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4 years ago

x-21<31 Solve the inequality and enter your solution as an inequality comparing the variable to a number

Mathematics

1 answer:

Radda [10]4 years ago

8 0

Answer:

x<52

Step-by-step explanation:

I added 21 to both sides

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Someone pls help I need this like now pls lots of points and brainliest

Serjik [45]

Answer:

See below

Step-by-step explanation:

5⁵(16²5³)³ = 5⁵(2⁴)³(5³)³ = 5⁵2¹²5⁹ = 5¹⁴2¹²

(8⁴5³/8⁵)² = (5³/8)² = (5³)²/(2³)² = 5⁶ / 2⁶

(5⁸3⁷/5⁴)¹⁰ = (5⁴3⁷)¹⁰ = (5⁴)¹⁰(3⁷)¹⁰ = 5⁴⁰3⁷⁰

<u>Multiplying powers with same base:</u>

$a^b*a^c=a^{b+c}$
Same base with powers added up

<u>Multiplying powers with same exponent but different base:</u>

$a^b*c^b=(ac)^b$
Same exponent with bases multiplied

8 0

3 years ago

Y-2=-4(x-1) in standard form

GalinKa [24]

Answer:

4x+y=6

Step-by-step explanation:

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2 years ago

Read 2 more answers

Select the correct answer.

Alex787 [66]

Answer:

Step-by-step explanation:

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3 years ago

I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.