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The inequality 10.45b + 56.50 < 292.67 is used to find the number of boxes (b) that can be loaded on a truck without exceedin

PSYCHO15rus [73]

We solve the inequality by subtracting 56.50 from both sides of the equation,
10.45b + 56.50 - 56.50 < 292.67 - 56.50
10.45b < 236.17
Then, divide both sides of the inequality by 10.45
b < 22.6
The solution suggests that the number of boxes than can be loaded on a truck without exceeding the weight limit of the truck should always be lesser than 22.6. Since we are talking about number of boxes, the maximum number of boxes that can be loaded should only be 22.

6 0

3 years ago

Read 2 more answers

In this system of linear equations with no solution

Ksivusya [100]

Two line with the same slope are parallel, if two parallel lines have a different y-intercept they will never cross each other.

(Any two lines with different slopes will always intersect at some point)

3 0

3 years ago

Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$

4 0

2 years ago

Complete the factorization. a^2+3a-28=(a-4)(a+ )

Murrr4er [49]

Answer:

Step-by-step explanation:

6 0

2 years ago

Find the surface area of the composite figure.

Sloan [31]

With the help of the area formulae of rectangles and triangles and the concept of surface area, the surface area of the composite figure is equal to 276 square centimeters.

<h3>What is the surface area of a truncated prism?</h3>

The surface area of the truncated prism is the sum of the areas of its six faces, which are combinations of the areas of rectangles and right triangles. Then, we proceed to determine the surface area:

A = (12 cm) · (4 cm) + 2 · (3 cm) · (4 cm) + 2 · (12 cm) · (3 cm) + 2 · 0.5 · (12 cm) · (5 cm) + (5 cm) · (4 cm) + (13 cm) · (4 cm)

A = 48 cm² + 24 cm² + 72 cm² + 60 cm² + 20 cm² + 52 cm²

A = 276 cm²

With the help of the area formulae of rectangles and triangles and the concept of surface area, the surface area of the composite figure is equal to 276 square centimeters.

To learn more on surface areas: brainly.com/question/2835293

#SPJ1

7 0

10 months ago