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3 years ago

Help plsssssssssssssssss

Mathematics

1 answer:

Sphinxa [80]3 years ago

3 0

Answer:

2020 edge

Step-by-step explanation:

use what u learned

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Given the function f(x)=x^2 and the transformed function g(x)=4x^2+1, explain how the parent function was transformed.

satela [25.4K]

Answer:

In the function g(x) = 4x² + 1, 4 means that all the points in the parent function were multiplied by 4. + 1 means that the parent graph will shift up by 1.

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3 years ago

Type the solution.y=–3x–7y=x–3

Andrej [43]

Answer:

(x,y)=(-1,-4)

Step-by-step explanation:

y=–3x–7

y=x–3

(y=)-3x-7=x-3

-3x-x=-3+7

-4x=4

x=4/(-4)

x=-1

y=x-3

y=-1-3

y=-4

(x,y)=(-1,-4)

6 0

3 years ago

N-4-9 I really need help with this

yKpoI14uk [10]

Answer:

n - 13

Step-by-step explanation:

n-4-9

We can combine like terms.

n - (4+9)

n - 13

8 0

4 years ago

Identify f (x)=3x2+8x+4

german

Im gona assume you mean

f(x) = 3x + 2 + 8x + 4. Add like-terms
Y = 11x + 6
0= 11x + 6
-6 = 11x
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7 0

3 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

3 years ago