All categories

3 years ago

How do how you cross cancel multiplying fractions

Mathematics

1 answer:

DENIUS [597]3 years ago

5 0

It's easy once you spot the ones that can cross cancel!

Say we have the fractions 8/10 and 20/23. $\frac{8}{10} \frac{20}{23}$ (it's easier to see on top of each other)

If you look diagonally , so 8 and 23 and 10 and 20, you can see that 10 and 20 have a common factor. So we divide it by the highest common factor to reduce those numbers, making it easier to multiply. 10 and 20 can become 1 and 2, dividing by 10. So now we are left with 8/1 and 2/23, and now we multiply normally going across so 16/23.

This works going both diagonals and simplifying both, but in that case it would be easier to try and simplify the fractions before cross multiplying them.

Basically: look for those diagonals and if they can be divided down by the highest common factor, go for it to make it easier to multiply normally afterwards.

Hope I helped!

You might be interested in

Solve two sixths times three fifths.

Artemon [7]

Answer: 2/6 times 3/5 = 1/5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

If the distance from Haley’s house to school is 2.5 cm on the map and the actual distance is 3 miles. What is the scale for the

Brrunno [24]

To find your answer you divide 3 by 2.5 to get 1.2. The scale for the map is 1.2.

Hope this helps,
♥Nikki♥
(P.S. I'm not completely sure about this so you might want to get a second opinion on this particular question.)

7 0

3 years ago

g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur

sdas [7]

Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

$\displaystyle V = \pi r^2h$

Substitute:

$\displaystyle (300) = \pi r^2 h$

Solve for h:

$\displaystyle \frac{300}{\pi r^2} = h$

Recall that the surface area of a cylinder is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for h.

$\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}$

Find its derivative:

$\displaystyle A' = 4\pi r - \frac{600}{r^2}$

Solve for its zero(s):

$\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}$

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

$\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}$