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3 years ago

I need help?!! Answer quicklyy

Mathematics

1 answer:

Kisachek [45]3 years ago

8 0

Answer:

-7

Step-by-step explanation:

To find f(-8), use the blue line labeled f(x) and go to -8 on the x axis and find its corresponding y value by moving vertically down from -8 to the blue line and find the y value of that point. You can see that the blue line hits (-8, -5) so the y value is -5 and f(-8) = -5.

Do the same thing for g(4) and you should find the red graph g(x) hits the point (4, 3). So g(4) = 3.

-1 * -5 - 4 * 3 = 5 - 12 = -7

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Help with this problem please !!!!!

Karolina [17]

Answer:

Step-by-step explanation:

Cross multiply

18k = 12 x 42

18 k = 504

k = 504/18

k = 28

3 0

3 years ago

Read 2 more answers

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

Answer choices: (5,6) (5,2) (3,6) (3,2).

Degger [83]

Answer:

(3,6)

Step-by-step explanation:

If the starting point is the dot then we start from there - (4,4) then we go up 2 units at (4,6) then moving one unit to the left our answer is (3,6).

6 0

3 years ago

Evaluate the expression \dfrac{7^2}{x^2-2} x 2 −2 7 2 start fraction, 7, squared, divided by, x, squared, minus, 2, end fracti

sertanlavr [38]

Answer:

Step-by-step explanation:

We want to evaluate the fraction below for x = 3. We will put the value of x to be 3:

$\dfrac{7^2}{x^2-2}\\\\= \dfrac{7^2}{3^2-2}\\\\= \dfrac{49}{9-2}\\\\= \dfrac{49}{7} = 7$