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A line passes through (2, –1) and (8, 4).Write an equation for the line in point-slope form.

Dima020 [189]

Hello : let A(2,-1) B(8,4)
the slope is : (YB - YA)/(XB -XA)
(4+1)/(8-2) = 5/6

an equation for the line in point-slope form is : y-(-1) =( 5/6)(x-2)
y+1 = (5/6)x -5/3
6y+6 = 5x -10
the equation in standard form is : 5x-6y = 16

3 0

4 years ago

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One number is 7 less than a second number. Twice the second number is 7 less than 5 times the first. Find the smaller of two num

aliya0001 [1]

N = x-7
2x = 5n-7

2x = 5(x-7)-7
2x=5x-35-7
-3x=-42
x=14
n=7

7 is the answer

8 0

4 years ago

A town has 5000 people in year t = 0. Calculate how long it takes for the population P to double once, twice, and three times, a

UNO [17]

Answer:

Step-by-step explanation:

At the time t = 0, population of the town = 5000

Rate of population increase = 500 per year

Therefore, the equation that will represent the population will be

$P_{t}=P_{0}+500t$

Where $P_{t}$ = Population after t years

$P_{0}$ = Initial population

t = Time in years

a). For double once the population will be 500×2 = 10000

By plugging in the values in the equation,

10000 = 5000 + 500t

500t = 10000 - 5000

500t = 5000

t = $\frac{5000}{500}$

t = 10 years

For Double twice,

Population will be = 10000×2 = 20000

Now we plug in the values in the equation again

20000 = 5000 + 500t

500t = 20000 - 5000

500t = 15000

t = $\frac{15000}{500}$

t = 30 years

For double thrice,

Population of the town = 20000×2 = 40000

Now we plug in the values in the equation,

40000 = 5000 + 500t

500t = 40000 - 5000

500t = 35000

t = $\frac{35000}{500}$

t = 70 years

b). If the population growth is 5%.

Then the growth will be exponential represented by

$T_{n}=T_{0}(1+\frac{r}{100})^{t}$

$T_{n}$ = Population after t years

$T_{0}$ = Initial population

t = time in years

For double once,

Population after t years = 10000

$10000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{10000}{5000}$

$(1.05)^{t}=2$

Take log on both the sides

$log(1.05)^{t}=log2$

tlog(1.05) = log2

t = $\frac{log2}{log1.05}$

t = 14.20 years

For double twice,

Population after t years = 20000

$20000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{20000}{5000}$

$(1.05)^{t}=4$

Take log on both the sides

$log(1.05)^{t}=log4$

tlog(1.05) = log4

t = $\frac{log4}{log1.05}$

t = 28.413 years

For double thrice

Population after t years = 40000

$40000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{40000}{5000}$

$(1.05)^{t}=8$

Take log on both the sides

$log(1.05)^{t}=log8$

tlog(1.05) = log8

t = $\frac{log8}{log1.05}$

t = 42.620 years

4 0

3 years ago

Jane has $5 more than three times as much money as Lonnie has.Bob has $1 less than four times as much money as Lonnie has. All t

kkurt [141]

Starting equation:
92=J+L+B

To find how much money Jane has, we use the equation:
J=5+3L
To find how much money Bob has, we use the equation:
B=4L-1

Enter those equations in the problem

92=(5+3L)+L+(4L-1)
Combine like terms
92=4+8L
Subtract 4 to isolate variable
88=8L
Divide by 8 to isolate variable
11=L

Answer:
Lonnie has $11
Jane has $38
Bob has $43

side note: Lonnie is that one poor friend we all have

5 0

3 years ago

What is the solution to the following equation? 12 + 3x+7 = 0

Diano4ka-milaya [45]

The answer is x=-6.33333 (-19/3)

First, add like terms (12+7), which would equal to 19.
Then isolate the variable by subtracting 19 and adding it to the other side of the equal sign.
Now with this new equation of 3x=-19, isolate again by dividing 3 from x and do the same thing to the other side (-19/3).
The answer should be -6.33 repeating, but in algebra you would give a more simplified answer. This would be -19/3

3 0

2 years ago

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