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3 years ago

7

URGENT!!!!!!! i really need help. can some help with this piecewise function? or at least explain it to me so i can figure it ou

t...
f(x)= {2 if x ≥ 4

1 answer:

USPshnik [31]3 years ago

3 0

Is it F = 1/2 Don't take my word for it, though, I am just trying to solve it in a way that would give you an answer because the problem itself, didn't make much sense. Good Luck!

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Use Gauss-Jordan elimination to solve the following linear system:

Olenka [21]

-3x + 4y = -6

20x - 4y = 40
-----------------------
17x = 34

x = 2

5(2) - y = 10

10 - y = 10

y = 0

the correct answer is D. (2,0)

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3 years ago

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A baseball pitcher won 60% of his games. How many did he win if he pitched 35 games?

Sergio039 [100]

Answer:

21 games

Step-by-step explanation:

Multiply the number of games by the percentage.

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3 years ago

Calculus 2. Please help

Anarel [89]

Answer:

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}}} \, dx = \infty$

General Formulas and Concepts:

<u>Algebra I</u>

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

<u>Calculus</u>

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integrals

Definite Integrals

Integration Constant C

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

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

U-Substitution

U-Solve

Improper Integrals

Exponential Integral Function: $\displaystyle \int {\frac{e^x}{x}} \, dx = Ei(x) + C$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

[Integral] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \int\limits^1_0 {\frac{e^{-x^2}}{x} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \int\limits^1_a {\frac{e^{-x^2}}{x} \, dx$

<u>Step 3: Integrate Pt. 2</u>

<em>Identify variables for u-substitution.</em>

Set: $\displaystyle u = -x^2$
Differentiate [Basic Power Rule]: $\displaystyle \frac{du}{dx} = -2x$
[Derivative] Rewrite: $\displaystyle du = -2x \ dx$

<em>Rewrite u-substitution to format u-solve.</em>

Rewrite <em>du</em>: $\displaystyle dx = \frac{-1}{2x} \ dx$

<u>Step 4: Integrate Pt. 3</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {-\frac{e^{-x^2}}{x} \, dx$
[Integral] Substitute in variables: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {\frac{e^{u}}{-2u} \, du$
[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}\int\limits^1_a {\frac{e^{u}}{u} \, du$
[Integral] Substitute [Exponential Integral Function]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(u)] \bigg| \limits^1_a$
Back-Substitute: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-x^2)] \bigg| \limits^1_a$
Evaluate [Integration Rule - FTC 1]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-1) - Ei(a)]$
Simplify: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{Ei(-1) - Ei(a)}{2}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \infty$

∴ $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$ diverges.

Topic: Multivariable Calculus

7 0

3 years ago

LITERALLY THE MOST ROCK HARD QUESTION I HAVE EVER TRIED TO SOLVE PLEASE HELP ME OUT BECAUSE MY EYES ARE BURNING FROM STARING AT

ivolga24 [154]

1) You will need to simplify both sides of the inequality.
$6x-13 \ \textless \ 6x-12
$

2) Now you will need to subtract the like terms which leaves you with
$-13\ \textless \ -12$

We cannot do anything else so the answer will be a<span>ll real numbers are solutions.</span>

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3 years ago

Bill is making a poster with an area of 30 square inches. Jeff is making a poster with an area of 30 square centimeters. Phil th

sukhopar [10]

No, because there are 2.54 centimeters in one inch which means 30cm=11.8111 inches therefore they won't be the same size.

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3 years ago