How to solve this multistep inequality (3f+9<21)

During his high school football career a running back averaged 81 yards/game with a standard

hodyreva [135]

The answer is ym=b your welcome

4 0

2 years ago

What is 1/2 + 4/7 equal to?

Mashcka [7]

15/14 1 and 1/14 is the answer

8 0

2 years ago

Read 2 more answers

A local food truck specializes in gourmet grilled cheese sandwiches. Each month, the owners of the truck have a constant total o

Serggg [28]

Answer:

A. (9 - 2.25)s ≥ 5,400

B. They will have to sell more than 800 sandwiches per month. I first started by solving the parentheses which is 6.75. Then I divided 5,400 by 6.75 to get 800.

C. It will now end up looking like this: (8 - 2.25)s ≥ 5,400. Then it will be 5.75. Then divide 5,400 by 5.75. Then you end up with 939.13. You have to round up when you are talking about inequalities with money so it will be 940. So they will have to sell 140 more sandwiches during that month.

Step-by-step explanation:

7 0

2 years ago

Help pls it’s due today *photo*

a_sh-v [17]

9514 1404 393

Answer:

ACBD-BDAC

Step-by-step explanation:

It is convenient to compare the values of each of the functions when h=1. On the graphs, find 1 on the horizontal axis, then read the value of the function at that point from the vertical axis.

For the equations, the value for h=1 will be its coefficient in the equation.

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1) a: 80,b: 50, c: 75, d: 30

Fastest to slowest is ACBD.

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2) a: 4, b: 2, c: 8, d: 3

Slowest to fastest is BDAC

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The code is these lists, separated by a dash: ACBD-BDAC.

4 0

2 years ago

Compute the differential of surface area for the surface S described by the given parametrization.

AysviL [449]

With $S$ parameterized by

$\vec r(u,v)=\langle e^u\cos v,e^u\sin v,uv\rangle$

the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

We have

$\dfrac{\partial\vec r}{\partial u}=\langle e^u\cos v,e^u\sin v,v\rangle$

$\dfrac{\partial\vec r}{\partial v}=\langle -e^u\sin v,e^u\cos v,u\rangle$

with cross product

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle ue^u\sin v-ve^u\cos v,-ve^u\sin v-ue^u\cos v,e^{2u}\cos^2v+e^{2u}\sin^2v\rangle$

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle e^u(u\sin v-v\cos v),-e^u(v\sin v+u\cos v),e^{2u}\rangle$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{e^{2u}(u\sin v-v\cos v)^2+e^{2u}(v\sin v+u\cos v)^2+e^{4u}}$

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=e^u\sqrt{u^2+v^2+e^{2u}}$

So we have

$\mathrm dS=\boxed{e^u\sqrt{u^2+v^2+e^{2u}}\,\mathrm du\,\mathrm dv}$

8 0

2 years ago