Answer:
1.35759 x 10 ⁹
Step-by-step explanation:
i hope this helped
1. The shape of cross-section is a circle.
2. The face parallel to ABCD is EFGH. Since this is a a rectangular shape,
A = L*H = 12*6 = 72 cm^2
3. The cross-section parallel to ABC is DEF with h = 12 ft, b= 5ft (where h is the height and b is the base of a right angled triangle).
Area, A = 1/2 *b*h = 1/2*5*12 =30 ft^2
4. Plane BDHF is a rectangle shape whose length is the diagonal of ABCD.
Diagonal BD = sqrt (AB^2+BD^2) = sqrt (8^2+7^2) = 10.63 cm.
Perimeter, P = 2(BD+DH) = 2(10.63+6) = 33.26 cm
Answer:
x . 1 . 3 . 5 . 7 . 9
y . 6 . 18 . 30 . 42 . 54
Step-by-step explanation:
X-values are the odd numbers
Y-values are the corresponding x-values multiplied by 6
You haven't told me what the question is. But I put the mouse
to my forehead, closed my eyes, took a deep breath, and I could
see it shimmering in my mind's eye. It was quite fuzzy, but I think
the question is
"What score does Andrew need on the next test
in order to raise his average to 72% ?"
The whole experience drew an incredible amount of energy
out of me, and the mouse is a total wreck. So we'll just go ahead
and answer that one. I hope it's the correct question.
The average score on 4 tests is
(1/4) (the sum of all the scores) .
In order for Andrew to have a 72% average on 4 tests,
the sum of the 4 scores must be
(4) x (72%) = 288% .
Out of that total that he needs, he already has
(64% + 69% + 73%) = 206%
on the first three tests.
So in order to average 72% for all 4 tests,
he'll need to score
(288% - 206%) = 82%
on the fourth one.
