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irinina [24]
3 years ago
8

Original price is $300; Discount is 25%

Mathematics
2 answers:
Svet_ta [14]3 years ago
5 0

Answer:

$225

Step-by-step explanation:

25% of 300 is also 1/4 of 300 which is 75

you take the discounted price away from the original price

300-75=225

mart [117]3 years ago
3 0
25% of 300

25x300/100 = 7500/100 = 75

25% of $300 is $75
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Jennifer is driving her boat West at a speed of 12 mph. The current is moving due South at 5 mph.
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Speed is the rate of distance over time.

<em>The speed of the boat is 13mph south-west</em>

Given

<em />v_1 = 12mph<em> --- west</em>

<em />v_2 = 5mph<em> -- south</em>

<em />

The speed is illustrated with the attached image

Because both speeds act at a diagonal, the speed (v) of the boat is calculated using Pythagoras theorem as follows:

v^2 = 12^2 + 5^2

v^2 = 144 + 25

v^2 = 169

Take square root

v = \sqrt{169

v = 13

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<em>Hence, the speed of the boat is 13mph south-west</em>

Read more about speed at:

brainly.com/question/22610586

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3 years ago
The population of a certain inner-city area is estimated to be declining according to the model P(t)=331,000e−0.017t, where t is
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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.

x\in(-\infty,0) \implies x = -1 \implies f''(-1) = 32e^2 > 0

x\in(0,1) \implies x = \dfrac12 \implies f''\left(\dfrac12\right) = \dfrac5{43} > 0

x\in(1,3) \implies x = 2 \implies f''(2) = -\dfrac{16}{e^4} < 0

x\in(3,\infty) \implies x = 4 \implies f''(4) = \dfrac{192}{e^8} > 0

The sign of f'' changes to either side of x=1 and x=3, but not x=0. This means only \boxed{x=1} and \boxed{x=3} are inflection points.

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1 year ago
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Answer:

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