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3 years ago

Can someone help me please

Mathematics

1 answer:

Vesna [10]3 years ago

8 0

Answer:

#5) Its line “k”

Well to solve this you need to know what negative slopes are.

They decrease from height from left to right

And also a slope is always calculated like this

rise -> run

Think of it like a bed.

You have to wake up which means you have to rise and then you run to do whatever it is you gotta do.

Rise is going vertical and run is going horizontal.

So for line K you have it because since its negative its -3/1 so technically its -3 rise(basically fall since its negative), 1 run (right).

#6) 1/2 is the slope of line N. Or (c)

Well n is a potsive slope so we can automatically remove a and b.

And we can count like this, it started from the orgin it looks like (but you can use any given point really) and then just follow it, rise 1 and run 2. Rise 2 and run 3 doesn’t work so it has to be 1.2 or (c)

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See attached picture, is there a way to enter equations or algebraic symbols?

coldgirl [10]

SOLUTIONS

Solve a given equations or algebraic symbols?

$\begin{gathered} f(x)=x^2+2x \\ g(x)=1-x^2 \end{gathered}$

(A)

$\begin{gathered} (f+g)(x)=(x^2+2x)+(1-x^2) \\ collect\text{ like terms} \\ x^2-x^2+2x+1 \\ (f+g)(x^)=2x+1 \end{gathered}$

(B)

$\begin{gathered} (f-g)(x)=(x^2+2x)-(1-x^2) \\ =x^2+2x-1+x^2 \\ =x^2+x^2+2x-1 \\ (f-g)(x)=2x^2+2x-1 \end{gathered}$

(C)

$\begin{gathered} fg(x)=(x^2+2x)(1-x^2) \\ =x^2-x^4+2x-2x^3 \\ fg(x)=-x^4-2x^3+x^2+2x \end{gathered}$

(D)

$\begin{gathered} \frac{f}{g}(x)=(x^2+2x)\div(1-x^2) \\ =\frac{x^2+2x}{1-x^2} \end{gathered}$

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1 year ago

Find the simple interest.

Mila [183]

900(1.03)^5
900*1.1593=1,043.37 answer

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3 years ago

there are 6 twenty pound boxes and 8 sixty pound boxes of produce on a truck. how many pounds of produce are on the truck

Klio2033 [76]

6 *20 = 120

8 *60 = 480

120 + 480 = 600 pounds total

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4 years ago

4TH TIME ASKING THIS!!! Please help me! Someone pleaseeee. I need the correct answers. I don’t want to fail

Alexeev081 [22]

Answer:

The functions are inverses; f(g(x)) = x ⇒ answer D

$h^{-1}(x)=\sqrt{\frac{x+1}{3}}$ ⇒ answer D

Step-by-step explanation:

* Lets explain how to find the inverse of a function

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = $f^{-1}(x)$

* Lets use these steps to solve the problems

∵ $f(x)=\sqrt{x-3}$

∵ f(x) = y

∴ $y=\sqrt{x-3}$

- Exchange x and y

∴ $x=\sqrt{y-3}$

- Square the two sides

∴ x² = y - 3

- Add 3 to both sides

∴ x² + 3 = y

- Change y by $f^{-1}(x)$

∴ $f^{-1}(x)=x^{2}+3$

∵ g(x) = x² + 3

∴ $f^{-1}(x)=g(x)$

∴ The functions are inverses to each other

* Now lets find f(g(x))

- To find f(g(x)) substitute x in f(x) by g(x)

∵ $f(x)=\sqrt{x-3}$

∵ g(x) = x² + 3

∴ $f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x$

∴ f(g(x)) = x

∴ The functions are inverses; f(g(x)) = x

* Lets find the inverse of h(x)

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

∴ x + 1 = 3y²

- Divide both sides by 3

∴ $\frac{x + 1}{3}=y^{2}$

- Take √ for both sides

∴ ± $\sqrt{\frac{x+1}{3}}=y$

∵ x ≥ 0

∴ We will chose the positive value of the square root

∴ $\sqrt{\frac{x+1}{3}}=y$

- replace y by $h^{-1}(x)$

∴ $h^{-1}(x)=\sqrt{\frac{x+1}{3}}$

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