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2 years ago

Can I combine “2x” and “-4x” because they both have x or can I not because ones positive and ones negative?

Mathematics

2 answers:

Vikki [24]2 years ago

7 0

Answer:

You can combine 2x and -4x even though one is negative. Just subtract 4 from 2 to get -2x. :)

Step-by-step explanation:

Harlamova29_29 [7]2 years ago

6 0

Answer:

You can because you would be combining like terms.

Step-by-step explanation:

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Steve paid 10% tax on a purchase of $40. Select the dollar amount of the tax and the total dollar amount Steve paid on the numbe

Shalnov [3]

10% of 40 =
10% × 40 =
.10 × 40 = 4.00

Tax is $4
Amount Paid is 40 + 4 = $44

6 0

2 years ago

Jill has 200 movies downloaded on her phone. Of these movies 24 are comedies. What percent of Jills movies are comedies?

katen-ka-za [31]

Answer:

12%

12% of Jill's movies are comedies.

Step-by-step explanation:

24/200 x 100/1= 12%

7 0

2 years ago

There were 7 pizzas. 5 people ate all of the pizzas. each person ate an equal amount of pizza. how much pizza did each person ea

shtirl [24]

Answer: D

Step-by-step explanation:

A seems complicated and doesn’t make an sense while D is simple

3 0

2 years ago

Find the diameter of a circle with a circumference of 443 feet. Use 3.14 for pi.

marishachu [46]

Answer:

$\huge\boxed{\sf d = 141\ feet}$

Step-by-step explanation:

<u>Given Data:</u>

Circumference = C = 443 ft

<u>Required:</u>

Diameter = d = ?

<u>Formula:</u>

d = C / π

<u>Solution:</u>

d = 443 / 3.14

d = 141.0 feet

d = 141 feet

$\rule[225]{225}{2}$

Hope this helped!

3 0

1 year ago

Read 2 more answers

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

2 years ago