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3 years ago

Simplify your answer and write it as a proper fraction improper fraction or whole number tan(b)

Mathematics

1 answer:

storchak [24]3 years ago

4 0

Answer:

where is the answer that needs to be simplified

Step-by-step explanation:

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X>-3 ; (-4,3) Tell wether this is the solution of the inequality.

IceJOKER [234]

Not sure what you mean by solution since that’s already simplified and inequalities don’t hav coordinate points as solution.
But -4<-3 so that point wouldn’t go into the “shaded area” of that inequality.
Hope this helps

3 0

3 years ago

Can some one help me find the 5 coordinates that's it. And I'll plug them into my equation thanks c:

avanturin [10]

(0,0), (-4,-2), (-4,3), (-9,-3), (-9,3)

7 0

3 years ago

Read 2 more answers

What is the value of y?

tatuchka [14]

X = 8
y = 40

the two base angles are equal to each other so you would do 5x = 40 and solve for x

a triangle has 180 total degrees. 40*2 = 80
180 - 80 = 100
2y + 20 = 100
2y = 80
y = 40

4 0

3 years ago

For the following integral, find the approximate value of the integral with 4 subdivisions using midpoint, trapezoid, and Simpso

PIT_PIT [208]

Answer:

$\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}$

$\textsf{Trapezium rule}: \quad \pi$

$\textsf{Simpson's rule}: \quad \dfrac{4 \pi}{3}$

Step-by-step explanation:

Midpoint rule

$\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

Trapezium rule

$\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}$

Simpson's rule

$\displaystyle \int_{a}^{b} y \:\:\text{d}x \approx \dfrac{1}{3}h\left(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_{n-2}+4y_{n-1}+y_n\right)\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

Given definite integral:

$\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x$

Therefore:

a = 0
b = 2π

Calculate the subdivisions:

$\implies h=\dfrac{2 \pi - 0}{4}=\dfrac{1}{2}\pi$

Midpoint rule

Sub-intervals are:

$\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right]$

The midpoints of these sub-intervals are:

$\dfrac{1}{4} \pi, \dfrac{3}{4} \pi, \dfrac{5}{4} \pi, \dfrac{7}{4} \pi$

Therefore:

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}$

Trapezium rule

$\begin{array}{| c | c | c | c | c | c |}\cline{1-6} &&&&&\\ x & 0 & \dfrac{1}{2}\pi & \pi & \dfrac{3}{2} \pi & 2 \pi \\ &&&&&\\\cline{1-6} &&&&& \\y & 0 & 1 & 0 & 1 & 0\\ &&&&&\\\cline{1-6}\end{array}$

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}$

Simpson's rule

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}$

6 0

2 years ago

What is the area of the obtuse angle ?

Zolol [24]

Answer:C = 49 sq. units

Step-by-step explanation:

Multiply 7 by 14 to get 98

Then divide 98 by 2 to get that obtuse triangle's area; 49 sq. units

4 0

3 years ago