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3 years ago

13+26 [(2.8 x 5) divided by 7] HELPPP

1 answer:

4 0

65
1. you work in the parentheses and multiply 2.8 by 5 which is 14
2. you divide 14 by 7 which is 2
3. you follow PEMDAS and multiply 26 by what is in the parentheses (2) which is 52
4. you add 13 with 52 which is 65
hope this helped!!

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Write an equation for the line below

zimovet [89]

Answer:

y = 1/3x - 3

Step-by-step explanation:

Two points are (0, -3) and (-6, -5)

Slope formula [ y2-y1/x2-x1 ]

-5-(-3)/-6-0

-2/-6

1/3

y = 1/3x - 3

Best of Luck!

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4 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.

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

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

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

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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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(-2,-6) and (-6,10) write the equation of the line with the given points in slope intercept form

Stels [109]

The coordinate points (-2,-6) and (-6,10) written in slope-intercept form is...

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3 years ago

What is porpotinal to 1/5

Irina-Kira [14]

An equation of the form y=ax+b, where y and x are variables, and a and b are constants, is called a linear equation.

The reason it is called linear is because the graph of the equation is a line.

The line passes through the origin only if b=0, as in our problem.

Any line (except vertical lines) represents a proportional relationship in that the change in y is proportional with the change in x. It is not a condition that the line passes through the origin.

In our specific case, y=(1/5)x, means that the change in y is always 1/5 of the change in x. That is, if x changes by 5 units, y changes by 1. If x changes by 10 units, y changes by 2, and so on.

So, the constant of proportionality is 1:5, or 0.2.

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Step-by-step explanation:

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3 years ago