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Georgia [21]
3 years ago
11

A triangular prism is shown below. What is the lateral surface area of this prism, in square meters?

Mathematics
2 answers:
Feliz [49]3 years ago
7 0

Answer:

10? I cant see it

Step-by-step explanation:

AlladinOne [14]3 years ago
4 0
Me neither here are some glasses person above me

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4.) You have saved $2000 for college and are saving at the rate of $250 per month. Assume
Veseljchak [2.6K]

Answer: 5,000

Step-by-step explanation: 2,000 saved 250 per month so 250 times

7 0
3 years ago
Show working please!
Nina [5.8K]
You do -2 plus 4. Answer is 2.
Then do 2-(-2) which is 0.
The final answer is 2.
7 0
3 years ago
On a number line what nomber is 3 units to the left of 3​
tatyana61 [14]

Answer:

Hello There!!

Step-by-step explanation:

The answer is 0. The number is 3 and you move 3 units to the left which is toward the negative numbers sides below zero which equals the answer as 0. [3-3=0.]

hope this helps,have a great day!!

~Pinky~

7 0
3 years ago
HELP HELP HELP HELP WOPFOPOPMPOMGOPRDAGS
Basile [38]

Answer:

Step-by-step explanation:

You can't ever let this go negative. At least not at the grade you are in.

It can be 0.

So the domain must start at x = 7

sqrt(5*7 - 35) = sqrt(0) = 0

x can have any value (including 7) between 7 and infinity. If you choose a number less than 7 (like 6) the square root will go negative and that's not to be done.

So the interval is

7 ≤ x < ∞

4 0
3 years ago
Here are the endpoints of the segments BC, FG, and JK.<br> B, −67
yulyashka [42]

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{[-4 - (-6)]^2 + [4 - 7]^2}\implies BC=\sqrt{(-4+6)^2+(-3)^2} \\\\\\ BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}

F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}

JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill

4 0
2 years ago
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