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2 years ago

Help on 22 and 23 please

Mathematics

1 answer:

hram777 [196]2 years ago

4 0

Answer:

22.

R= -8, 15

S= -5, 9

T= -8,9

23.

J = 4, 4

K = 4, 3

L = 1, 1

M = 1, 4

Step-by-step explanation:

rotation of -90= y, -x

so R (-7, -5) = -5,7

S (-1, -2) = -2, 1

T (-1, -5) = -5, 1

now we translate 3 left (x-3) and 8 up (y+8)

R (-5, 7) = -8, 15

S (-2, 1) = -5, 9

T (-5, 1) = -8, 9

reflection of x-axis= x, -y

J (-4, 4) = -4, -4

K (-3, 4) = -3, -4

L (-1, 1) = -1, -1

M (-4, 1) = -4, -1

Then a 180° rotation is (-y, -x)

J (-4, -4) = 4, 4

K (-3, -4) = 4, 3

L (-1, -1) = 1, 1

M (-4, -1) = 1, 4

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Perform the indicated operation. (-7) • (-5)<br> -12 <br> 35<br> 12<br> -35

Flauer [41]

Hi,

Equation:

$- 7 \times (- 5)$

Multiplying two negatives equals a positive.

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Now equation looks like:

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r3t40

6 0

2 years ago

Read 2 more answers

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

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