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Dahasolnce [82]
3 years ago
13

What is the equation of a line that is perpendicular to - x + 2y = 4 and passes through the point (-2,1)

Mathematics
1 answer:
Damm [24]3 years ago
3 0
Perpendicular has slope that multiplies to -1

the slope for ax+by=c is -a/b
-1x+2y=4
slope=-(-1)/2=1/2
1/2 times what=-1?
-2

y=mx+b
m=slope
we need slope is -2
y=-2x+b
find b
given
(-2,1)
(x,y)
x=-2 and y=1 is a solution

1=-2(-2)+b
1=4+b
mius 4 both sides
-3=b


y=-2x-3 is equation

first one
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kozerog [31]

Answer:

Domain = {-2, 2, 3, 4, 5}

Range= {6, 0, 6, -1, 3)

It is a function

Step-by-step explanation:

It is a function because there is only one y value for each x value

4 0
3 years ago
Diego wants to have $1,000,000 in 5 years. He plans to invest $250,000 to start and make yearly payments of $125,000 to the acco
julia-pushkina [17]

Answer:

Yes

Step-by-step explanation:

With the $875,000 from his input alone he is really close to $1,000,000 and doing simple intrest on the $250,000 is $105,000 a year is he earns 3.5% intrest a month.

So in 5 years he'll have about $1,662,500 using simple intrest but with compound it'd be more like 2mil

Hope this helps!

5 0
2 years ago
If f:X is 3x + b and ff(2) = 12, find the value of b​
Tamiku [17]

Answer:

b =6

Step-by-step explanation:

Given

f(x) =3x + b

f(2) = 12

Required

Find b

f(2) = 12 implies that:

12 = 3 * 2 + b

12 = 6 + b

Collect like terms

b = 12 - 6

b =6

5 0
3 years ago
1/7+2/3= plz help ive been stuck on the question
Romashka [77]

Answer:

1/7+2/3

3+14/21

17/21

3 0
3 years ago
Read 2 more answers
Suppose there are two lakes located on a stream. Clean water flows into the first lake, then the water from the first lake flows
never [62]

Answer:

a.  For the first lake;

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m·t/200 - m·t²/80000

b. t ≈ 1.00505 hours

c. 200 hours

Step-by-step explanation:

The flow rate of water in and out of the lakes = 500 liters/hour

The volume of water in the first lake = 100 thousand liters

The volume of water in the second lake = 200 thousand liters

The mass of toxic substances that entered into the first lake =  500 kg

The concentration of toxic substance in the first lake = m₁/(100000)

Therefore, we have;

The quantity of fresh water supplied at t hours = 500 × t

The change

The change in the mass of the toxic substance with time is given as follows

dm/dt = (m - m/100000 × 500 × t)/100000

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m/100000 × 500 × t - (m/100000 × 500 × t)/200000 × 500 × t

Using an online tool, we have;

c = m·t/200 - m·t²/80000

b. When c < 0.001 kg per liter, we have m < 0.001 × 100000, which gives m < 100

Substituting gives;

0.001 = (100·t - 0.005·100·t²)/100000, solving with an online tool, gives;

t ≈ 1.00505 hours

c. For maximum concentration, we have;

c = m·t/200 - m·t²/80000

m/200000 = m·t/200 - m·t²/80000

1/200000 = t/200 - t²/80000

dc/dt = d(t/200 - t²/80000)/dt = 0

Solving with an online tool gives t = 200 hours

6 0
2 years ago
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