Answer:
A cube
Step-by-step explanation:
If you fold it, you'll see
It can help you by checking your answer
<span>If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
</span><span><span><span>[e^(<span>2+h) </span></span>− <span>e^2]/</span></span>h </span>= [<span><span><span>e^2</span>(<span>e^h</span>−1)]/</span>h
</span><span>so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:
</span><span>
=<span>e^2</span></span>
B) 2/1
find a point, you go up 2 and then over 1!
hope this helps, feel free to mark brainliest!!
Answer:
(-8,4)
Step-by-step explanation:
Given: 2x-3y=-28 and x+6y=16
solve for a variable and then substitute back into other equation
x + 6y = 16
x = 16 - 6y; now use this in the other equation
2x - 3y = -28; substitute into x
2(16 - 6y) - 3y = -28; distribute 2
32 - 12y - 3y = -28; combine y's
32 - 15y = -28; isolate 15y
32 + 28 = 15y; add the numbers
60 = 15y; divide by 15
y = 4
Now use this to plug into other equation
x + 6y = 16; y=4
x + 6(4) = 16;
x + 24 = 16; subtract 24 from both sides
x = -8
