All categories

3 years ago

Please somebody this is due by 11:59 pm

Mathematics

2 answers:

AVprozaik [17]3 years ago

4 0

Answer:

x = -15, angle b = 115, angle c = 30.

Step-by-step explanation:

The 3 sides of a triangle add up to 180, so the equation will be

35 - 5 - 8x - 2x = 180

30 - 10x = 180

Subtract 30 from both sides.

-10x = 150

Divide by -10.

x = -15.

Finding angle b:

-5 - 8x

-5 - 8(-15)

-5 + 120

115

Finding angle c:

-2x

-2(-15)

To check it, we add angle a, angle b, and angle c together.

35 + 115 + 30

150 + 30

180.

FinnZ [79.3K]3 years ago

3 0

Answer:

The horizontal value in a pair of coordinates: how far along the point is. The X Coordinate is always written first in an ordered pair of coordinates (x,y), such as (12,5). In this example, the value "12" is the X Coordinate.

Angel b=angle a+angle c-180

We have now proven that all three angles of ΔABC have a measure of 60 degrees. Therfore, ∠C must have a measure of 60 degrees

You might be interested in

Solve for X and Y using substitution.<br> 1)<br> 12x +y = 36<br> x+4y=24

vovikov84 [41]

{x,y}={24,12} I think that is the answer to this question....

8 0

3 years ago

The domain of a relation is

loris [4]

The x value is the domain

4 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

In a topographical map of a city, the vertices of the city limits are A(10,9),B(18,9),C(18,2),D(14,4.5),E(10,4.5). The coordinat

Viefleur [7K]

I think it’s D??????

3 0

3 years ago

Marie made the box plot below to display a set of data. Which set of data could the box plot represent?

Assoli18 [71]

Answer:

see below

Step-by-step explanation:

First of all, you want to find the data set the matches the extreme values of 5 and 35. That eliminates the 2nd and 4th choices.

Then you want to find the data set that has a median of 15. The first data set has a middle value (median) of 20, so that choice is eliminated.

The data set of the 3rd choice matches the box plot extremes, median, and quartile values.

7 0

3 years ago

Please somebody this is due by 11:59 pm​

Please somebody this is due by 11:59 pm