Answer:
pqs and onl
Step-by-step explanation:
pqs onl
Answer:
<h2>48</h2>
Step-by-step explanation:
(1)
apple × apple × 2apple = 54
2apple³ = 54 <em>divide both sides by 2</em>
apple³ = 27 → apple = ∛27
apple = 3
(2)
apple + apple × green_apple = 24 <em>substituite apple = 3</em>
3 + 3 × green_apple = 24 <em>subtract 3 from both sides</em>
3 × green_apple = 21 <em>divide both sides by 3</em>
green_apple = 7
(3)
strawberry - bannana - banana = 0 → strawberry = 2banana
(4)
strawberry + banana + cherry = 24 <em>substitute from (3)</em>
2banana + banana + cherry = 24
3banana + cherry = 24
(5)
apple + banana - cherry = -1 <em>substitute apple = 3</em>
3 + banana - cherry = -1 <em>subtr3 from both sides</em>
banana - cherry = -4
Add both sides of (4) and (5)
3banana + cherry = 24
<u>+banana - cherry = -4 </u>
4banana = 20 <em>divide both sides by 4</em>
banana = 5
Substitute it to (4):
3(5) + cherry = 24
15 + cherry = 24 <em>subtract 15 from both sides</em>
cherry = 9
Substitute to the last equation:
3 + 5 × 9 = 3 + 45 = 48
/USED PEMDAS/
According to the calculations made, 410 gumballs will be needed to fill the box.
Since you are trying to figure out how many gumballs you need to fill a 7.3 x 5.0 x 9.4 rectangular box for Halloween, and each gumball has a radius of 1/2 in, if the packing density for spheres is 5/8 of the volume will be filled with gumballs while the rest will be air how many gumballs will be needed, to determine this amount the following calculation must be performed:
- (Volume of box x 5/8) / volume of gumballs = Amount of gumballs
- Volume of a sphere = 4/3 x 3.14 x (radius x radius x radius)
- Volume of a gumball = 4/3 x 3.14 x (0.5 x 0.5 x 0.5) = 0.5235 inches
- ((7.3 x 5 x 9.4) x 5/8) / 0.523 = X
- (343.1 x 5/8) / 0.523 = X
- 214.4375 / 0.523 = X
- 410 = X
Therefore, 410 gumballs will be needed to fill the box.
Learn more in brainly.com/question/1578538
Answer:
9
Step-by-step explanation:
f(x) = x − 7
g(x) = x^2
Find g(4) = 4^2 = 16
Then find f(g(4)) = f(16) = 16 -7 = 9
A=bh
A=(38)(35)
A=1,330 squared inches
The side length of 44in is irrelevant unless you are calculating the perimeter, or distance around the paralellogram.
