What is 242 divided by 57 worked out

A restaurant manager states the number of customers that enter the

konstantin123 [22]

Yes it is true because if 3x +15=2x+60 it is true

6 0

3 years ago

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Find the sum. (–7b + 8c) – (12a + 14) + (5a + 5b)

Semmy [17]

Combine like terms
Then you’ll get -7a-2b+8c-14 which is the answer.

3 0

3 years ago

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Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):

$y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1 \right)} \, dx$

$y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx$

$y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C$

$C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0$

$C = 0$

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

5 0

4 years ago

Whats 23x+34=x<br> u ain't r e a l l y got tu

MatroZZZ [7]

Answer:

x= -17/11

Step-by-step explanation:

7 0

3 years ago

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Which of the following is an improper integral?

guapka [62]

Answer:

A) $\displaystyle \int\limits^3_0 {\frac{x + 1}{3x - 2}} \, dx$

General Formulas and Concepts:

<u>Calculus</u>

Discontinuities

Removable (Hole)
Jump
Infinite (Asymptote)

Integration

Integrals
Definite Integrals
Integration Constant C
Improper Integrals

Step-by-step explanation:

Let's define our answer choices:

A) $\displaystyle \int\limits^3_0 {\frac{x + 1}{3x - 2}} \, dx$

B) $\displaystyle \int\limits^3_1 {\frac{x + 1}{3x - 2}} \, dx$

C) $\displaystyle \int\limits^0_{-1} {\frac{x + 1}{3x - 2}} \, dx$

D) None of these

We can see that we would have a infinite discontinuity if x = 2/3, as it would make the denominator 0 and we cannot divide by 0. Therefore, any interval that includes the value 2/3 would have to be rewritten and evaluated as an improper integral.

Of all the answer choices, we can see that A's bounds of integration (interval) includes x = 2/3.

∴ our answer is A.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

6 0

3 years ago