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3 years ago

Write the equation that describes the line with slope = 1 and y-intercept= 5/2 in slope-intercept form.

Mathematics

1 answer:

Nonamiya [84]3 years ago

3 0

Answer: y=x+5/2

Step-by-step explanation:

using the slope intercept formula, y=mx+b where m is the slope and b is the y-intercept, you can substitute the two values to get y=x+5/2. Please note that the slope, m, is one so it's not shown.

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What set/sets of real numbers does -54/19 belong to?

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That is a rational number since you can write it as a ratio (=fraction).

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4 years ago

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Ray Of Light [21]

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3 years ago

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Please help and show work :)

Mila [183]

Answer:

MK = 14

Step-by-step explanation:

MK = ML + LK , substitute values

4x - 2 = 8 + x + 2

4x - 2 = x + 10 ( subtract x from both sides )

3x - 2 = 10 ( add 2 to both sides )

3x = 12 ( divide both sides by 3 )

x = 4

Thus

MK = 4x - 2 = 4(4) - 2 = 16 - 2 = 14

8 0

3 years ago

Find the equation of the tangent line to the curve when x has the given value.

Leviafan [203]

Answer:

14) The equation of the tangent line to the curve $f(x)=-2-x^2$ at x = -1 is $y=2x-1$

15) The rate of learning at the end of eight hours of instruction is $w'(8) = 5 \frac{items}{hour}$

Step-by-step explanation:

14) To find the equation of a tangent line to a curve at an indicated point you must:

1. Find the first derivative of f(x)

$f(x)=-2-x^2\\\\\frac{d}{dx} f(x)=\frac{d}{dx}(-2-x^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx} f(x)=\frac{d}{dx}\left(-2\right)-\frac{d}{dx}\left(x^2\right)\\\\f'(x)=-2x$

2. Plug x value of the indicated point, x = -1, into f '(x) to find the slope at x.

$f'(-1)=-2(-1)=2$

3. Plug x value into f(x) to find the y coordinate of the tangent point

$f(-1)=-2-(-1)^2=-3$

4. Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line

$y-y_1=m(x-x_1)\\y+3=2(x+1)\\y=2x-1$

5. Graph your function and the equation of the tangent line to check the results.

15) To find the rate of learning at the end of eight hours of instruction you must:

1. Find the first derivative of f(x)

$w(t)=15\sqrt[3]{t^2} \\\\\frac{d}{dt}w= \frac{d}{dt}(15\sqrt[3]{t^2})\\\\w'(t)=15\frac{d}{dt}\left(\sqrt[3]{t^2}\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\f=\sqrt[3]{u},\:\:u=\left(t^2\right)\\\\w'(t)=15\frac{d}{du}\left(\sqrt[3]{u}\right)\frac{d}{dt}\left(t^2\right)\\\\w'(t)=15\cdot \frac{1}{3u^{\frac{2}{3}}}\cdot \:2t\\\\\mathrm{Substitute\:back}\:u=\left(t^2\right)$

$w'(t)=15\cdot \frac{1}{3(t^2)^{\frac{2}{3}}}\cdot \:2t\\w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}$

2. Evaluate the derivative a t = 8

$w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}\\\\w'(8)=\frac{10\cdot 8}{\left(8^2\right)^{\frac{2}{3}}}\\\\=\frac{80}{\left(8^2\right)^{\frac{2}{3}}}\\\\\left(8^2\right)^{\frac{2}{3}}=16\\\\=\frac{80}{16}\\\\w'(8) = 5 \frac{items}{hour}$

The rate of learning at the end of eight hours of instruction is $w'(8) = 5 \frac{items}{hour}$

7 0

4 years ago

Please find the length x.

Feliz [49]

It should be 6...

Notice how the triangles are the exact same shape.

The side 2.5 is 1 less than 3.5

Therefore x should be 1 less than 7 which is 6

7 0

3 years ago