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raketka [301]
3 years ago
5

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

<u>Conclusion</u>

  • 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:

A

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)}

where m_i is the midpoint of the i-th subinterval,

m_i=\dfrac{\ell_i+r_i}2

Then the integral I is equal to the sum of the integrals of each interpolation over the corresponding i-th subinterval.

I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt

It's easy to show that

\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)

so that the value of the overall integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}

4 0
3 years ago
PLEASE HELP ME LINK IS BELOW
n200080 [17]

Answer:

55

Step-by-step explanation:

Since there is a right angle at the bottom, we know that x and angle 35 degrees form a right angle so 90 minus 35 is 55.

8 0
3 years ago
Read 2 more answers
According to the distributive property, 6 (a + b) =
lilavasa [31]
<span>6 (a + b) = 6a +  6b

answer is </span><span>B) 6a + 6b </span>
7 0
3 years ago
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