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3 years ago

What is 2x2? thank you.

Mathematics

2 answers:

diamong [38]3 years ago

6 0

The answer is uhmm let me see 2 4 6 8 10 12 14 16 18 20. Oh the answer is 4

IrinaVladis [17]3 years ago

5 0

Step-by-step explanation:

it's 4 bro ;) Thanksss

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Three 24-hour clocks show the time to be 12 noon. One of the clocks is always correct one loses a minute every 24 hours and one

Ganezh [65]

There is one clock that shows the right time so we do not have to worry about the one which is always correct.

Talking about the second clock that loses a minutes in every 24 hours (or in a day), so after 60 days (since it has lost 60 minutes because it is losing 1 minute everyday) it will show 11:00 a.m when it is exactly the noon.

So this way, in total it will take $60\times 24=1440$ days before it shows the correct noon.

Now, the third clock gains a minute every 24 hours (or in a day) , after 60 days (when it has gained 60 minutes or a complete hour) it will show 1:00 p.m when it is exactly the noon.

This way, it will take $60\times 24=1440$ days (since it has gained a minute everyday) when it shows the correct noon.

Therefore, it will take 1440 days before all the three clocks show the correct time again.

5 0

4 years ago

A number n plus 7 is less than or equal to 9

devlian [24]

N ≤ 2, if n plus 7 is less than or equal to 9 (n+7 ≤ 9) then carry the 7 over to create n ≤ 9-7 giving the answer n ≤ 2 (n less than or equal to 2)

7 0

3 years ago

Help plzzzzzzzzzzzzzzzz

agasfer [191]

Answer:

One pound of raw chicken will serve approximately four people. or 12 ounces

so A

Step-by-step explanation:

5 0

3 years ago

Experts/ace/geniuses helpp

Alex Ar [27]

Square root 144 = 12 so this is the rational number

8 0

4 years ago

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

3 years ago