PLEASE HELP GEOMETRY!!!

Calculating the Surface Area of a Triangular Prism The triangular prism shown has 1235 triangular face(s) and 1235 lateral face(

GenaCL600 [577]

Corrected Question

The triangular prism shown has 1, 2, 3 or 5 triangular faces and 1,2,3, or 5 lateral faces.

The area of one triangular face is 7.5,8.125,15,or 17 $mm^2$

The surface area of the triangular prism is 55.5,127.5,135,or 150 $mm^2$

Answer:

(B)2 triangular faces

(C)3 lateral faces.

(A)Area of one Triangular Face $=7.5mm^2$

(C)Total Surface Area of the Triangular prism $=135mm^2$

Step-by-step explanation:

The triangular prism is attached.

The triangular prism shown has 2 triangular faces and 3 lateral faces.

Base of the triangle =2.5mm

Height of the Triangle =6mm

Area of one Triangular Face $=\frac{1}{2}*6*2.5=7.5mm^2$

The dimensions of the lateral rectangles are:

2.5 mm by 8mm
6 mm by 8mm
6.5 mm by 8mm

Therefore, total surface area of the triangular prism

=2(Area of one Triangular Face)+Area of 3 rectangular faces

$=2(7.5)+ (2.5 X 8+ 6 X 8 + 6.5 X 8)\\=15+120\\=135mm^2$

Total Surface Area of the Triangular prism $=135mm^2$

8 0

3 years 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

Which expression represent a number that is one fourth as great as 10-2

Nataliya [291]

I think the answer to the expression would be D. (10 - 2) x 4 because (10 - 2) would be multiplied 4 times, which means it would be four times greater. I hope this made sense and also helped you in some way.

5 0

3 years ago

Read 2 more answers

How many do u wanna bet nobody on this app is not going to help me on this question

Vika [28.1K]

The last one :))))))))))

5 0

3 years ago

PLEASE HELP WILL MARK YOU BRAINLIEST !!!!! D; I SWEARR

Setler79 [48]

Whattttttt!!!!!!!!!!!!!!!!!

7 0

2 years ago