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3 years ago

What is the equation of the line that passes through

1 answer:

7 0

Solve equation for y.

5x - y = 3

-y = -5x + 3

y = (-5x + 3)/(-1)

y = 5x - 3

The slope for this equation is 5.

The slope of the equation we want is the negative reciprocal of 5.

Let m = slope of equation we want.

m = -1/5

We are given the point (10,4).

Plug both into the point-slope formula.

y - 4 = (-1/5)(x - 10)

y - 4 = (-1/5)x + 2

y = (-1/5)x + 2 + 4

The equation we want is

y = (-1/5)x + 6

Did you follow?

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Choose the symbol that correctly compares the fractions below. 1/6 ? 4/5

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Answer:

The value of 1/6 is less than the value of4\5.

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Answer:

3 and 5 is the correct answer in this question

Step-by-step explanation:

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3 years ago

Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu

kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

$f(x) = {x}^{5} - {x}^{3} + 6$

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

$f'(x) = 5 {x}^{4} - 3 {x}^{2}$

At turning point f'(x)=0.

$5 {x}^{4} - 3 {x}^{2} = 0$

This implies that,

${x}^{2} (5 {x}^{2} - 3) = 0$

${x}^{2} = 0 \: or \: 5 {x}^{2} - 3 = 0$

$x = - \frac{ \sqrt{15} }{5} \: or \: x = 0 \: \: or \: x =\frac{ \sqrt{15} }{5}$

We plug this values into the original function to obtain the y-values of the turning points

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} +750)) \:and \: (0, - 6) \: and\: ( \frac{ \sqrt{15} }{5} , \frac{1}{125} ( - 6 \sqrt{15} +750))$