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3 years ago

What is the area of the triangle?

Mathematics

1 answer:

docker41 [41]3 years ago

6 0

Answer:

A = 54cm^2

Step-by-step explanation:

A = (base × height)/2

base = 9cm

height= 12cm

A = (9 × 12)/2

A = 108/2

A = 54

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I need help with this question

sergiy2304 [10]

Answer:

Step-by-step explanation:

There all in the same order so just put them into boxes like that

7 0

3 years ago

How much more money a week does Juanita save than Daniel?

babunello [35]

Answer:

Step-by-step explanation:

Juanita makes 15 an hour and Daniel makes 10 so 15-10 = 5

4 0

3 years ago

Chapter 8 Written Homework 1. A hypothesis test is conducted to test the claim that the proportion of people with dark hair at M

UNO [17]

Answer:

we reject H₀

Step-by-step explanation: Se annex

The test is one tail-test (greater than)

Using α = 0,05 (critical value ) from z- table we get

z(c) = 1,64

And Test hypothesis is:

H₀ Null hypothesis μ = μ₀

Hₐ Alternate hypothesis μ > μ₀

Which we need to compare with z(s) = 2,19 (from problem statement)

The annex shows z(c), z(s), rejection and acceptance regions, and as we can see z(s) > z(c) and it is in the rejection region

So base on our drawing we will reject H₀

3 0

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.