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USPshnik [31]
2 years ago
12

The price of mailing a small package is 0.32 $ for the 1st ounce and $0.21 for each additional ounce. Sandra put $1.16 to mail h

er package. How much did it weigh?
A. 4 ounces B. 5 ounces C. 6 ounces D. 7 ounces
Mathematics
2 answers:
ELEN [110]2 years ago
4 0
So you would subtract 0.32 from 1.16 because it is the initial amount. Then divide what you get from that (0.84) by .21. You're gonna get .04- but that's not the answer because you have to add the initial ounce that is worth $0.32. So The answer is B
Neporo4naja [7]2 years ago
3 0
The answer would be B because you would first subtract 0.32 from 1.16 so that would equal to one ounce and for the remaining of 1.16 you would divide or subtract by 0.21
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I think the answer is 12
8 0
2 years ago
What is the slope of the line given by the following equation? Give an exact number.  -13x 5y=-7
Black_prince [1.1K]
The slope should be 13/5x
8 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
The two figures shown are congruent. Which statement is true?
zimovet [89]

9514 1404 393

Answer:

  (b) One figure is a rotation image of the other

Step-by-step explanation:

The axis of symmetry is horizontal for the left figure, and vertical for the right figure. The right figure has been rotated 90° CW from the left figure.

6 0
3 years ago
Which value must be added to the expression x2 – 3x to make it a perfect-square trinomial?
Vitek1552 [10]

Answer:

\frac{9}{4}

Step-by-step explanation:

Given

x² - 3x

To make the expression a perfect square

add ( half the coefficient of the x- term )²

x² + 2( - \frac{3}{2} )x + \frac{9}{4}, thus

x² - 3x + \frac{9}{4}

= (x - \frac{3}{2})² ← a perfect square

7 0
2 years ago
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