Choose the correct slope of the line that passes through the points (−4, 8) and (−3,−6).

Mathematics

2 answers:

Papessa [141]2 years ago

6 0

the answer is B. -14

horrorfan [7]2 years ago

5 0

-6-8=-14

-3+4=1

14/1=14

You might be interested in

What is the sixth term in the addition pattern that begins with 15,29,43,57?

WITCHER [35]

Answer:

C is your answer

Step-by-step explanation:

Add every number by 14

5 0

3 years ago

four apples and one banana cost £1.40 seven apples and one banana cost £2.00 work out the cost of an apple and the cost of a ban

beks73 [17]

Answer:

apple: £0.20
banana: £0.60

Step-by-step explanation:

Let "a" and "b" represent the costs of one apple and one banana, respectively. Then the purchases can be written ...

4a +b = 1.40

7a +b = 2.00

Subtracting the first equation from the second gives ...

(7a +b) -(4a +b) = (2.00) -(1.40)

3a = 0.60 . . . . simplify

a = 0.20 . . . . . .divide by 3

Using this in the first equation, we have ...

4(0.20) +b = 1.40

b = 0.60 . . . . . subtract 0.80

The cost of an apple is £0.20; the cost of a banana is £0.60.

8 0

3 years ago

A water tank is being drained at a steady rate. The initial level of the water in the tank was 5.5 feet. The graph below shows t

alisha [4.7K]

The answer would be C. to find this use the rise over run method

5 0

2 years ago

Read 2 more answers

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

zloy xaker [14]

Answer:

$t=-20ln\left(\dfrac{1}{3}\right)$

Step-by-step explanation:

The relationship between A, the area of the glacier in square kilometers, and t, the number of years the glacier has been melting, is modeled by the equation.:

$A=45e^{-0.05t}$

We want to determine the value of t for which the area, A(t)=15 square kilometers.

$15=45e^{-0.05t}\\$Divide both sides by 45\\\dfrac{15}{45} =\dfrac{45e^{-0.05t}}{45}\\\dfrac{1}{3}=e^{-0.05t}\\$Take the natural logarithm of both sides\\ln\left(\dfrac{1}{3}\right)=ln\left(e^{-0.05t}\right)\\ln\left(\dfrac{1}{3}\right)=-0.05t\\$Divide both sides by -0.05$\\t=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05} \\=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05}\\t=-20ln\left(\dfrac{1}{3}\right)$