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3 years ago

Please Help Me!!!!!!!

Mathematics

2 answers:

Svetradugi [14.3K]3 years ago

6 0

Answer:

-3/4

Step-by-step explanation:

Point (-4, 4), and point (0,1)

m = (y2 - y1)/ (x2 - x1) = (4 -1)/(-4-(0)) = 3/(-4) = - 3/4

soldier1979 [14.2K]3 years ago

6 0

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Linda purchased a birthday gift from Bed Bath and Beyond for her mother. After applying the

3241004551 [841]

Answer:

Before discount = $61.25

After discount = $49.00

Step-by-step explanation:

Given information:

Discount = $12.25
Coupon = 20%

The price of the gift before the discount is 100%.

If the $12.25 discount was 20% of the purchase price, then:

⇒ original price (before discount) = (12.25 / 20) × 100 = $61.25

To find the price after the discount, simply subtract the discount from the found original price:

⇒ $61.25 - $12.25 = $49.00

Conclusion

Before discount price = $61.25
After discount price = $49.00

5 0

2 years ago

find the difference of each number and 4. Complete the table to support your answers. The first example is provided. Number Subt

Gelneren [198K]

Answer:

Step-by-step explanation:

Just took the quiz

5 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.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$