PLEASE HELP I NEED THIS

My new Labrador puppy weighs 3 pounds more than 0.25 of his mom's weight His mom weighs 60 pounds what is kaiser weight?

Delicious77 [7]

Ok, first you have to know that 0.25 is one fourth. Finding one fourth of 60 is the first step. 1/4 of 60 is 15. Now, add 3 to 15, and you have your answer! The answer is 18 pounds.

7 0

3 years ago

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A driver works 40 hours a week for the first two months

kakasveta [241]

Not sure but you can figure it out.

7 0

3 years ago

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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.

4 0

1 year ago

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(-5)x (-5) x (-5) x (-5) = <br> What would the answer be <br><br> 20 points

Natasha_Volkova [10]

Answer:

$625x^{3}$

Step-by-step explanation:

4 0

3 years ago

BRAINLIEST AND FIVE STARS TO CORRECT ANSWER

Llana [10]

The answer is the first option, <-20, -42>.

We can find this by first finding what -2u would equal by multiplying <5, 6> by -2. This gives us <-10, -12>.

Then we need to find out what 5v is equal to, by multiplying <-2, -6> by 5 to get <-10, -30>.

Now that we know what -2u and 5v are, we can substitute them into the equation and get
<-10, -12> + <-10, -30>, which we can split up into -10 - 10 = -20, and -12 - 30 = -42, so your final answer is <-20, -42>.

I hope this helps!

8 0

3 years ago