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2 years ago

7

Rob had a 15% off coupon for the sporting goods store. He bought a tennis racket that had a regular ticket price of $94.00. How

much did Rob spend on the racket after using his coupon?

1 answer:

Mnenie [13.5K]2 years ago

3 0

So I’m guessing there’s no tax so here’s the answer without tax.

Work;
15% of 94.00 = 14.1

94.00 - 14.1 = 79.9

So after the coupon used, Rob spent 79.9 on the tennis racket.

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Find the value of x in the triangle shown below

Georgia [21]

Answer:

the length of the hypotenuse must be 10.

Step-by-step explanation:

This is a right triangle, so we can apply the Pythagorean Theorem.

6² + 8² = 10² so the length of the hypotenuse must be 10.

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2 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:

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

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2 years ago

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olasank [31]

I think b=6. so try that ok

8 0

2 years ago

Please answer properly.

madreJ [45]

Answer:

$1) y=-6$

$2)x=3$

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Step-by-step explanation:

$4y+24=0$

subtract 24 from both sides

$4y=-24$

divide both sides by 4

$y=-6$

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$15+3=6x$

combine like terms

$18=6x$

divide both sides by 6

$3=x$

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$9y-72=0$

add 72 to both sides

$9y+72$

divide both sides by 9

$y=8$

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$8a=75-11$

combine like terms

$8a=64$

divide both sides by 8

$a=8$

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Hope this is helpful.

8 0

3 years ago

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What is the equation of the line? Please help asap!!!! First Correct answer will get brainestttttt !!

faust18 [17]

Answer:

3y= -2x-6

Step-by-step explanation:

The two points on the line are (-3,0) and (0,-2)

so you first get the gradient;

gradient= <u>change in y</u>

change in x

= <u>-2-0</u>

0-(-3)

=<u> -2</u>

3

so the answer above is the gradient

then pick one point that you used to get the gradient with, so as for me I'll pick (-3,0) and then a general point which is always (x, y)

since you have the gradient you can easily get the equation by doing this

<u>y-0</u><u>. </u><u> </u> = <u>-2</u>

x-(-3). 3

then crossmultiply to get the equation of the line

3y= -2(x+3)

3y= -2x-6

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2 years ago