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3 years ago

Please help me solve problem #12 for my son

Mathematics

1 answer:

inna [77]3 years ago

8 0

There would be 36/12 if you were talking about 3 wholes but if you were just talking about 1/3 you would have 3/12

now for the example I don't know

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Mrs pender shows her students 2 sticks measuring 6 inches and 2 sticks measuring 8 inches. She asked the student to arrange the

MrMuchimi

Answer:

The answer is A got it correct on the test

Step-by-step explanation:

have a good day!

3 0

2 years ago

An object is thrown upward from the top of a 192 ft building with an initial velocity of 64 feet

Airida [17]

Answer:

t=6

Step-by-step explanation:

ground height = 0

(are you sure your formula is correct? isn't it - 16t²?)

if h=16t² +64t+192 is true then

16t² +64t+192 = 0

t² + 4t + 12 = 0

t = (-4 ± √(4² - 4*12)) / 2*1 = (-4 ± √-32) / 2 = -2 ± 2√-2

There is no solution of t

if it is h= - 16t² +64t+192

0 =- 16t² + 64t + 192

t² - 4t - 12 = 0

(t + 2) (t -6) = 0

t should be positive

t = 6 sec

7 0

2 years ago

2. Find the product of each of the following:

Kruka [31]

Step-by-step explanation:

$i) \: \frac{3}{5} \times ( \frac{ - 7}{8} ) = - \frac{21}{40} \\ \\ ii) \: \frac{ - 9}{2} \times \frac{5}{4} = - \frac{45}{8} \\ \\ iii) \: \frac{ - 6}{11} \times \frac{5}{3} = - \frac{10}{11} \\ \\ iv) \: \frac{ - 2}{3} \times \frac{6}{7} = - \frac{4}{7} \\$

6 0

2 years ago

100 points

lbvjy [14]

Answer:

Step-by-step explanation:

IQR is the difference between Q3 and Q1

<u>According to the top box plot:</u>

Q1 = 4
Q3 = 14

IQR = 14 - 4 = 10

4 0

2 years ago

Read 2 more answers

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

2 years ago