Which graph represents the solution to this inequality?

Mathematics

1 answer:

morpeh [17]3 years ago

7 0

In the second to last step, you end up with -x ≤ 13, but x can't be negative, so the sign is flipped, turning x positive and 13 negative, leaving us with x ≥ -13, which is why the line goes right.

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Which of the following is equal to the expression below?

I am Lyosha [343]

Answer:

cuteboy979409

Step-by-step explanation:

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UkoKoshka [18]

Answer:

5/11

Step-by-step explanation:

<u>Corresponding side have same ratio:</u>

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Anettt [7]

This problem can be readily solved if we are familiar with the point-slope form of straight lines:
y-y0=m(x-x0) ...................................(1)
where
m=slope of line
(x0,y0) is a point through which the line passes.

We know that the line passes through A(3,-6), B(1,2)

All options have a slope of -4, so that should not be a problem. In fact, if we check the slope=(yb-ya)/(xb-xa), we do find that the slope m=-4.

So we can check which line passes through which point:

a. y+6=-4(x-3)
Rearrange to the form of equation (1) above,
y-(-6)=-4(x-3) means that line passes through A(3,-6) => ok

b. y-1=-4(x-2) means line passes through (2,1), which is neither A nor B
****** this equation is not the line passing through A & B *****

c. y=-4x+6 subtract 2 from both sides (to make the y-coordinate 2)
y-2 = -4x+4, rearrange
y-2 = -4(x-1)
which means that it passes through B(1,2), so ok

d. y-2=-4(x-1)
this is the same as the previous equation, so it passes through B(1,2),
this equation is ok.

Answer: the equation y-1=-4(x-2) does NOT pass through both A and B.

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Answer:

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Step-by-step explanation:

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Find the critical numbers of the function f(x) = x6(x − 1)5 what does the first derivative test tell you that the second derivat

patriot [66]

The given function is f(x) = x⁶(x-1)⁵

The first derivative is
f'(x) = 6x⁵(x-1)⁵ + 5x⁶(x-1)⁴
= x⁵(x-1)⁴(6x - 6 + 5x)
= x⁵(x-1)⁴(11x - 6)
The critical values are the zeros of f'(x). They are
x = 0, 6/11, and 1.
The critical values indicate that turning points exist for f(x) at the critical points. However, we do not know the nature of the turning points.

Write the first derivative in the form f'(x) = (x-1)⁴(11x⁶ - 6x⁵).
The second derivative is
f''(x) = 4(x-1)³(11x⁶ - 6x⁵) + (x-1)⁴(66x⁵ - 30x⁴)
= (x-1)³(44x⁶ - 24x⁵ + 66x⁶ - 30x⁵ - 66x⁵ + 30x⁴)
= (x-1)³(110x⁶ - 120x⁵ + 30x⁴)
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The sign of f''(x) at the critical values tell us the nature of the turning point.

f''(0) = 0, therefore a point of inflection exists at x = 0.
f''(6/11) > 0, therefore a local minumum exists at x = 6/11.
f''(1) = 0, therefore a point of inflection exists at x=1.

The graphs shown below confirm these results.

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3 years ago