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Travka [436]
3 years ago
12

What is the slope of the line represented by the equation 3x+2y=6?

Mathematics
2 answers:
fiasKO [112]3 years ago
8 0
Make it into y = mx + b
2y = -3x + 6
Divide by 2
Y = -3/2x + 3
The slope is -3/2
spayn [35]3 years ago
7 0

SLOPE: negative slope

ANSWER: x = -2/3y + 2

hope this helped

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If g(x) = 1 - ¾ x <br> find: -5g(4) - 1
cricket20 [7]

Answer:

<u>-</u><u>5</u><u>g</u><u>(</u><u>4</u><u>)</u><u> </u><u>-</u><u> </u><u>1</u><u> </u><u>is</u><u> </u><u>9</u>

Step-by-step explanation:

g(x) = 1 -  \frac{3}{4} x

when x is 4:

g(4) = 1 -  \frac{3}{4}  \times 4 \\  = 1 - 3 \\  =  - 2

therefore:

- 5g(4) - 1 =  - 5( - 2) - 1 \\  = 10 - 1 \\  = 9

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3 years ago
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Solve the system of linear equations by graphing
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The answer to this problem is (x,y) = (-4,2)
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3 years ago
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1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

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Damm [24]

the standard form of a quadratic formula is

y = ax^2 + bx + c

in this case you will solve using foil method

(× - 4)(x + 3)

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<em>x</em><em>^</em><em>2</em><em> </em><em>+</em><em> </em><em>3x</em><em> </em><em>-</em><em> </em><em>4x</em><em> </em><em>-12</em>

<em>x</em><em>^</em><em>2</em><em> </em><em>-</em><em> </em><em>x</em><em> </em><em>-</em><em> </em><em>1</em><em>2</em>

<em>therefore</em><em> </em>

<em>y</em><em> </em><em>=</em><em> </em><em>x^</em><em>2-</em><em> </em><em>x</em><em> </em><em><u>-</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u> </u></em>

5 0
3 years ago
Pls help ASAP due tonight math
g100num [7]

Answer:

The equation of the line is y=1.4x+6.8

Step-by-step explanation:

Given 2 points on the graph, you can find the slope using the equation: m=\frac{y2-x2}{y1-x1} . The 2 points are (-7, -3) and (-2, 4). X1 is -7, X2 is -2, Y1 is -3, Y2 is 4. The slope would be 1.4 once you plug in the numbers to the equation to finding the slope. Once you find the slope, use one of the points to find the y-intercept along with the slope you found and the y-intercept will be 6.8.

4 0
3 years ago
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