What is the solution to the system of equations below. Y=3/4x-12and y=4x-31 (-4,-15) (-4,-12). (4,-9). (4,-47)

Mathematics

1 answer:

andriy [413]2 years ago

3 0

Answer:

(4, –47)

Step-by-step explanations

I think this is the right answer but sorry if im wrong.

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Answer:

D. A formula

Step-by-step explanation:

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3 years ago

Determine the area enclosed by y=2x+3, the x-axis and the ordinates x=3 and x=4

jok3333 [9.3K]

Answer:

$\displaystyle \int\limits^4_3 {2x + 3} \, dx = 10$

General Formulas and Concepts:
<u>Calculus</u>

Integration

Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

y = 2x + 3

<em>x</em>-interval [3, 4]

<em>x</em>-axis

<em>See attachment for graph.</em>

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^4_3 {2x + 3} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^4_3 {2x} \, dx + \int\limits^4_3 {3} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 2 \int\limits^4_3 {x} \, dx + 3 \int\limits^4_3 {} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = 2 \bigg( \frac{x^2}{2} \bigg) \bigg| \limits^4_3 + 3(x) \bigg| \limits^4_3$
[Integrals] Integrate [Integration Rule - FTC 1]: $\displaystyle A = 2 \bigg( \frac{7}{2} \bigg) + 3(1)$
Simplify: $\displaystyle A = 10$