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3 years ago

10

Jennifer got a box of chocolates. The box is a right triangular prism shaped box. It is 7 inches long, and the triangular base m

easures 2 in x 3in x 4in. What is the surface area of the box of chocolates?

1 answer:

Dafna1 [17]3 years ago

4 0

The surface area would be 68 squared inches. I hope this helps!

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The volume of air inside a rubber ball with radius r can be found using the function V(r) = four-thirds pi r cubed. What does V

Nat2105 [25]

Corrected Question

The volume of air inside a rubber ball with radius r can be found using the function V(r) = four-thirds pi r cubed. What does V(5/7) represent?

Answer:

(B)the volume of the rubber ball when the radius equals five-sevenths feet

Step-by-step explanation:

The Volume of a sphere of radius r can be found using the formula:

$V(r)=\dfrac43 \pi r^3$

Therefore comparing the expression: $V(\dfrac57)$ with V(r):

$Radius, r=\dfrac57$

Thus, $V(\dfrac57)$ is the volume of a ball of radius $\dfrac57$ feet.

The correct option is B.

4 0

3 years ago

Read 2 more answers

For graph f what is f(0)?

Darina [25.2K]

Answer:

2

Step-by-step explanation:

The function shows a point at 2 where x = 0.

3 0

2 years ago

3. Today, Dahlia folded 29 paper cranes. Now there are 41 paper cranes in class. How many cranes were there in class yesterday?

ohaa [14]

Answer:

There were 12 cranes in class yesterday. For all the cranes, there were a total of 290 centimeters.

Step-by-step explanation:

41-29=12.

29*10=290

4 0

1 year ago

*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0

3 years ago

Solve: -1/4 (2x + 8) = -6

vredina [299]

Answer:

x = 8

Step-by-step explanation:

-1/4 (2x + 8) = -6

Multiply both sides by -4.

2x + 8 = 24

Subtract 8 from both sides.

2x = 16

Divide both sides by 2.

x = 8

3 0

3 years ago