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Bogdan [553]
2 years ago
13

Write the equation of the line that passes through the points (-6,-3) and (-3,1)

Mathematics
1 answer:
Ivahew [28]2 years ago
5 0

Answer:

y =  \frac{4}{3} x  +  5

Step-by-step explanation:

The slope is

\frac{ - 3 - 1}{ - 6 - ( - 3)}  =  \frac{4}{3}

Substituting into point-slope form,

y - 1 =  \frac{4}{3} (x +  3) \\  \\ y - 1 =  \frac{4}{3} x  + 4 \\  \\ y =  \frac{4}{3} x  +  5

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Select the correct answer.
insens350 [35]

Answer:

3,158

Step-by-step explanation:

The formula is :

h(x)=-49(x)-125

h(-67)=(-49)*(-67)-125

h(-67)=3283-125

h(-67)=3158

6 0
3 years ago
2а + 20 = -3а - 5 pls help
sammy [17]

Answer: A= -5

Step-by-step explanation:

4 0
2 years ago
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Determine what type of model bet fits the given situation: Every day, half a bacteria population dies. A. Linear B.quadratic C.
makvit [3.9K]

Answer:

Exponential

Step-by-step explanation:

<em>Starter:</em>

According to the title:

"Every day, half a bacteria population dies."

Exponential: y = ab^x ( b > 0 , b ≠ 1)

So the answer is C.

<em>Calculations:</em>

An exponential model is of the form y = a • b^t.

If you start with population of 500 million bacteria

t = 0

so 500 million = a • b^0 = a

Since every day half the population dies then in 1 day population will be 250 million

so a = 500 million, y = 250 million and t = 1

250 million = 500 million • b^1

b = 0.5

Therefore your equation would be

y = 500 million (or whatever the pop) • 0.5^t

Where t = number of days

This could also be written in exponential form ( e = 2.73)

4 0
2 years ago
Alex needs to save at least $5.80 per day for a week, in order to have enough money to buy the new video game that he wants. Det
Alexxx [7]

Answer:

the answer would be v ≥ $5.80

Step-by-step explanation:

he doesn't need exactly 5.80 a day so that marks out A and C and since he needs v to be at least 5.80 it means B would be the answer

3 0
3 years ago
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ella [17]

Answer:

b. 4ft

Step-by-step explanation:

thank you for the points, have an amazing day! :)

4 0
3 years ago
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