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3 years ago

How do I multiply fractions by whole numbers

2 answers:

8 0

First write the whole number as fraction and simply place the whole number in order one

And then multiply the numbers of two fractions and multiply the denominators of the two numbers and then you have to simplify....

Volgvan3 years ago

5 0

Multiplying fractions by whole numbers:
1. You have to convert the whole number into an improper fraction.
2. You then multiply the fractions straight across.
Ex: 2/3 × 6 =2/3 × 6/1 = 2×6/3×1= 12/3= 4

You might be interested in

An elf ate 15 of your muffins. that was 5/8 of all of them. how many are left?

zavuch27 [327]

Answer: 9

Step-by-step explanation:

An elf ate 15 of your muffins and that was 5/8 of all of them. To get the number of muffins left goes thus:

We can first calculate the total number of muffins the person had. Let the total number of muffins be y. That means the elf ate 5/8 of y.

5/8 of y = 15

5/8 × y = 15

0.625 × y = 15

0.625y = 15

Divide both side by 0.625

0.625y/0.625 = 15/0.625

y = 24

The total amount of muffins is 24. Since the elf has eaten 15, the amount left will be: 24-15 = 9

6 0

3 years ago

Richard can make 8 dough balls in two hours. If the amount of time is directly proportional to the number of dough balls, how ma

Sergio039 [100]

Answer:

4.5 hours

Step-by-step explanation:

8:2h = 18:x

8x = 18×2h

x = 36h/8

x = 4.5

hope this helps :)

8 0

3 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .