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BlackZzzverrR [31]
3 years ago
13

Is 2|12 true or false

Mathematics
1 answer:
Mademuasel [1]3 years ago
6 0

Answer:

true

Step-by-step explanation:

Because it is

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If you used 1 drop of your sample and 4 drops of diluent what is your dilution ratio?
frez [133]

Answer:

1/5

Step-by-step explanation:

1 + 4 = 5

1/5

7 0
3 years ago
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Grade 12<br>mathematics<br>(√98-√50)² ​
allsm [11]

Answer:

8

Step-by-step explanation:

Two different approaches:

<u>Method 1</u>

Apply radical rule √(ab) = √a√b to simplify the radicals:

√98 = √(49 x 2) = √49√2 = 7√2

√50 = √(25 x 2) = √25√2 = 5√2

Therefore,  (√98 - √50)² = (7√2 - 5√2)²

                                         = (2√2)²

                                         = 4 x 2

                                         = 8

<u>Method 2</u>

Use the perfect square formula: (a - b)² = a² - 2ab + b²

where a = √98   and   b = √50

So            (√98 - √50)² = (√98)² - 2√98√50 + (√50)²

                                      = 98 - 2√98√50 + 50

                                      = 148 - 2√98√50

Apply radical rule √(ab) = √a√b to simplify radicals:

√98 = √(49 x 2) = √49√2 = 7√2

√50 = √(25 x 2) = √25√2 = 5√2

Therefore,     148 - 2√98√50 = 148 - (2 × 7√2 × 5√2)

                                                 = 148 - 140

                                                 = 8

4 0
3 years ago
The following data points represent the number of classes that each teacher at Broxin High School teaches.
nika2105 [10]

Answer:

3

Step-by-step explanation:

4 0
3 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
3 years ago
Two forces with the magnitude of 100and 59 pounds act on an object at angles of 50° and 160°, respectively. Find the direction a
OLga [1]

Answer:

that is absolutely question my prother

8 0
2 years ago
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