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3 years ago

Write a matrix to represent the following system 2x + 6y = -12 5x - 5y = 10

Mathematics

2 answers:

kari74 [83]3 years ago

7 0

1)(D) (3,1)

2)(C)(-3,-5)

3)(B)(0,-2)

4)(B)

100%right On Connexus Lesson 3 Unit 1

trapecia [35]3 years ago

6 0

Your answer will be:
[2 6 -12]

[5 -5 10]

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What is 15/4 + (−4/13) as a fraction in simplest form.

OLga [1]

Step-by-step explanation:

15/4 + (−4/13) can be rewritten as 15/4-4/13.

15/4-4/13

15 x13

4 x13

195/52

4 x4

13 x4

16/52

195/52-16/52=179/52 or 3 23/52

hope it helps!

7 0

3 years ago

If f(x)=x-3, g(x)=x^2+1 and h(x) = -4x+1 , find f(g(2))?

natka813 [3]

F(g(2))
g(2)=2^2+1=4+1=5
f(5)=5-3=2
f(g(2))=2

8 0

4 years ago

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