Please help! Will give brainliest.

Mathematics

1 answer:

Deffense [45]2 years ago

7 0

$4.15n - ($250 + $2.45n)

Now that we have our expression, let's set it greater than or equal to 0

$4.15n - ($250 + $2.45n) >= 0

Let's simplify the left side.

$4.15n - $250 - $2.45n >= 0.

Combine like terms.

$1.7n - $250 >= 0

Add $250 to both sides.

$1.7n >= $250

Divide both sides by $1.7

n >= 148

You would need to make 148 light sabers to make a positive profit.

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Use the Shell Method to find the volume of the solid obtained by rotating region under the graph of f(x)=x^2+2 for 0≤x≤5 about x

Irina18 [472]

For each x in the interval 0 ≤ x ≤ 5, the shell at that point has

• radius = 5 - x, which is the distance from x to x = 5

• height = x ² + 2

• thickness = dx

and hence contributes a volume of 2π (5 - x) (x ² + 2) dx.

Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:

$\displaystyle 2\pi \int_0^5 (5-x)(x^2+2)\,\mathrm dx=2\pi\int_0^5 (10-2x+5x^2-x^3)\,\mathrm dx=\boxed{\frac{925\pi}6}$

4 0

2 years ago

The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p

Juliette [100K]

Check the picture below.

based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,

$\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)$

$\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}$

now, we can plug those values on A = (1/2)bh,

$\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5$