I attached a picture please tell me where she made a mistake i am really struggling thanks.

Which of the following values are in the range of the function graphed below? Check all that apply

a_sh-v [17]

Answer:

B, C, D

Step-by-step explanation:

In this problem, the range is what the output, or y, can be. The origin, or the middie of the graph, is when x=0 and y=0. From the 10s on the screen, we can gather that 5 lines = a distance of 10 on the graph. Using this information, we can say

5 lines = distance of 10

divide both sides by 5 to find the distance for each line

1 line = distance of 2

The function goes from y=0 to three lines down, for a distance of 6. The range is therefore [-6,0] as all values from -6 to 0 on the y axis are included on the graph, including 0 and -6. In this range, -6, -2, and -1 are all included.

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=e%20%3D%20mc%20%7B2%7D" id="TexFormula1" title="e = mc {2}" alt="e = mc {2}" align="absmiddle"

serg [7]

Answer:

The equation — E = mc2 — means "energy equals mass times the speed of light squared." It shows that energy (E) and mass (m) are interchangeable.

Hope this helped

6 0

3 years ago

A test consists of 100 multiple choice questions, each with five possible answers, only one of which is correct. Find the mean a

TEA [102]

Answer: Mean = 20 and standard deviation = 4

Step-by-step explanation:

Let x represents the success of getting the correct answer.

Here, the total number of trials(n) = 100

The probability of getting the correct answer, p = 1/5

Thus, the mean of the number of correct answer,

$\mu_x= np$

$\mu_x= 100\times 1/5$

$\mu_x= 20$

Now, the standard deviation, $\sigma_x= \sqrt{n\times p\times (1-p)}$

$\sigma_x= \sqrt{100\times 1/5\times 4/5}$

$\sigma_x= \sqrt{16}$

$\sigma_x= 4$

Thus, First option is correct.

6 0

3 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago

Please somebody else help me to do . I will mark the correct answer as a brainiest

vodomira [7]

there is your answer follow and mark me as <em><u>brainliest </u></em>

4 0

3 years ago