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iogann1982 [59]
3 years ago
15

Ken has 4.66 pounds of walnuts, 2.1 pounds of cashews, and 8 pounds of peanuts. He mixes them together and divides them equally

among 18 bags. How many pounds of nuts are in each bag?
Mathematics
1 answer:
Nat2105 [25]3 years ago
5 0
7.20 pounds in each... IM SRRY IF WRONG
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3 years ago
What is 5 1/4 divided by 7?
NARA [144]
ANSWER: I think it’s 3/4
4 0
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Read 2 more answers
If two angles are complementary. One angle measures 22 degrees. And the other angle measures 2x degrees, what is the value of x?
levacccp [35]

Answer:

x=34

Step-by-step explanation:

34*2=68

68+22=90

A complementary angle adds up to 90 degrees.

5 0
3 years ago
While sitting on a rooftop 95 feet off the ground Mary flicks a twig up and off of the ledge with the initial vertical velocity
cestrela7 [59]

Answer:

The maximum height of the twig is 96.6 feet off the ground

Step-by-step explanation:

To determine the maximum height of the twig,

First, we will determine the height distance covered by the twig after Mary flicks the twig up.

From the question,

Mary flicks a twig up and off of the ledge with the initial vertical velocity of 10 feet per second, that is

the initial velocity of the twig is 10 feet/second

At maximum height, the final velocity is 0

From on the equations of motions for bodies moving upwards,

v² = u² - 2gh

Where v is the final velocity

u is the initial velocity

g is the acceleration due to gravity (Take g = 9.8m/s² = 32.17 ft/s²)

and h is the height

From the question

u = 10 feet/second

and v = 0 feet/second

Putting the values into the equation

v² = u² - 2gh

0² = 10² - 2(32.17)h

0 = 100 - 64.34h

64.34h = 100

h = 100/64.34

h = 1.6 feet

This is the height distance covered by the twig after Mary flicks it up.

Now, the maximum height of the twig will be the sum of the height of the rooftop from the ground and the height distance covered by the twig.

That is,

Maximum height = 95 feet + 1.6 feet

Maximum height = 96.6 feet

Hence, the maximum height of the twig is 96.6 feet off the ground.

7 0
3 years ago
Suppose an airline policy states that all baggage must be box shaped with a sum of​ length, width, and height not exceeding 114
NISA [10]

Answer:

Step-by-step explanation:

Represent the length of one side of the base be s and the height by h.  Then the volume of the box is V = s^2*h; this is to be maximized.

The constraints are as follows:  2s + h = 114 in.  Solving for h, we get 114 - 2s = h.

Substituting 114 - 2s for h in the volume formula, we obtain:

V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)

This is to be maximized.  To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:

dV

----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2

ds

Simplifying this, we get dV/ds = -4s^2 + 114s = 0.  Then either s = 28.5 or s = 0.

Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in

and the volume is V = s^2(h) = 46,298.25 in^3

7 0
3 years ago
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