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Lelu [443]
3 years ago
7

Kayla knows that 1/16 means "1 divided by 16." She uses this to find the decimal equivalent for 116. Enter a digit into each box

to complete her work. CLEAR CHECK 0 . | 16 1 . 0 0 0 0 – 0 1 0 0 – 9 6 0 – 3 2 8 0 – 0

Mathematics
1 answer:
lesantik [10]3 years ago
6 0

Answer:

0.0625

Step-by-step explanation:

We are given that a number

\frac{1}{16}

We have to find the decimal equivalent for \frac{1}{16}

To find the decimal equivalent to given fraction we will use division method.

By using  division method

We get that

\frac{1}{16}=0.0625

In division ,first box filled by

Second box filled by 8

In quotient

First box filled by 0

Second box filled by 6

Third box filled by 2

Fourth box filled  by 5.

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–1 =n + 5 <br> PLSS!!!! HELP!! I DONT KNOW THIS ONE!
navik [9.2K]

I guess the would help you

8 0
2 years ago
If the sum of three consecutive multiples of 7 is 945. Find the smallest among them​
Firlakuza [10]

consider first number is = x

formula,

(n / 2)(first number + last number) = sum,

7/2(x+x+6)= 945

2x+6= 945×2/7

2x+6 = 270

2x= 270-6

x=264/2= 132

Answer the smallest number is 132.

8 0
3 years ago
I need help (please show your work)
oksano4ka [1.4K]

Answer:

the solution is (-4, 3)

Step-by-step explanation:

if you were to graph y = -7/4x - 4 you could plot your first point at (0, -4) and get your second point by going up 7 units (from (0, -4) and to the left by 4

if you were to graph y = -1/4x + 2 you could plot your first point at (0, 2) and get your second point by going up 1 unit (from (0, 2) and to the left by 4

the second point plotted for each equation will be the point of intersection, (-4, 3)

6 0
3 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
Help me plzplzplzplz1
mezya [45]

Answer:

The answer is 400 when you round it off to the nearest hundred.

4 0
3 years ago
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