Hi there!
The formula for the lateral area of a cylinder is LA = 2 x pi x r x h. Using this formula, we can plug in the given values and solve for the lateral area.
WORK:
LA = 2pi x r x h
LA = 2pi x 4.5 x 17
LA = 153pi mm^2
The correct answer is the first option - 153pi mm^2
Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
Step 1: Factor out variable m.<span><span>m<span>(<span><span>−<span>50n</span></span>+35</span>)</span></span>=<span>3p</span></span>Step 2: Divide both sides by -50n+35.<span><span><span>m<span>(<span><span>−<span>50n</span></span>+35</span>)</span></span><span><span>−<span>50n</span></span>+35</span></span>=<span><span>3p</span><span><span>−<span>50n</span></span>+35</span></span></span><span>m=<span><span>3p</span><span><span>−<span>50n</span></span>+35</span></span></span>Answer:<span>m=<span><span><span>3p</span><span><span>−<span>50n</span></span>+35</span></span></span></span>
Answer: 26 13/30 feet
Step-by-step explanation:
The hall is in the shape of a rectangle. The perimeter of a rectangle is:
= 2(length + width)
= 2l + 2w
Since perimeter = 71 2/3 ft and Width = 9 2/5 ft wide, the length will be calculated thus:
Perimeter = 2l + 2w
71⅔ = 2l + 2(9 2/5)
71 2/3 = 2l + (2 × 47/5)
71 2/3 = 2l + 94/5
71 2/3 = 2l + 18 4/5
Collect like terms
2l = 71 2/3 - 18 4/5
2l = 71 10/15 - 18 12/15
2l = 52 13/15
l = (52 13/15 ÷ 2)
l = 793/15 × 1/2
l = 793/30
l = 26 13/30
Therefore, the length is 26 13/30ft
She has $55. She spent $22.50 on rides so now she has $32.50 . Then she spends $14.25 on lunch and snack so now she has $18.25.
A. would be wrong because Courtney IS correct, she has $18.25 left.
C. is also wrong because, again, Courtney IS correct.
So the only option left is B and it explains why Courtney is correct.
Option B is the correct answer.
Hope this helps.
9514 1404 393
Explanation:
Here's one way to go at it.
Draw segments AB and CO. Define angles as follows. (The triangles with sides that are radii are all isosceles, so their base angles are congruent.)
x = angle OAB = angle OBA
y = angle OAC = angle OCA
z = angle OBC = angle OCB
Consider the angles at each of the points A, B, C.
At A, we have ...
angle CAB = x + y
At B, we have ...
angle CBA = x + z
At C, we have ...
angle ACB = y + z
The sum of the angles of triangle ABC is 180°, as is the sum of angles in triangle ABO. This gives ...
x + x + ∠AOB = (x+y) +(x+z) +(y+z)
∠AOB = 2(y+z) = 2∠ACB
This shows ∠AOB = 2×∠C, as required.
