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3 years ago

A farmer packs 12 apples in each box.If the farmer has 4272 apples,how many boxes does she needs?

Mathematics

2 answers:

sertanlavr [38]3 years ago

7 0

Answer:

356

Step-by-step explanation:

Since this is a division problem we need to set it up. Since 4272 is the total number it will be on the top of the fraction.

4272/12

Then you solve just like any other division problem to get 356

Alex_Xolod [135]3 years ago

4 0

Answer:

356 boxes

Step-by-step explanation:

Just divide 4272 by 12

you do this because that is how many apples are needed in each box

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Graph the linear equation 3x - y = 12. Write the equation in slope-intercept form,

S_A_V [24]

Answer: y=3x-12

Step-by-step explanation:

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3 years ago

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Scrat [10]

Answer:

the answer is a-11

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3 years ago

In a class of 23 student the girls were 7 more than the boys how many boys were in the class

Softa [21]

Answer:

x=# of boys

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x+x+7=25

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If you don't understand variables use trial and error.

Boys Girls

1 8

2 9

3 10

4 11

5 12

6 13

7 14

8 15

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You could start this way...

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12 13

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10 15

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5 0

3 years ago

Percy plotted the points (5,6)

Annette [7]

Answer:

1,6 and 6,6

Step-by-step explanation:

since they are on the same y-axis, changing the x-value shouldn't matter.

3 0

3 years ago

Read 2 more answers

Derivative, by first principle <img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

4 years ago