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3 years ago

11

In Mr. Romero's class is 25 students, 13 of them participate in school sports. What percent of students participate in school sp

orts?

1 answer:

never [62]3 years ago

7 0

Just do this to find the percentage:

13/25 * 100%

= 0.52 *100%

= 52%

You might be interested in

X2<br> -<br> 1<br> In<br> Out<br> 5<br> 9<br> -3<br> -7<br> 7<br> -20

DerKrebs [107]

Answer:

|4| = |7|

this is the answer of this question

4 0

3 years ago

compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.

Nina [5.8K]

$\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$

We have been given two vectors $\vec{a}$ and $\vec{b}$ , we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of $\vec{b}$ onto $\vec{a}$ means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

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3 0

1 year ago

Graph the linear function. f(x) = -3x + 1

Anastaziya [24]

Here’s the graph. I hope this is right

6 0

3 years ago

Read 2 more answers

Graphing linear equations

ella [17]

Answer:

To ensure uniformity on an exam

Or

To test whether you can distinguish between the two formats

Step-by-step explanation:

Standard form is when a straight line equation is rearranged in the form:

$ax + by = c$

Therefore y=2x+4 in standard form is

$2x - y = - 4$

The slope-intercept form is when a a straight line equation is written in the form:

$y = mx + c$

where m is the slope and c is the y-intercept.

The given equation is

$y = - \frac{1}{2} x - 5$

This is already in slope-intercept form:

The standard form and slope-intercept forms are just formats.

Your instructor may restrict you to leave your answer in one of these formats maybe for uniformity on a test.

You may also decide to rewrite an equation in slope-intercept form, so that you can easily identify the slope and y-intercept easily for graphing purpose.

6 0

4 years ago

Point T(-9,5) lies on the perpendicular bisector of UV. If the

Flura [38]

The coordinate of point V is at (-16, 9)

If Point T(-9,5) lies on the perpendicular bisector of UV, this means that the point divides the line UV into two equal parts

Given the following coordinates

Midpoint T = (-9, 5)

U = (-2, 1)

Required

coordinate of point V

Using the midpoint formulas;

$T(x, y) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \\T(-9, 5) = (\frac{-2+x_2}{2}, \frac{1+y_2}{2})$

Get the value if x₂ and y₂

$-9 = \frac{-2+x_2}{2}\\-18 = -2+x_2\\x_2 = -18+2\\x_2 = -16\\$

Similarly;

$5 = \frac{1+y_2}{2}\\10 = 1+y_2\\y_2 = 10-1\\y_2 = 9\\$

Hence the coordinate of point V is at (-16, 9)

Learn more here: brainly.com/question/18049211

5 0

3 years ago