The net for a triangular prism consists of
2 identical triangles, here all nets have sides 5,12, and 13.
3 rectangles, all with a common dimension: the height H=14 of the prism.
Each of the 3 rectangles should have dimensions
H=14 plus one of the following:
each of 5, 12, 13 corresponding to one side of the base (triangle).
So the dimensions of the rectangles are
5x14, 12x14, 13x14
And dimensions of the triangles are
5,12 and 13.
The total surface area is therefore
(5+12+13)*14 + 2*(5*12/2)
=420+60
=480.
There is only one net that satisfies this condition.
Answer:
The answer would be 2nd top choice and 1st bottom choice
Step-by-step explanation:
Answer and Step-by-step explanation:
<u>How to solve:</u>
Use the Pythagorean Theorem to find the length of side <em>BC</em>.
<u>Pythagorean Theorem Formula:</u>
<u />
<u />
<u />
Side AB will be A, side AC will be B, and side BC will be C.
<u>Plug in values.</u>

<u>This means the answer is A. 17.</u>
<u></u>
<u>I hope this helps!</u>
<em><u>#teamtrees #PAW (Plant And Water)</u></em>
We are given with a limit and we need to find it's value so let's start !!!!
But , before starting , let's recall an identity which is the <em>main key</em> to answer this question
Consider The limit ;
Now as directly putting the limit will lead to <em>indeterminate form 0/0.</em> So , <em>Rationalizing</em> the <em>numerator</em> i.e multiplying both numerator and denominator by the <em>conjugate of numerator </em>

Using the above algebraic identity ;


Now , here we <em>need</em> to <em>eliminate (√x-2)</em> from the denominator somehow , or the limit will again be <em>indeterminate </em>,so if you think <em>carefully</em> as <em>I thought</em> after <em>seeing the question</em> i.e what if we <em>add 4 and subtract 4</em> in <em>numerator</em> ? So let's try !


Now , using the same above identity ;


Now , take minus sign common in <em>numerator</em> from 2nd term , so that we can <em>take (√x-2) common</em> from both terms

Now , take<em> (√x-2) common</em> in numerator ;

Cancelling the <em>radical</em> that makes our <em>limit again and again</em> <em>indeterminate</em> ;

Now , <em>putting the limit ;</em>
