A line passes through the points ( - 9, - 24) and (8, – 9).Calculate the slope of the line. Write your answer as a fraction.
(If the slope is undefined, enter "DNE")

Answers

Answer 1
Answer:

Answer:

15 / 17

Step-by-step explanation:

slope = (y2 - y1) / (x2 - x1)

  = (-9 + 24) / (8 + 9)

  = 15 / 17


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Solve 2cos^2x+3sinx=0

Answers

Answer:

x =  {\sin^( - 1)  2}  \: \: or \:   \:  x =    (7\pi)/(6),   \:  \:  (5\pi)/(3)

Step-by-step explanation:

2 { \cos}^(2) x + 3 \sin x = 0 \n 2 {(1 -  \sin}^(2) x) + 3 \sin x = 0 \n 2 - 2 \sin^(2) x+ 3 \sin x = 0 \n  2 \sin^(2) x -  3 \sin x - 2 = 0  \n 2 \sin^(2) x -  4\sin x +  \sin x- 2 = 0  \n 2\sin x(\sin x - 2) + 1(\sin x - 2) = 0 \n (\sin x - 2)(2\sin x + 1) = 0 \n (\sin x - 2) = 0 \: or \: (2\sin x + 1) = 0  \n \sin x = 2 \: or \: 2\sin x =  - 1 \n x =  {\sin^( - 1)  2}  \: \: or \:   \: \sin x =   - (1)/(2) \n x =  {\sin^( - 1)  2}  \: \: or \:   \:  x =    (7\pi)/(6),  \:  \:  (5\pi)/(3)

Three 3.0 g balls are tied to 80-cm-long threads and hung from a single fixed point. Each of the balls is given the same charge q. At equilibrium, the three balls form an equilateral triangle in a horizontal plane with 20 cm sides. What is q?

Answers

Answer:

q = 0.105uC

Step-by-step explanation:

We can determine the force on one ball by assuming two balls are stationary, finding the E field at the lower right vertex and calculate q from that.

Considering the horizontal and vertical components.

First find the directions of the fields at the lower right vertex. From the lower left vertex the field will be at 0° and from the top vertex, the field will be at -60° or 300° because + charge fields point radially outward in all directions. The distances from both charges are the same since this is an equilateral triangle. The fields have the same magnitude:

E=kq/r²

Where r = 20cm

= 20/100

= 0.2m

K = 9.0×10^9

9.0×10^9 × q /0.2²

9.0×10^9/0.04

2.25×10^11 q

These are vector fields of course

Sum the horizontal components

Ecos0 + Ecos300 = E+0.5E

= 1.5E

Sum the vertical components

Esin0 + Esin300 = -E√3/2

Resultant = √3E at -30° or 330°

So the force on q at the lower right corner is q√3×E

The balls have two forces, horizontal = √3×E×q

and vertical = mg, therefore if θ is the angle the string makes with the vertical tanθ = q√3E/mg

mg×tanθ = q√3E.

..1

Then θ will be...

Since the hypotenuse = 80cm

80cm/100

= 0.8m

The distance from the centroid to the lower right vertex is 0.1/cos30 =

0.1/0.866

= 0.1155m

Hence,

0.8×sinθ = 0.1155

Sinθ = 0.1155/0.8

Sin θ = 0.144375

θ = arch sin 0.144375

θ = 8.3°

From equation 1

mg×tanθ = q√3E

g = 9.8m/s^2

m = 3.0g = 0.003kg

0.003×9.8×tan(8.3)

0.00428 = q√3E

0.00428 = q×1.7320×E

Where E=kq/r²

Where r = 0.2m

0.0428 = kq^2/r² × 1.7320

K = 9.0×10^9

0.0428/1.7320 = 9.0×10^9 × q² / 0.2²

0.02471×0.04 = 9.0×10^9 × q²

0.0009884 = 9.0×10^9 × q²

0.0009884/9.0×10^9 = q²

q² = 109822.223

q = √109822.223

q = 0.105uC

PLEASE help its for an exam i would really appreciate it , at the top it says “Drag the tiles to solve the given equation for x and justify step in your solution. you must fill every slot in the table in order for your answer to be scored correctly.”

Answers

Answer:

2A. 2x-7 =3

B. Subtraction property

3A. 2x = 10

B. Addition property

4A. x=5

B. Division property

A survey by the American Veterinary Medical Association found that out of 120 families surveyed, 18 owned a bird. Fifteen years ago, a similar study found that only 5 families out of 100 surveyed had a bird. Are the validity conditions for a two-sample z-test/interval adequately met

Answers

Answer:

Step-by-step explanation:

The assumptions include:

The samples from each population must be independent of one another: this was adequately met.

The populations from which the samples are taken must be normally distributed: this was not talked about

The population standard deviations must be known: not known

or the sample sizes must be large (i.e. n1≥30 and n2≥30: true

The validity conditions for a two-sample z-test/interval were not all adequately met

at the same time a 15 foot pole casts a 7.5 foot shadow a nearby tree casts an 11 foot shadow how tall is the tree​

Answers

answer: 22

15/7.5 = x/11

cross multiply

7.5x = 165

divide by 7.5 on both sides to get the x alone

7.5x/7.5 = 165/7.5

x = 22

the tree is 22 feet

sorry if it’s wrong

Solve for the area of ΔABC to the nearest whole number. A) 13 cm2 B) 26 cm2 C) 53 cm2 D) 106 cm2

Answers

Answer:

The answer is 53 cm2 or c.