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3 years ago

How do you solve this?

Mathematics

2 answers:

dolphi86 [110]3 years ago

3 0

You X to the second power.. 4 to the second power is 16 and 16-16=0 so x=4

Lina20 [59]3 years ago

3 0

Add 16
x² - 16 = 0
+ 16 +16

x² = 16
Take square root

$\sqrt{x^2} = \sqrt{16}$

Remember when you take the square root of a perfect square you get two values. +ve and -ve

x = -4
x = 4

because (-4)² = 16
( 4)² = 16

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3 years ago

Use the Fundamental Theorem of Calculus to find the "area under curve" of

lozanna [386]

Answer:

$\displaystyle A = 300$

General Formulas and Concepts:

Calculus

Integrals

Definite Integrals
Area under the curve
Integration Constant C

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:

Step 1: Define

Identify

f(x) = 6x + 19

Interval [12, 15]

Step 2: Find Area

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^{15}_{12} {(6x + 19)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^{15}_{12} {6x} \, dx + \int\limits^{15}_{12} {19} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 6\int\limits^{15}_{12} {x} \, dx + 19\int\limits^{15}_{12} {} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = 6(\frac{x^2}{2}) \bigg| \limits^{15}_{12} + 19(x) \bigg| \limits^{15}_{12}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle A = 6(\frac{81}{2}) + 19(3)$
Simplify: $\displaystyle A = 300$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

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3 years ago