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kow [346]
3 years ago
14

Solve for x: 3x − 5 = 2x + 6. (1 point) Group of answer choices 1 −1 11 −11

Mathematics
1 answer:
Ivahew [28]3 years ago
4 0

Answer:

x = 11

Step-by-step explanation:

3x - 5 = 2x + 6

Add 5 to each side:

3x = 2x + 11

subtract 2x from each side:

x = 11

Substitute to check:

3x - 5 = 2x + 6

3 ( 11 ) - 5 = 2 ( 11 ) + 6

33 - 5 = 22 + 6

28 = 28

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3 nineteens, tripled
34kurt

Answer:

19 x 3 = 57

19 plus 19 plus 19 = 57

19 to the power of 3 squared = 6859

Step-by-step explanation:

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3 years ago
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If a line passes thru the points (4,5) and (2,9) the slope of this line is
Allisa [31]

Answer:

m=-2

Explanation:

If a line passes through the points (4,5) and (2,9): to calculate the slope, use the formula below:

\text{Slope,m}=\frac{Change\text{ in y-axis}}{Change\text{ in x-axis}}

Substituting the points, we have:

\begin{gathered} m=\frac{9-5}{2-4} \\ =\frac{4}{-2} \\ =-2 \end{gathered}

The slope of this line is​ -2.

8 0
1 year ago
Compare these two functions. Which<br> has the greater rate of change?<br> y= -6x + 7
Diano4ka-milaya [45]

Answer:

y = -6x + 7 has greater rate of change

Step-by-step explanation:

for the first one, the rate of change is -6x and for the second one, the rate of change is -5x so the rate of change for y = -6x + 7 is greater.

Use the formula y2 - y1 / x2 - x1 to find the slop for the  second equation. the second equation is y = -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.
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
What is the area of a wll lwith 8ft sq by 8ft sq
kotegsom [21]

Answer:

64 feet

Step-by-step explanation:

L × W= Answer

Therefore

8×8=64

6 0
3 years ago
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