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2 years ago

For what values of $x$ is it true that $x^2 - 5x - 4 \le 10$? Express your answer in interval notation.

Mathematics

1 answer:

dybincka [34]2 years ago

5 0

Answer:

x ∈ [-2, 7]

Step-by-step explanation:

The given equation ...

x^2 -5x -4 ≤ 10

can be rewritten as ...

x^2 -5x -14 ≤ 0

and factored as ...

(x +2)(x -7) ≤ 0

Clearly, the "or equal to" condition will be met when x=-2 and x=7. For values of x between these numbers, one factor is negative and the other is positive. Hence the product will be negative. So, numbers in that interval are the solution set.

x ∈ [-2, 7]

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The point (Negative StartFraction StartRoot 2 EndRoot Over 2 EndFraction, StartFraction StartRoot 2 EndRoot Over 2 EndFraction)

sveticcg [70]

The values of cosine Ф and cotangent Ф are $\frac{-\sqrt{2} }{2}$ and -1

Step-by-step explanation:

When a terminal side of an angle intersect the unit circle at

point (x , y), then:

The x-coordinate is equal to cosine the angle between the positive part of x-axis and the terminal side
The y-coordinate is equal to sine the angle between the positive part of x-axis and the terminal side
If x and y coordinates are positive, then the angle lies in the 1st quadrant
If x-coordinate is negative and y-coordinate is positive, then the angle lies in the 2nd quadrant
If x and y coordinates are negative, then the angle lies in the 3rd quadrant
If x-coordinate is positive and y-coordinate is negative, then the angle lies in the 4th quadrant

∵ The terminal ray of angle Ф intersects the unit circle at point $(\frac{-\sqrt{2} }{2},\frac{\sqrt{2} }{2})$

- According to the 1st and 2nd notes above

∴ cosФ = x-coordinate of the point

∴ sinФ = y-coordinate of the point

∵ The x-coordinate of the point is negative

∵ They-coordinate of the point is positive

- According the the 4th note above

∴ Angle Ф lies in the 2nd quadrant

∵ x-coordinate = $\frac{-\sqrt{2} }{2}$

∴ cosФ = $\frac{-\sqrt{2} }{2}$

∵ y-coordinate = $\frac{\sqrt{2} }{2}$

∴ sinФ = $\frac{\sqrt{2} }{2}$

- cotФ is the reciprocal of tanФ

∵ tanФ = sinФ ÷ cosФ

∴ cotФ = cosФ ÷ sinФ

∴ cotФ = $\frac{-\sqrt{2} }{2}$ ÷ $\frac{\sqrt{2} }{2}$

∴ cotФ = -1

The values of cosine Ф and cotangent Ф are $\frac{-\sqrt{2} }{2}$ and -1

Learn more:

You can learn more about the trigonometry function in brainly.com/question/4924817

#LearnwithBrainly

4 0

2 years ago

Read 2 more answers

What is the ratio for tan (C)? (Hint: SOHCAHTOA)

Sever21 [200]

Answer:

D. 3/4

Step-by-step explanation:

tan(C) = opposite / adjacent = 3/4

4 0

2 years ago

Find a point on the curve x^3+y^3=11xy other than the origin at which the tangent line is horizontal.

attashe74 [19]

Compute the derivative dy/dx using the power, product, and chain rules. Given

x³ + y³ = 11xy

differentiate both sides with respect to x to get

3x² + 3y² dy/dx = 11y + 11x dy/dx

Solve for dy/dx :

(3y² - 11x) dy/dx = 11y - 3x²

dy/dx = (11y - 3x²)/(3y² - 11x)

The tangent line to the curve is horizontal when the slope dy/dx = 0; this happens when

11y - 3x² = 0

y = 3/11 x²

(provided that 3y² - 11x ≠ 0)

Substitute y into into the original equation:

x³ + (3/11 x²)³ = 11x (3/11 x²)

x³ + (3/11)³ x⁶ = 3x³

(3/11)³ x⁶ - 2x³ = 0

x³ ((3/11)³ x³ - 2) = 0

One (actually three) of the solutions is x = 0, which corresponds to the origin (0,0). This leaves us with

(3/11)³ x³ - 2 = 0

(3/11 x)³ - 2 = 0

(3/11 x)³ = 2

3/11 x = ³√2

x = (11•³√2)/3

Solving for y gives

y = 3/11 x²

y = 3/11 ((11•³√2)/3)²

y = (11•³√4)/3

So the only other point where the tangent line is horizontal is ((11•³√2)/3, (11•³√4)/3).

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1 year ago

Guys hi am i correct ??

Sloan [31]

Yes, you are correct.

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