Someone pls help me look at the picture

The graph of f is given in the figure to the right. Let A(x)equals=Integral from 0 to x f left parenthesis t right parenthesis

Tpy6a [65]

Answer:

$A(4)=-4\pi$

$A(8)=-4\pi +8$

$A(12)=-4\pi +16$

$A(14)=-4\pi +15$

Step-by-step explanation:

we are given

$A(x)=\int\limits^x_0 f{x} \, dx$

Calculation of A(4):

we can plug x=4

$A(4)=\int\limits^4_0 f{x} \, dx$

Since, this curve is below x-axis

so, the value of integral must be negative

and it is quarter of circle

so, we can find area of quarter circle

radius =4

$A(4)=-\frac{1}{4}\times \pi \times (4)^2$

$A(4)=-4\pi$

Calculation of A(8):

we can plug x=8

$A(8)=\int\limits^8_0 f{x} \, dx$

we can break into two parts

$A(8)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx$

now, we can find area and then combine them

$A(8)=-4\pi +\frac{1}{2}\times 4\times 4$

$A(8)=-4\pi +8$

Calculation of A(12):

we can plug x=12

$A(12)=\int\limits^12_0 f{x} \, dx$

we can break into two parts

$A(12)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx$

now, we can find area and then combine them

$A(12)=-4\pi +\frac{1}{2}\times 8\times 4$

$A(12)=-4\pi +16$

Calculation of A(14):

we can plug x=14

$A(14)=\int\limits^14_0 f{x} \, dx$

we can break into two parts

$A(14)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx+\int\limits^14_12 f{x} \, dx$

now, we can find area and then combine them

$A(14)=-4\pi +\frac{1}{2}\times 8\times 4-\frac{1}{2}\times 1\times 2$

$A(14)=-4\pi +16-1$

$A(14)=-4\pi +15$

8 0

2 years ago

Suppose that you ask three friends to go to the mall. each one has a 0.80 chance of saying yes. what is the probability that all

blondinia [14]

The probability is 0.512

5 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

I picked a and c correct me if I'm wrong I have until the 19th to bring my stuff up

k0ka [10]

Answer:

I believe they are correct. Sorry this is late

Step-by-step explanation:

4 0

2 years ago

A coordinate grid is placed over a map. City A is located at (- 3, 2) , and City B is located at (4, 8) . If City C is at the mi

Andreyy89

Answer:

City A is the closest

Step-by-step explanation:

Given

City A = (-3,2)

City B = (4,8)

First, we calculate the coordinates of City C

If City C is located at the midpoint between A and B, then it's coordinates is

(x1 + x2)/2 , (y1 + y2)/2

where

x1 = x coordinate of City A = -3

x2 = x coordinate of City B = 4

y1 = y coordinate of City A = -2

y2 = y coordinate of City B = 8

So, the coordinates of City C = (-3+4)/2 , (-2+8)/2

City C = (½,6/2)

City C = (½,3)

We then calculate the distance between City C and City A

And

We also calculate the distance between City C and City B

Then we compare both results

Formulation for distance = √(x2-x1)² + (y2-y1)²

For City C and City A

City A coordinatates = (x1,y1) = (-3,2)

City C coordinates = (x2,y2) = (½,3)

Distance between A and C = √(½-(-3))² + (3-2)² -------- Simplify the bracket

Distance = √(½+3)² + (1)² ---------- Solve the fraction

Distance = √(7/2)² + (1)² ------------ Open all brackets

Distance = √49/4 + 1

Distance = √53/4

Distance = 7.280109889280518/2

Distance = 3.640054944640259

Distance between City A and C = 3.64 (approximated)

For City C and City B

City B coordinatates = (x1,y1) = (4,8)

City C coordinates = (x2,y2) = (½,3)

Distance between B and C = √(½-(-3))² + (8-4)² -------- Simplify the bracket

Distance = √(½+3)² + (4)² ---------- Solve the fraction

Distance = √(7/2)² + (4)² ------------ Open all brackets

Distance = √49/4 + 16

Distance = √113/4

Distance = 10.63014581273464/2

Distance = 5.315072906367324

Distance between City B and C = 5.32 (approximated)

Comparing the results (3.64) and (5.32),

We conclude that City C is closer to City A than City B

7 0

2 years ago