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3 years ago

Why did I get this question wrong?

Mathematics

1 answer:

lord [1]3 years ago

4 0

Step-by-step explanation:

∫₀³⁰ (r/V C₀ e^(-rt/V)) dt

If u = -rt/V, then du = -r/V dt.

∫ -C₀ e^u du

-C₀ ∫ e^u du

-C₀ e^u + C

-C₀ e^(-rt/V) + C

Evaluate between t=0 and t=30.

-C₀ e^(-30r/V) − -C₀ e^(-0r/V)

-C₀ e^(-30r/V) + C₀

C₀ (-e^(-30r/V) + 1)

I got the same answer. Try changing the lowercase v to an uppercase V.

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What is the area of a sector with a central angle of 4π5 radians and a radius of 11 cm?

WARRIOR [948]

Answer: 151.98 I just took the test!

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the area of the region under the graph of f on [a, b].<br> f(x) = x2 − 8x + 17; [−1, 2]

Harman [31]

Answer:

The area is 42 area units.

Step-by-step explanation:

The area of a function f(x) on an interval [a,b] is given by:

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

Applying to this question:

$A = \int\limits^-1_2 {x^2 - 8x + 17} \, dx$

Then

$F(x) = \frac{x^{3}}{3} - 4x^{2} + 17x$

$A = F(2) - F(-1)$

$F(2) = frac{2^{3}}{3} - 4*2^{2} + 17*2 = \frac{8}{3} + 18 = \frac{8 + 3*18}{3} = \frac{62}[3}$

$F(-1) = frac{(-1)^{3}}{3} - 4*(-1)^{2} + 17*(-1) = -\frac{1}{3} - 21 = \frac{-1 - 21*3}{3} = -\frac{64}[3}$

$A = F(2) - F(-1) = \frac{62}[3} - (-\frac{64}[3}) = \frac{62+64}{3} = 42$

The area is 42 area units.

5 0

3 years ago

Lexi said that two lines

omeli [17]

Answer:

Step-by-step explanation:

The easy way to make 2 perpendicular lines is let one be x = 3 and let y = 2.

The third line can be anything that has a minus sign in front of the x. See the graph below. The third line cannot be perpendicular to the other 2.

Sorry, the bottom of the graph got cut off. If you print out the graph and extend the vertical line and the slanted line, you will see how this works.

The three lines are

x = 3
y = 2
y = -x - 2

7 0

2 years ago

Can someone help me out with this

babymother [125]

The answer would be x= 68 degrees

7 0

2 years ago

The variable y is directly proportional to the variable x. If y = 32 when x = 20, what is the value of x when y = 40?

elena55 [62]

Answer:

The Correct option is C ) 25

Therefore the value of x is 25 when y =40.

Step-by-step explanation:

Given:

Variable 'y' is directly proportional to the variable 'x'.

$\therefore y=kx$ ......Direct Variation

Where,

k = Constant of proportionality

To Find:

value of x = ? when y = 40

Solution:

First we need to find Constant of proportionality

When x = 20 and y = 32

Substituting the values we get

$32=k\times 20\\k=\dfrac{32}{20}=1.6\\\\k=1.6$

Now when k =1.6 , y = 40 then x will be

$40=1.6\times x\\\\x=\dfrac{40}{1.6}=25\\\\x=25$

Therefore the value of x is 25 when y =40.

6 0

3 years ago