All categories

3 years ago

Do all horizontal lines have the same slope

Mathematics

2 answers:

DENIUS [597]3 years ago

6 0

Yes, all horizontal lines have the same slope of 0, as it has no change in the y-positions.

zavuch27 [327]3 years ago

3 0

Answer: Yes

Step-by-step explanation: For any horizontal or flat line, the slope will always be the same which simplify to 0.

Let's look at an example which I have provided in the first image attached. I have labeled two points on the line and they are point A and point B.

To get from point A to point B along this line, our rise is 0 units and our run is 3 units so our slope or rise over run is 0/3. However, 0/3 can be simplified to 0. So for this line, our slope or m equals 0.

You might be interested in

Answer correctly and i give you brainliest

enot [183]

Answer:

Step-by-step explanation: The answer is b because 65.04 is the answer we get from finding the difference of those irregular triangles and the circle.

P.S. They must be done seperately.

5 0

2 years ago

1+4=5 2+5=12 3+6=21 8+11=?

g100num [7]

I think the pattern is that you multiply the first number by the second , then add the first number. SO:
e.g. for the first one 1 x 4 = 4 , 4 + 1 = 5
for the second one 2 x 5 = 10, 10 + 2 = 12...
So for 8+11:
you do 8 x 11 = 88 , 88 + 8 = 96

7 0

3 years ago

The first thing you should do in solving a word problem is estimate what you think the answer would be.

Lelu [443]

Answer:

The answer is True.

Step-by-step explanation:

Estimate the Answer Before Solving Having a general idea of a ballpark answer .
Or the problem lets students know if their actual answer is reasonable or not.

6 0

2 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

Someone help please and please make sure it’s right :)

Anuta_ua [19.1K]

Answer:

Step-by-step explanation:

5x + 5 = 4x + 16 ( being alternate angles)

5x - 4x = 16 - 5

x = 11

Hope it will help :)

6 0

2 years ago