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3 years ago

Factor the expression below. x2-10x+25

Mathematics

1 answer:

docker41 [41]3 years ago

3 0

Answer:

Step-by-step explanation:

(x-5)(x-5)

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Answer the question below

Lena [83]

To find surface area, you would find the area of each side of the shape. Since this is a cube, each side is a square. So area=side2, So each side is 9, so 9x9=81 for each face. There are 6 faces to a cube, so 6x81=486 in2 is your answer

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3 years ago

3/4(-10+2/5) please answer asap

Tju [1.3M]

Answer:

3/4(-10+2+2/5)

answer -5.7

Step-by-step explanation:

5 0

3 years ago

Can someone answer this for me?

erik [133]

Answer:

sin = opposite / hypotenuse

cos = adjacent/hypotenuse

sin U = 4/5

cos U = 3/5

sin T = 3/5

cos T= 4/5

3 0

3 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: $\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C$