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3 years ago

Matt and Ash are selling fruit for a school fundraiser. Customers can buy small boxes of fruit or

Mathematics

1 answer:

aliina [53]3 years ago

5 0

Answer: The cost of a small box is $4. The cost of a large box is $7.

Step-by-step explanation:

Let s = cost of a small box and L = cost of a large box

3s + 5L = 47

11s + 10L = 114

Multiply the top equation by -2 to combine with the second equation

-6s - 10L = -94

+11s +10L = 114

L cancels out, now solve for s

5s = 20

s = 4

Now, plug the value of s back into an equation to find the value of L

3(4) + 5L =47

12 + 5L = 47

5L = 35

L = 7

You might be interested in

Which answer choice correctly identifies the missing information in the problem? Joey and Kirk are in the 6th grade. Kirk is 5 f

user100 [1]

Answer:Kirk is taller than Joey by 8 inches

Step-by-step explanation:

Step 1

Kirk's height is given as =5 feet 4 inches tall.

Joey's height 4 feet 8 inches tall

Difference in height = 5 feet 4 inches

- 4 feet 8 inches

Step 2

Bear in mind that 12 inches = 1 feet . Now,8 cannot be subtracted from 4 , so we borrow 1 feet from 5feet, which is equivalent to 12 inches and add to inches so it becomes

4 feet (12+4)inches = 4 feet 16inches

- 4 feet 8 inches - 4 feet 8 inches

= 8 inches

Therefore Kirk is taller than Joey by 8 inches

4 0

3 years ago

I need help on this question

kotegsom [21]

Answer:

My guess would probably be $370.

Explanation:

The reason I say that is because since 200 euros was 270 American dollars, why not try the same thing for 300 euros? If you do the process of elimination, it wouldn't be the $540. The same for the second one so, you have 2 remaining options left. The third or the fourth.

Hope that helps!!

3 0

3 years ago

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

7 0

3 years ago

SOMEONE HELP?!! ASAP !! I’ll give brainliest!! Please and ty !!

dsp73

Answer:

$(\pi 4^2 (9))-(\pi 2^2 (9))$

and

$339.292 cm^3$

Step-by-step explanation:

This question is simple once you break it down.

Let's concentrate on the outer cylinder.

First, find the volume of the outer cylinder.

The volume of a cylinder is expressed as: $V = \pi r^2 h$

Gathering information from the diagram..

d = 8 cm

h = 9 cm

You can get the radius from the diameter by just dividing the diameter by 2.

$r = \frac{8}{2}=4$

r = 4

Plug everything into the volume equation.

$V = \pi 4^2 (9)$

Don't solve just yet!

Now, find the volume of the inner cylinder.

Do the exact same thing as you did for the outer cylinder.

d = 4 cm

h = 9 cm

$r = \frac{4}{2}=2$

r = 2

Plug everything into the volume equation..

$V = \pi 2^2 (9)$

The question asks for an expression to represent the volume of the large cylinder after the cylindrical hole is removed. This means you need to subtract the volume of the cylindrical hole from the volume for the large cylinder.

Again, the volume for the large cylinder is $V = \pi 4^2 (9)$ and the volume for the cylindrical hole is $V = \pi 2^2 (9)$

This means..

$(\pi 4^2 (9))-(\pi 2^2 (9))$

That is your answer to the first part of the question.

For the second part, just solve.

$(\pi 4^2 (9))-(\pi 2^2 (9))$

$(\pi*16*9)-(\pi*4*9)$