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3 years ago

Put these numbers from least to greatest. I'm too lazy to do it. it's for trigonometry.

Mathematics

2 answers:

blagie [28]3 years ago

8 0

Answer:

15, 20, 25, 40, 50, 55, 65, 70, 75, 80, 85, 95

Hope this was all the numbers.

Strike441 [17]3 years ago

7 0

Answer:

15, 20, 25, 40, 50, 55, 65, 70, 75, 80, 85, 95

Hope this helps

-Amelia

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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:

Calculus

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

PLEASE HELP!!!!!!!!!!!

Temka [501]

Answer:

the answer is A

Step-by-step explanation:

8 0

3 years ago

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The number of pieces of chalk in a classroom deceased by 10 throughout the school year. Is this a geometric sequence?

VARVARA [1.3K]

No, the geometric sequence is when the number is multiplied by a value that increases.

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3 years ago

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dlinn [17]

A 3x with an exponet of 3

3 0

3 years ago

Consider functions f and g.1 + 12f(1) = 12 + 4. – 12for * # 2 and 7 -64.2 – 16. + 1641 +48for a # -12 Which expression is equal

lisabon 2012 [21]

Given the following functions below,

$\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}\text{ and} \\ g(x)=\frac{4x^2-16x+16}{4x+48} \end{gathered}$

Factorising the denominators of both functions,

Factorising the denominator of f(x),

$\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}$

Factorising the denominator of g(x),

$\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}$

Multiplying both functions,

$undefined$

4 0

1 year ago