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3 years ago

Isosceles triangle LMN is graphed with vertices L(0, 1), M(3, 5), and N(6,1). What is the slope of side LN?

2 answers:

9 0

L(0, 1)
N(6, 1)

slope formula;
m = y2 - y1/ x2 - x1

m = 1 - 1/ 6 - 1

m = 0/5

m = 0

hope this helped, God bless!

Aleksandr-060686 [28]3 years ago

8 0

Don't really know this math but I know it's m=0

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Given parallelogram MNOP, MA = x + 5, AO = y + 2, PA = 3x and AN = 2y

erma4kov [3.2K]

Answer:

The values of x and y are x = 6 and y = 9

Step-by-step explanation:

MNOP is a parallelogram its diagonal MO and PN intersected at point A

In any parallelogram diagonals:

Bisect each other
Meet each other at their mid-point

In parallelogram MNOP

∵ MO and NP are its diagonal

∵ MO intersect NP at point A

- Point A is the mid-point pf them

∴ MO and NP bisect each other

∴ MA = AO

∴ PA = AN

∵ MA = x + 5

∵ AO = y + 2

- Equate them

∴ x + 5 = y + 2 ⇒ (1)

∵ PA = 3x

∵ AN = 2y

- Equate them

∴ 2y = 3x

- Divide both sides by 2

∴ y = 1.5x ⇒ (2)

Now we have a system of equations to solve it

Substitute y in equation (1) by equation (2)

∴ x + 5 = 1.5x + 2

- Subtract 1.5x from both sides

∴ - 0.5x + 5 = 2

- Subtract 5 from both sides

∴ - 0.5x = -3

- Divide both sides by - 0.5

∴ x = 6

- Substitute the value of x in equation (2) to find y

∵ y = 1.5(6)

∴ y = 9

The values of x and y are x = 6 and y = 9

4 0

2 years ago

Which expression has a value greater than 43 ?<br><br> PLEASEEE HELP!

lozanna [386]

Answer:

7^2

Step-by-step explanation:

7*7=49

4^3=43

49>43

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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

14.2 is 35.5% of what number W

bonufazy [111]

14.2/x=100/35.5
5.041=x
Final
35.5% of 14.2=5.041

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3 years ago

5+(−4)+(−7)+2 jijujuijuijuijujijuijuijuijiji

Maurinko [17]

Combine the terms. Note that: one positive and one negative sign results in a negative sign.

Simplify: 5 + (-4) = 5 - 4 = 1

1 + (-7) = 1 - 7 = -6

-6 + 2 = -4

-4 is your answer

8 0

2 years ago

Read 2 more answers