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2 years ago

13

Mrs. Harris took 6 oranges out of the refrigerator and cut them into wedges, each wedge represents 1/6 of the entire orange. Her

children at 3/4 of the wedges.
A) how many wedges did her children eat

B) how many wedges are left?

1 answer:

lesantik [10]2 years ago

7 0

A = 27
B = 9

Total orange = 6
One orange wedge = 6
Total wedge = 6*6 = 36
36 / 4 = 9
9*3 = 27
36-27 = 9
If it is helpful t

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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

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ICE Princess25 [194]

<h3>Answer</h3>

$\dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)$

<h3>Explanation</h3>

By the product rule $(d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$ , we have

$\begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}$

By the chain rule:

$\begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}$

By the power rule:

$(x^3)' = 3x^2$

thus

$\begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}$

Nothing to do to simplify any further, other than factoring out $x^2$ .

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