square roots

$b = \frac{3}{2}$

Solving (2):

Considering

$\theta = 60^o$

We have:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

This gives:

$\sin(60^o) = \frac{a}{2\sqrt 2}$

Solve for a

$a = 2\sqrt 2 * \sin(60^o)$

$\sin(60^o) = \frac{\sqrt 3}{2}$

So:

$a = 2\sqrt 2 * \frac{\sqrt 3}{2}$

$a = \sqrt 2 * \sqrt 3$

$a = \sqrt 6$

To solve for b, we make use of Pythagoras theorem

$(2\sqrt 2)^2 = a^2 + b^2$

This gives

$(2\sqrt 2)^2 = (\sqrt 6)^2 + b^2$

$8 = 6 + b^2$

Collect like terms

$b^2 = 8 - 6$

$b^2 = 2$

Take square roots

$b = \sqrt 2$

3 0

3 years ago

Let f(x) = 3x ­- 6 and g(x) = x ­- 2. Find f(g(5)) - ie. (f o g)(5) and its domain

Let f(x) = 3x - 6 and g(x) = x - 2. Find f(g(5)) - ie. (f o g)(5) and its domain