All categories

3 years ago

What is the slope of a line that goes through the point (-7,-2) and (-6,7)

Mathematics

1 answer:

navik [9.2K]3 years ago

5 0

y=mx+b

y2-y1/x2-x1

the answe would be 9

You might be interested in

The quotient of a number and two imcreased by five is 10 equation

morpeh [17]

x/2 + 5 = 10

Subtract 5 from both sides.

x/2 = 5

Multiply both sides by 2.

x = 10 (This is your answer.)

3 0

3 years ago

What is the value of x in the following equation (2y+x)+(2y+x)=(-2y+x)+(2x-15)

adoni [48]

(2y + x) + (2y + x) = (-2y + x) + (2x - 15)

2y + x + 2y + x = -2y + x + 2x - 15

4y + 2x = -2y + 3x - 15
+2y + 2y

6y + 2x = 3x - 15
-2x -2x

6y = x - 15
+15 +15

15 + 6y = x

5 0

3 years ago

What is the least positive integer? What is the greatest negative integer? worth: 16 points

Ganezh [65]

The least positive integer is 1. The greatest negative integer is -1.

3 0

3 years ago

Read 2 more answers

The shape has which of the following coordinates

Rasek [7]

Step-by-step explanation:

1a. ( 3, -4)

1b. ( 8, -2)

1c. ( 10, -10)

hope it helps!

4 0

3 years ago

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and t

WINSTONCH [101]

Answer:

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.

How long will it take for this population to grow to a hundred rodents? To a thousand rodents?

Step-by-step explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

$\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}$

Seperate the differential equation and solve for the constant C.

$\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2} }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}$

You have 100 rodents when:

$100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months$

You have 1000 rodents when:

$1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months$

7 0

3 years ago