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3 years ago

I will mark you brainiest if you can answer this

Mathematics

1 answer:

Gennadij [26K]3 years ago

5 0

Step-by-step explanation:

A = - 5-1/2-0 =-6/2 = - 3

Y=ax+b

1=-3 x 0 +b

1=b

Y=-3x+1

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3 years ago

What are 6 whole numbers -3.5 and 5.5

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Answer:

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3 years ago

Loren drove 200 miles at a certain rate, and his wife, Lois, drove 100 miles at a rate 10 mph slower. If Loren had driven for th

NeX [460]

As long as Loren drove, the law of motion was

$200 = st_1 \implies t_1 = \dfrac{200}{s}$

As long as Loid drove, the law of motion was

$100 = (s-10)t_2 \implies t_2 = \dfrac{100}{s-10}$

So, the total time they took is

$t_1+t_2=\dfrac{200}{s}+\dfrac{100}{s-10}$

Had Loren driven the whole time, the law of motion would have been

$300=st_3 \implies t_3 = \dfrac{300}{s}$

And we know that this time would have been 30 minutes (i.e. 0.5 hours) faster. So, we have

$t_3 = t_1+t_2-0.5$

This translates into

$\dfrac{300}{s}=\dfrac{200}{s}+\dfrac{100}{s-10}-\dfrac{1}{2}$

If we subtract 200/s from both sides, we have

$\dfrac{100}{s}=\dfrac{100}{s-10}-\dfrac{1}{2}$

We can simplify the right hand side by summing the two fractions:

$\dfrac{100}{s-10}-\dfrac{1}{2} = \dfrac{200-(s-10)}{2(s-10)}=\dfrac{210-s}{2(s-10)}$

So, we have to solve

$\dfrac{100}{s}=\dfrac{210-s}{2(s-10)}$

If we cross multiply the denominators, we have

$200(s-10)=s(210-s) \iff 200s-2000=210s-s^2 \iff s^2-10s-2000=0$

Which yields the solutions

$s=-40,\quad s=50$

We accept the positive solution, because the negative would mean to travel backwards, so Loren's rate was 50mph

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3 years ago

Read 2 more answers

Pls help me! Pls pls pls pls

k0ka [10]

Answer:

5. A.

6. F

Step-by-step explanation:

5. Perpendicular lines have slopes that are negative reciprocals of one another, so slope is 3

y = 3x + b

using coordinates (-6,3)

3 = 3(-6) + b

b = 21

y = 3x + 21

6. I think F is the right answer

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2 years ago

How do you get the numbers 4 5 13 to equal 5

MAVERICK [17]

You can subtract 4 from 13 then subtract 4 again

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4 years ago