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2 years ago

Pls help

Mathematics

1 answer:

galina1969 [7]2 years ago

8 0

Answer:

van: 8
bus: 22

Step-by-step explanation:

If we let v and b represent the seating capacity of vans and buses, respectively, then the two trips can be described by ...

8v +8b = 240

4v +b = 54

Dividing the first equation by 8 gives ...

v +b = 30

Subtracting that from the second equation gives ...

3v = 24

v = 8 . . . . . . . . . . . divide by 3

b = 30 -8 = 22

The number of students in each van is 8; the number in each bus is 22.

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A real estate agent earned $7200 commission on a property sale of $240,000. What is her rate of commission?

cluponka [151]

The real estate agent has a commission rate of 33.3%

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which would mean she has a (1/3) commission rate. hope this helps you out!

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3 years ago

Eva wants to make two pieces of pottery.She needs 3/5 pound of clay for one piece and 7/10 pound of clay for the other piece.She

marta [7]

Answer:

a) Eva will need 2 bags of clay to make both pieces of pottery.

b) She will have leftover $\frac{11}{10}$ pounds of clay.

Step-by-step explanation:

a) We have to find the total amount of clay, as she needs $\frac{3}{5}pound=P_{1}$ and $\frac{7}{10}pound=P_{2}$ , so $P_{1}+P_{2}= \frac{3}{5} +\frac{7}{10} =\frac{6+7}{10}=\frac{13}{10}pounds$ , is the total amount of clay. Then as we Know that one bag of clay has $\frac{4}{5} =\frac{8}{10} pounds$ , It isn´t enough, but with two bags $2*\frac{8}{10}=\frac{16}{10} pounds$ , It is enough clay.

b) She will have leftover $\frac{16}{10} -\frac{13}{10} =\frac{3}{10}pounds$ of the second bag, then we add the third one, so $\frac{3}{10} +\frac{8}{10} =\frac{11}{10}pounds$ are leftover.

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2 years ago

2. If the present salary of an employee is $36,500 and they are to receive a 7% raise next year, what will be the new salary?

Natasha_Volkova [10]

Answer:

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2 years ago

3. Samantha earns $7.80 per hour. If her regular hours are 40

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Answer:

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Step-by-step explanation:

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3 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

3 years ago