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2 years ago

Chloe earned a grade of 87% on her multiple choice math final that had a total of 200

Mathematics

2 answers:

JulijaS [17]2 years ago

7 0

200 x 0.87 = 174

She answered 174 questions correctly.

7nadin3 [17]2 years ago

3 0

Answer:

Chloe answered <u>174</u> problems correctly.

Explanation:

To find percentage, you would multiply the percentages decimal to the total.

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Find the equation of the linear function represented by the table below in the slope-intercept form.

MAXImum [283]

Answer:

y=x-32

Step-by-step explanation:

y=mx+b where m is the slope and b is the y-intercept.

slope=change in y/change in x

=5/5

y=5x+b --> choose a random point

(7,3) ---> 3=5(7)+b=35+b

b= -32

y=x-32

6 0

2 years ago

Read 2 more answers

Estimate the area under the graph of f(x) = 2 cos(x) from x = 0 to x = π/2 using four approximating rectangles and right and lef

I am Lyosha [343]

So hmm notice the picture below

you're simply being asked to, approximate the area, by using 4 rectangles, like in the graph, it doesn't have to be an exact value, just an approximation, based on the width and height of the rectangles, and getting their area by that

those rectangles are 0.39 radian long, since if you split π/2 in 4 pieces, you get around that much, the last rectangle there is a bit smaller than 0.39.... but they don't have to be 4 rectangles of equal length

so, get the area of the rectangles, based on their width and height, and sum them up

7 0

2 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$