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3 years ago

How to solve 9 * 4 - 8 / 2 *4

Mathematics

2 answers:

egoroff_w [7]3 years ago

8 0

Answer: the answer is 20

Step-by-step explanation:

Use order of operations and divide 8 by 2 and you get 4. Then use the PEMDAS (parentheses,exponents,multiplacation,division,addition, subtraction) method. Go left to right. And solve. :) I hope this helps

r-ruslan [8.4K]3 years ago

3 0

You follow the rules of order of operations multiplication and division the add or subtract in this case start from the left do 9 times 4 which is 36 then 8 divided by 2 times 4 once you get that then subtract those numbers hope that helps

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480kg of concrete mix. sand and gravel in the ratio 1:3:4

Anestetic [448]

Answer:

Step-by-step explanation:

1+3+4 = 8

480 / 8 = 60

60 x 1 = 60 (concrete)

60 x 3 = 180 (sand)

60 x 4 = 240 (gravel)

60:180:240

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3 years ago

F(x) = 4x2 – 3x + 5x3 – 7 . Decide if the function is a polynomial

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3 years ago

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3 years ago

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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

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

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1 year ago

How many times does 15 go into 46

Irina18 [472]

15 goes into 46 3 times

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3 years ago

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