All categories

3 years ago

find the length of each isosceles right triangle when the hypotenuse is of the given measure given 8cm

1 answer:

4 0

It is given that the triangle is isosceles right triangle . And in an isosceles right triangle the legs are equal . Let each leg be of x units. Now we use pythagorean identity, which is

$a^2 +b ^2=c^2$

$x^2 +x^2 = 8^2$

$2x^2 = 64$

$x^2 = 32$

$x = 4 \sqrt 2$

So the length of each leg is $4 \sqrt 2 .$

You might be interested in

Daniel hosted a party in which 15 to 45 people could attend and he charged $7 per person for admission. What is the range of adm

irina [24]

Answer:

c. [105,315]

Step-by-step explanation:

6 0

4 years ago

What is the coefficient in the expression? 3(mx)4 + 5

Goryan [66]

3 is the answer to your question

7 0

3 years ago

Read 2 more answers

What is 2+2 i nede hep

Ede4ka [16]

People don't actually know this, but it isn't 4, it's 22. If Base+ball is Baseball, then 2+2 is 22

I hope this helps!

7 0

3 years ago

There are between 24 and 40 students in a class

defon

So,whats your question????

8 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:

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

In conclusion, 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.

7 0

3 years ago