All categories

1 year ago

Does this graph show a function? Explain how you know.

Mathematics

1 answer:

igomit [66]1 year ago

6 0

Answer:

C. Yes; the graph passes the vertical line test.

Step-by-step explanation:

Reasons the others are wrong:

A. It does pass the vertical line test.

B. Y-values can have any x values. It is x-values that can't have multiple y-values.

D. There are y-values with more than one x-value on this graph so that's just false. Even if it were true, that still doesn't invalidate it due to the reasons given in Answer B.

You might be interested in

Rohan buys 9 tangerines for the same price each, and 1 watermelon for $4. He spends less than $14 altogether. Which inequality c

Ludmilka [50]

9x + 4 < 14 is the required inequality.

8 0

2 years ago

Find the value of PT PT = 5x + 3 and TQ 7x - 9

77julia77 [94]

Is there anything more to this question? Like are PT and TQ equal?

5 0

2 years ago

Read 2 more answers

a bowl of candies contains the following makeup of colors: 4 red, 2 yellow, 5 green, and 1 purple. taina is selecting a candy at

Ivanshal [37]

<u>Candies</u>

4 red

2 yellow

5 green

1 purple

- Taina doesn't like red candies.

4/4+2+5+1 = 4/12

4/12 is the probability Taina <em>will</em> get a red candy, but we're looking for the opposite.

12/12 - 4/12 = 8/12 = 2/3

The probability Taina will not select a red candy is 2/3.

8 0

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:

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

5 0

2 years ago

Plz help this is due in 39 minutes

ale4655 [162]

Answer:

x=50

y=50

Step-by-step explanation:

its right because sas

8 0

2 years ago

Read 2 more answers