MATHEMATICS COLLEGE

A parking lot is 100 yards long. What is the length of 3/4 of the parking lot, in feet? (Not in words)

Answers

Answer 1

Answer:

1 yard = 3 feet.

100 yards x 3 = 300 feet.

300 feet x 3/4 = 225 feet.

The answer is 225 feet.

Answer 2

Answer:

Answer:

225 ft hope it helps

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A beluga whale is 5 yards and 3 inches long, and a gray whale is 15 yards and 5 inches long. What is the difference in length between the two whales?

Answers

Answer:

Difference in length between two whales is 362 inches or 10 yards and 2 inches.

Step-by-step explanation:

Lets convert the values to inches:

1 yard = 36 inches

Length of a beluga whale in inches:

5 yards = $5*36$ inches = $180$ inches

Therefore total length of a beluga whale = 180 inches + 3 inches

=183 inches

Length of a gray whale in inches:

15 yards = $15*36$ inches = $540$ inches

Therefore total length of a beluga whale = 540 inches + 5 inches

=545 inches

Difference in length between two whales = 545 inches - 183 inches

= 362 inches

To represent this in yards and inches we can divide the value by 36.

$(362)/(36) =10.06$ yards

=10 yards and 2 inches

Difference in length between two whales is 10 yards and 2 inches.

Simplify:
4n + 6 -(5n+3)
A. -n +3
B. -n +9
C. 9n + 3
D. 9n +9

Answers

Answer:

Answer is A.

-n+3

Step-by-step explanation:

Let f(x)=3x+5, g(x)=2^x
Find f(g(x))

Answers

Answer:

The expression will be 3(2^x) + 5

Step-by-step explanation:

In order to find the new expression, you have to let g(x) be the x :

$f(x) = 3x + 5$

$g(x) = {2}^(x)$

$fg(x) = 3( {2}^(x) ) + 5$

Determine whether the improper integral converges or diverges, and find the value of each that converges.∫^0_-[infinity] 5e^60x dx

Answers

Answer:

The improper integral converges.

$\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)$

General Formulas and Concepts:
Calculus

Limit

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_(x \to c) x = c$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Method: U-Substitution

Improper Integral: $\displaystyle \int\limits^(\infty)_a {f(x)} \, dx = \lim_(b \to \infty) \int\limits^b_a {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = 5 \int\limits^0_(- \infty) {e^(60x)} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) 5 \int\limits^0_(a) {e^(60x)} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 60x$
[u] Differentiate [Derivative Properties and Rules]: $\displaystyle du = 60 \ dx$
[Bounds] Swap: $\displaystyle \left \{ {{x = 0 \rightarrow u = 0} \atop {x = a \rightarrow u = 60a}} \right.$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(a) {60e^(60x)} \, dx$
[Integral] Apply Integration Method [U-Substitution]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(60a) {e^(u)} \, du$
[Integral] Apply Exponential Integration: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) e^u \bigg| \limits^0_(60a)$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1 - e^(60a))/(12)$
[Limit] Evaluate [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1 - e^(60(-\infty)))/(12)$
Rewrite: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12) - (1)/(12e^(60(\infty)))$
Simplify: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)$

∴ the improper integral equals $\displaystyle \bold{(1)/(12)}$ and is convergent.

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Learn more about improper integrals: brainly.com/question/14413972

Learn more about calculus: brainly.com/question/23558817

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration

Answer:

$\int_(-\infty)^0 5 e^(60x) dx = (1)/(12)[e^0 -0]= (1)/(12)$

Step-by-step explanation:

Assuming this integral:

$\int_(-\infty)^0 5 e^(60x) dx$

We can do this as the first step:

$5 \int_(-\infty)^0 e^(60x) dx$

Now we can solve the integral and we got:

$5 (e^(60x))/(60) \Big|_(-\infty)^0$

$\int_(-\infty)^0 5 e^(60x) dx = (e^(60x))/(12)\Big|_(-\infty)^0 = (1)/(12) [e^(60*0) -e^(-\infty)]$

$\int_(-\infty)^0 5 e^(60x) dx = (1)/(12)[e^0 -0]= (1)/(12)$

So then we see that the integral on this case converges amd the values is 1/12 on this case.

Explain how to write 2 dividid 5 as a fraction

Answers

Answer:

2/5

Step-by-step explanation:

Two divided by 5 is simply written as 2/5.

Hope I helped!

2/5 is how it’s written as a fraction, 2 is over 5, 2 goes inside and 5 outside when you write the division expression

both cylinders are emptied, and water is poured into the narrow cylinder up to the 11th mark. how high would this water rise if it were poured into the empty wide cylinder

Answers

The height of the water when poured into the empty wide cylinder is;

Option A; To the 7¹/₃ mark

Formula for volume of a cylinder is;

V = πr²h

where;

r is radius

h is height

We are told that water is poured into the wide cylinder up to the 4th mark. Thus, for the wide cylinder, h = 4. Thus;

V_wide = 4πR²

Similarly, we are told that water is poured into the narrow cylinder up to the 6th mark. Thus, for the narrow cylinder, h = 6. Thus;

V_narrow = 6πr²

Now, the volume of the water will be the same since it was the same quantity that was poured. Thus;

V_wide = V_narrow

4πR² = 6πr²

⇒ R²/r² = 6/4

simplifies to get; R²/r² = ³/₂

Now both cylinder were emptied and water poured rises to the 11th mark for the narrow cylinder. Thus;

πR²H = 11πr²

R²/r² = 11/H

Earlier, we saw that R²/r² = ³/₂. Thus;

11/H = ³/₂

H = 22/3

H = 7¹/₃

The complete question is;

Attached are the drawings of a wide and a narrow cylinder. The cylinders have equally spaced marks on them. Water is poured into the wide cylinder up to the 4th mark (see A). This water rises to the 6th mark when poured into the narrow cylinder (see B). Both cylinders are emptied, and water is poured into the narrow cylinder up to the 11th mark. How high would this water rise if it were poured into the empty wide cylinder?

A) To the 7¹/₃ mark

B)To the 8th mark

C) To the 7¹/₂mark

D)To the 9th mark

E) To the 11th mark

Read more at; brainly.com/question/16760517

Answer:

COMPLETE QUESTION:

To the right are drawings of a wide and a narrow cylinder. The cylinders have equally spaced marks on them. Water is poured into the wide cylinder up to the 4th mark (see A). This water rises to the 6th mark when poured into the narrow cylinder (see B). Both cylinders are emptied, and water is poured into the narrow cylinder up to the 11th mark. How high would this water rise if it were poured into the empty wide cylinder?

a)To the 7 1/2 mark b)To the 9th mark c)To the 8th mark d)To the 7 1/3 mark

e)To the 11th mark

ANSWER : Option D (To the 7 1/3 mark)

Step-by-step explanation:

First part of the question enables us to get the relationship between the radius of the wider cylinder (R) and the narrow cylinder(r) i.e

Volume of cylinders

π x R² x 4 = πxr²x 6

R²/r² = 6/4

after both cylinder were emptied

π x R² x h = π x r² x 11

R²/r² = 6/4 = 11/h

h = (4 x 11) /6 = 22/3 = 7 1/3 mark