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3 years ago

Brad is 12 years older than Sam. If Brad were 8 years older than he is now, he would

Mathematics

1 answer:

MatroZZZ [7]3 years ago

5 0

Answer:

Sam is 20 years old.

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klio [65]

Answer:

Step-by-step explanation:

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2 years ago

87432 divided by 24 need help

Julli [10]

Answer:

3643

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the area of the triangle with the given

LekaFEV [45]

Answer:

616.2442

area to the nearest whole number=616

Step-by-step explanation:

using formula 1/2absinx

where a =44,b=29 ,x=105

1/2x44x29xsin105

44x29=1276

1276÷2=638

638 x sin 105

the sin of 105 is 0.9659

if u are using a four figure table where u can't find 105 under sin of angle

u simply subtract 105 from 180=75

638 x 0.9659 =616.2442

approx.616

6 0

3 years ago

I'm going to try this one more time...

irina [24]

Answer and Step-by-step explanation:

C and D is the answer.

It needs a lot of mass so that it can collapse on itself to form a black hole, and when it dies, it also collapses on itself, forming a black hole.

A is incorrect because the mass of our sun hasn't made it to collapse, and it still has some billions of years before the mass increases.

B is incorrect because the hydrogen needs to be depleted in order for the sun/star to start the nuclear process.

#teamtrees #PAW (Plant And Water)

3 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

1 year ago