U th th V 3 D F Geometry

The vertex of the parabola below is at the point (1.-3). Which of the equations below could be the one for this parabola?

Ilia_Sergeevich [38]

Answer: Choice D. y = (x-1)^2 - 3

The vertex is (h,k) = (1,-3). So h = 1 and k = -3.

We have a = 1 as the leading coefficient.

Plug those values into the equation below

y = a(x-h)^2 + k

y = 1(x - 1)^2 + (-3)

y = (x - 1)^2 - 3

3 0

2 years ago

Read 2 more answers

Solve the Anti derivative.

Alex Ar [27]

Answer:

$\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C$

General Formulas and Concepts:

Algebra I

Factoring

Calculus

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

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

Trig Integration: $\displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C$

Step-by-step explanation:

Step 1: Define

$\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Factor fraction denominator: $\displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx$
[Integral] Integration Property - Multiplied Constant: $\displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx$

Step 3: Identify Variables

Set up u-substitution for the arctan trig integration.

$\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx$

Step 4: Integrate Pt. 2

[Integral] Substitute u-du: $\displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du$
[Integral] Trig Integration: $\displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C$
[Integral] Simplify: $\displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C$
[integral] Multiply: $\displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C$
[Integral] Back-Substitute: $\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C$

Topic: AP Calculus AB

Unit: Integrals - Arctrig

Book: College Calculus 10e

7 0

2 years ago

True or false: Every sequence is either arithmetic or geometric. If this is true, explain. If false,

tiny-mole [99]

In an arithmetic sequence, the difference between consecutive terms is constant. In formulas, there exists a number $r$ such that

$a_{n+1}-a_n=r\quad \forall n\geq 1$

In an geometric sequence, the ratio between consecutive terms is constant. In formulas, there exists a number $r$ such that

$\dfrac{a_{n+1}}{a_n}=r\quad \forall n\geq 1$

So, there exists infinite sequences that are not arithmetic nor geometric. Simply choose a sequence where neither the difference nor the ratio between consecutive terms is constant.

For example, any sequence starting with

$1, 15, -3,\ldots$

Won't be arithmetic nor geometric. It's not arithmetic (no matter how you continue it, indefinitely), because the difference between the first two numbers is 14, and between the second and the third is -18, and thus it's not constant. It's not geometric either, because the ratio between the first two numbers is 15, and between the second and the third is -1/5, and thus it's not constant.

5 0

3 years ago

Factor this polynomial expression.

tankabanditka [31]

Answer:

D: 3(x + 5)(x + 5)

Step-by-step explanation:

Try multiplying the last answer choice (D); the result is 3(x^2 + 10x + 25), or

3(x + 5)(x + 5).

5 0

3 years ago

Rena wrote all the integers from 1 to 19. How many digits did she write?

harkovskaia [24]

She wrote 19 digits

8 0

2 years ago

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