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3 years ago

Given the graph above find an equation for the graph and write your anwser in slope intercept form

Mathematics

1 answer:

lys-0071 [83]3 years ago

5 0

Given:

Given that the graph of a linear function.

We need to determine the equation of the graph in slope intercept form.

Slope:

Consider any two coordinates from the equation of the line to determine the slope.

Let the two coordinates are (-4,3) and (4,-1)

The slope can be determined using the formula,

$m=\frac{y_2-y_1}{x_2-x_1}$

Substituting the coordinates in the above formula, we get;

$m=\frac{-1-3}{4+4}$

$m=\frac{-4}{8}$

$m=-\frac{1}{2}$

Thus, the slope of the equation is $m=-\frac{1}{2}$

Value of y - intercept:

The value of y - intercept is the value of y when x = 0.

Hence, from the graph, it is obvious that the value of y when x = 0 is 1.

Hence, the value of y - intercept is 1.

Thus, y - intercept is $b=1$

Equation of the line:

The equation of the line can be determined using the formula,

$y=mx+b$

Substituting the slope $m=-\frac{1}{2}$ and the y - intercept $b=1$

Thus, we get;

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

Thus, the equation of the line is $y=-\frac{1}{2}x+1$

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Use trigonometric substitution to evaluate the integral 13 + 12x − x2 dx . First, write the expression under the radical in an a

SCORPION-xisa [38]

I don't see a square root sign anywhere, so I'll assume the integral is

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx$

First complete the square:

$13+12x-x^2=49-(6-x)^2=7^2-(6-x)^2$

Now in the integral, substitute

$6-x=7\sin t\implies\mathrm dx=-7\cos t\,\mathrm dt$

so that

$t=\sin^{-1}\left(\dfrac{6-x}7\right)$

Under this change of variables, we have

$7^2-(6-x)^2=7^2-7^2\sin^2t=7^2(1-\sin^2t)=7^2\cos^2t$

so that

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx=-7\int\sqrt{7^2\cos^2t}\,\cos t\,\mathrm dt=-49\int|\cos t|\cos t\,\mathrm dt$

Under the right conditions, namely that cos(t) > 0, we can further reduce the integrand to

$|\cos t|\cos t=\cos^2t=\dfrac{1+\cos(2t)}2$

$\displaystyle-49\int|\cos t|\cos t\,\mathrm dt=-\frac{49}2\int(1+\cos(2t))\,\mathrm dt=-\frac{49}2\left(t+\frac12\sin(2t)\right)+C$

Expand the sine term as

$\dfrac12\sin(2t)}=\sin t\cos t$

Then

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \sin t=\dfrac{6-x}7$

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \cos t=\sqrt{7^2-(6-x)^2}=\sqrt{13+12x-x^2}$

So the integral is

$\displaystyle-\frac{49}2\left(\sin^{-1}\left(\dfrac{6-x}7\right)+\dfrac{6-x}7\sqrt{13+12x-x^2}\right)+C$

3 0

3 years ago

I need help please!!

mina [271]

Step-by-step explanation:

$\frac{ - 1}{5} \times \frac{4}{7}$

$\: \: \: \: \: \frac{ - 4}{35}$

$\frac{ - 1}{2} \times \frac{4}{5}$

$\frac{ - 2}{5}$

$\frac{ - 3}{2} \times \frac{7}{ - 10}$

$\frac{21}{20}$

$\frac{1}{2} \times \frac{7}{8}$

$\frac{7}{16}$

$\frac{ - 9}{5} \times \frac{1}{2}$

$\frac{ - 9}{10}$

$\frac{ - 32}{9} \times \frac{1}{3}$

$\frac{ - 32}{27}$

$\frac{ - 2}{1} \div \frac{ - 19}{5}$

$\: \: \: \: \: \: \frac{ - 2}{1} \times \frac{5}{ - 19}$

$\: \: \: \: \frac{ - 10}{19}$

$\frac{1}{9} \div \frac{ - 4}{3}$

$\frac{1}{9} \times \frac{3}{ - 4}$

$\frac{ - 1}{12}$

$\frac{13}{7} \div \frac{23}{4}$

$\frac{13}{7} \times \frac{4}{23}$

$\frac{52}{161}$

$\frac{ - 37}{10} \div \frac{9}{4}$

$\frac{ - 37}{10} \times \frac{4}{9}$

$\frac{ - 74}{45}$

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3 years ago

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tankabanditka [31]

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Now that all the numbers have the same denominator, you would add them to get 41/60.
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3 years ago

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vova2212 [387]

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3 years ago

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svlad2 [7]

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For example, 3/4=6/8 because 3x2=6 and 4x2=8, 6/8.

Hope this helps!

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2 years ago

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