All categories

3 years ago

Can someone please help with this I am confused about how to know which side is a and c

Mathematics

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

b=4

Step-by-step explanation:

The side opposite the right angle is the hypotenuse. The sides touching the right angle are the legs

Since this is right triangle we can use the Pythagorean theorem

a^2+b^2 = c^2 where a and b are the legs and c is the hypotenuse

3^2+b^2 = 5^2

9+b^2 = 25

Subtract 9 from each side

9-9+b^2 = 25-9

b^2 =16

Take the square root of each side

sqrt(b^2) = sqrt(16)

b=4

You might be interested in

I need help on this question

Cloud [144]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Find the percent of the number. 20% of 60

Svetradugi [14.3K]

Answer:

12 is 20% of 60

7 0

3 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

I need Serious help here!

marishachu [46]

Answer:

Step-by-step explanation:

0.5

7 0

2 years ago

Read 2 more answers

What does why y equal

Novay_Z [31]

Answer:

y=2/3x+4/3

Step-by-step explanation:

4 0

3 years ago