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3 years ago

Is it first or second? Please help me.

Mathematics

2 answers:

Rasek [7]3 years ago

6 0

Im pretty sure its first person

Colt1911 [192]3 years ago

3 0

This is in first person.

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Sketch the absolute value function f(x) = -|-2x - 3| + 5. and way to find domain and range

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3 years ago

Which expression will produce a negative product?

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3 years ago

Find the area of a triangle bounded by the y-axis, the line f(x)=9−4/7x, and the line perpendicular to f(x) that passes through

Setler79 [48]

<u>ANSWER: </u>

The area of the triangle bounded by the y-axis is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

<u>SOLUTION:</u>

Given, $f(x)=9-\frac{-4}{7} x$

Consider f(x) = y. Hence we get

$f(x)=9-\frac{-4}{7} x$ --- eqn 1

$y=9-\frac{4}{7} x$

On rewriting the terms we get

4x + 7y – 63 = 0

As the triangle is bounded by two perpendicular lines, it is an right angle triangle with y-axis as hypotenuse.

Area of right angle triangle = $\frac{1}{ab}$ where a, b are lengths of sides other than hypotenuse.

So, we need find length of f(x) and its perpendicular line.

First let us find perpendicular line equation.

Slope of f(x) = $\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}$

So, slope of perpendicular line = $\frac{-1}{\text {slope of } f(x)}=\frac{7}{4}$

Perpendicular line is passing through origin(0,0).So by using point slope formula,

$y-y_{1}=m\left(x-x_{1}\right)$

Where m is the slope and $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$

$y-0=\frac{7}{4}(x-0)$

$y=\frac{7}{4} x$ --- eqn 2

4y = 7x

7x – 4y = 0

now, let us find the vertices of triangle, one of them is origin, second one is point of intersection of y-axis and f(x)

for points on y-axis x will be zero, to get y value, put x =0 int f(x)

0 + 7y – 63 = 0

7y = 63

y = 9

Hence, the point of intersection is (0, 9)

Third vertex is point of intersection of f(x) and its perpendicular line.

So, solve (1) and (2)

$\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}$

Put x value in (2)

$\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}$

So, the point of intersection is $\left(\frac{252}{65}, \frac{441}{65}\right)$

Length of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right)$ and (0,9)

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}$

Now, length of perpendicular of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)$

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}$

Now, area of right angle triangle = $\frac{1}{2} \times \frac{252}{65} \times \frac{63}{65} \sqrt{65}$

$=\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

Hence, the area of the triangle is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

8 0

3 years ago

What's perimeter and area?

aleksklad [387]

Perimeter is the boundary/outline/ surrounding
For example if you measure around your house that would be the perimeter.
Area is the size of something.
Like if you wanted to know the amount of space in a box.

3 0

3 years ago

Suppose that at a state college, a random sample of 41 students is drawn, and each of the 41 students in the sample is asked to

ra1l [238]

Answer:

The confidence interval for 90% confidence would be narrower than the 95% confidence

Step-by-step explanation:

From the question we are told that

The sample size is n = 41

For a 95% confidence the level of significance is $\alpha = [100 - 95]\% = 0.05$ and

the critical value of $\frac{\alpha }{2}$ is $Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }= 1.96$

For a 90% confidence the level of significance is $\alpha = [100 - 90]\% = 0.10$ and

the critical value of $\frac{\alpha }{2}$ is $Z_{\frac{\alpha }{2} } =Z_{\frac{0.10 }{2} }= 1.645$

So we see with decreasing confidence level the critical value decrease

Now the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }$

given that other values are constant and only $Z_{\frac{\alpha }{2} }$ is varying we have that

$E\ \ \alpha \ \ Z_{\frac{\alpha }{2} }$

Hence for reducing confidence level the margin of error will be reducing

The confidence interval is mathematically represented as

$\= x - E < \mu < \= x + E$

Now looking at the above formula and information that we have deduced so far we can infer that as the confidence level reduces , the critical value reduces, the margin of error reduces and the confidence interval becomes narrower

3 0

3 years ago