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lidiya [134]
3 years ago
13

A. 3 (x + 2) = 18

Mathematics
1 answer:
Artyom0805 [142]3 years ago
8 0

Answer:

c

Step-by-step explanation:

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zzz [600]

Answer:

Step-by-step explanation:

3*10^10 / 6*10^10 = 0.5

The 10^10 cancel out. 3/6 = 0.5

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3 years ago
Catherine sketched the slope triangle on the previous slide. Notices that the slope triangle has a 2 and 4, so she thinks the sl
Liula [17]

Answer:

1/2

Step-by-step explanation:

3 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
4 years ago
Marvin and some of his classmates are painting their classroom at school. They only have gallon of paint left. How much of the p
RideAnS [48]
1/4 of a gallon or 1 quart each
3 0
3 years ago
Y = 3x+8 3x+ 2y =20 solving systems by substitution
konstantin123 [22]
Y = 3x + 8
3x + 2y = 20
STEP ONE:
Substitute the value for y
3x + 2(3x + 8) = 20
STEP TWO:
Multiply because of PEMDAS
3x + 6x + 16 = 20
9x + 16 = 20
STEP THREE:
Subtract 16 from 20
9x = 4
STEP FOUR:
Divide
9x/9 4/9
FINAL ANSWER:
x = 4/9
SOLVE FOR Y:
y = 3(4/9) + 8
y = 11 and 4/9
5 0
3 years ago
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