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erma4kov [3.2K]
3 years ago
9

At an apple orchard, Margaret picked 19 1 2 pounds of apples. The cashier put the apples into 3 bags with the same weight. How m

any pounds of apples are in each bag?
Mathematics
1 answer:
True [87]3 years ago
5 0

6.5 pounds of apples are there in each bag

Step-by-step explanation:

  • Step 1: Find the pounds of apple in each bag

Total weight of apples = 19 1/2 = 39/2

Number of bags = 3

Weight of each bag = 39/2 ÷ 3

                                 = 39/2 × 1/3 (∵ a÷b = a × 1/b)

                                 = 13/2 = 6.5 pounds

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2.5 kg cost £1.40<br><br>work out the cost of 4.25kg
Mkey [24]

Calculating the cost of 4.25 kilogram(kg) is £2.38, by multiplying it by 4.25 kg by 1 kg.

<h3>Cost</h3>

Costs are the expenses incurred in the manufacture, marketing, or preparation of a good or asset for regular use. In other terms, it's the cost incurred to produce a good, buy inventories, sell goods, or prepare equipment for use in a commercial activity.

Given,

2.5 kg cost  £1.40 i.e,

2.5 kg = £1.40

For 1 kg= \frac{1.40}{2.5}

= £0.56

To find the cost of 4.25 kg:

Then,

1 kg = £0.56

4.25 kg = 0.56 x 4.25

= £2.38

£2.38 is the cost of 4.25kg after calculating it.

To know more about calculating cost visit here:

brainly.com/question/11871927

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8 0
1 year ago
A rectangle has a height of 7 and a width of a^4+5a^2+4a .
boyakko [2]

Answer:

  7a^4 +35a^2 +28a

Step-by-step explanation:

Area is the product of height and width:

  area = height × width

  = 7 × (a^4 +5a^2 +4a)

  area = 7a^4 +35a^2 +28a . . . . . . use the distributive property

5 0
3 years ago
Read 2 more answers
Does the quadratic function
NeX [460]

Answer:

no real zeros

Step-by-step explanation:

We can use b^2-4ac which is 25-16*29.

We know that is negative meaning there is no real zeros.

By the way, can you follow me on Brainly?

Thanks!

6 0
2 years ago
In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track
Aloiza [94]

Answer:

Approximately 0.31\; \rm m, assuming that g = 9.81\; \rm N \cdot kg^{-1}.

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned} .

Weight of the slide:

\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}.

Normal force between the sled and the slope:

\begin{aligned}F_{\rm N} &= W\cdot  \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}.

Calculate the kinetic friction between the sled and the slope:

\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}.

Assume that the sled and its passenger has reached a height of h meters relative to the base of the slope.

Gain in gravitational potential energy:

\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}.

Distance travelled along the slope:

\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}.

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}.

In other words, the sled and its passenger would have lost (approximately) ((137 + 68.1)\, h)\; {\rm J} of energy when it is at a height of h\; {\rm m}.

The initial amount of energy that the sled and its passenger possessed was \text{KE} = 63\; {\rm J}. All that much energy would have been converted when the sled is at its maximum height. Therefore, when h\; {\rm m} is the maximum height of the sled, the following equation would hold.

((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}.

Solve for h:

(137 + 68.1)\, h = 63.

\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}.

Therefore, the maximum height that this sled would reach would be approximately 0.31\; \rm m.

7 0
2 years ago
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3 years ago
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