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Blababa [14]
2 years ago
6

Identify the transformations needed to graph the cosine function y = –0.5cos(x) – 3 from the parent cosine function. Check all t

hat apply.
vertical compression by a factor of 0.5

reflection across the y-axis

vertical translation 3 units down

vertical stretch by a factor of 0.5

reflection across the x-axis

vertical translation 3 units up

vertical translation 0.5 units down
Mathematics
2 answers:
Ksenya-84 [330]2 years ago
7 0

1,3 and 5 on ed.genuity

Zinaida [17]2 years ago
4 0
Vertical translation 3 units down, vertical compression by a factor of 0.5, and r<span>eflection across the x-axis. This is because since the number -3 is outside of the parenthesis, it is a vertical translation and is since it is negative it is 3 down. If the number multiplied by the function is less than 1 it is a compression and if it is greater than 1 it is a stretch. And the negative represents the reflection across the x axis.
</span>
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Evaluate the line integral by the two following methods. xy dx + x2y3 dy C is counterclockwise around the triangle with vertices
nadezda [96]

Answer:

a)

\frac{2}{3}

b)

\frac{2}{3}

Step-by-step explanation:

a) The first part requires that we use line integral to evaluate directly.

The line integral is

\int_C xydx +  {x}^{2}  {y}^{3} dy

where C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2)

The boundary of integration is shown in the attachment.

Our first line integral is

L_1 = \int_ {(0,0)}^{(1,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is y=0, x varies from 0 to 1.

When we substitute y=0 every becomes zero.

\therefore \: L_1 =0

Our second line integral is

L_2 = \int_ {(1,0)}^{(1,2)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is:

x = 0 \implies \: dx = 0

y varies from 1 to 2.

We substitute the boundary and the values to get:

L_2 = \int_ {1}^{2}1 \cdot y(0) +  {1}^{2}   \cdot \: {y}^{3} dy

L_2 = \int_ {1}^2 {y}^{3} dy =  \frac{8}{3}

The 3rd line integral is:

L_3 = \int_ {(1,2)}^{(0,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is

y = 2x \implies \: dy = 2dx

x varies from 0 to 1.

We substitute to get:

L_3 = \int_ {1}^{0} x \cdot \: 2xdx +  {x}^{2}  {(2x)}^{3}(2 dx)

L_3 = \int_ {1}^{0} 8 {x}^{5}  + 2 {x}^{2} dx  =  - 2

The value of the line integral is

L = L_1 + L_2 + L_3

L = 0 +  \frac{8}{3}  +  - 2 =  \frac{2}{3}

b) The second part requires the use of Green's Theorem to evaluate:

\int_C xydx +  {x}^{2}  {y}^{3} dy

Since C is a closed curve with counterclockwise orientation, we can apply the Green's Theorem.

This is given by:

\int_C \: Pdx +Q  \: dy =  \int \int_ R \: Q_y -  P_x \: dA

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int \int_ R \: 3 {x}^{2}  {y}^{2}  -  y \: dA

We choose our region of integration parallel to the y-axis.

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \int_ 0^{2x}  \: 3 {x}^{2}  {y}^{2}  -  y \: dydx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  {x}^{2}  {y}^{3}  -   \frac{1}{2}  {y}^{2} |_ 0^{2x}  dx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  8{x}^{5} -  2 {x}^{2}   dx =  \frac{2}{3}

8 0
2 years ago
An animal shelter wants their ratio of dogs to cats to be 3:2. If the animal shelter has 78 dogs, how many cats should they have
OleMash [197]
The simplest way to do this is to set up equivalent fractions. So, you'd do:

\frac{3}{2} = \frac{78}{x} and then solve for x. 

x=52
4 0
2 years ago
on monday, it took 3 builders 5 1/2 hours to build a wall. an identical wall needs to be built on tuesday and 5 builders are ava
castortr0y [4]

Answer:

£29.37

Step-by-step explanation:

→ First step is to find the amount of hours it takes for 5 builders

\frac{3*\frac{11}{2} }{5} =\frac{33}{2} /5=\frac{33}{2} *\frac{1}{5} =\frac{33}{10} =3\frac{3}{10}

→ Now we know how long 5 builder takes we need to multiply the hourly rate by their time worked

3\frac{3}{10} *8.90=\frac{33}{10} *8.90=3.3*8.90 = 29.37

5 0
2 years ago
Read 2 more answers
Two trains leave New York at the same time heading in opposite directions. Train A travels at 4/5 the speed of train one. After
grandymaker [24]
<h3>Answer:  </h3><h3>speed of train A = 44 mph</h3>

=============================================

Work Shown:

x = speed of train A

y = speed of train B

"train A travels 4/5 the speed of train B" (I'm assuming "train one" is supposed to read "train B"). So this means x = (4/5)y

distance = rate*time

d = x*7

d = (4/5)y*7 = (28/5)y represents the distance train A travels

d = y*7 = 7y represents the distance train B travels

summing those distances will give us 693

(28/5)y + 7y = 693

5*(  (28/5)y + 7y ) = 5*693

28y + 35y = 3465

63y = 3465

y = 3465/63

y = 55

Train B's speed is 55 mph

4/5 of that is (4/5)y = (4/5)*55 = 4*11 = 44 mph

Train A's speed is 44 mph

8 0
2 years ago
Suppose that your business is operating at the 4.5-Sigma quality level. If projects have an average improvement rate of 50% annu
Luda [366]

Answer:

  11.75 years

Step-by-step explanation:

If we ignore the fact that "6-sigma" quality means the error rate corresponds to about -4.5σ (3.4 ppm) and simply go with ...

  P(z ≤ -6) ≈ 9.86588×10^-10

and

  P(z ≤ -4.5) ≈ 3.39767×10^-6

the ratio of these error rates is about 0.000290372. We're multiplying the error rate by 0.5 each year, so we want to find the power of 0.50 that gives this value:

  0.50^t = 0.000290372

  t·log(0.50) = log(0.00290372) . . . . take logarithms

  t = log(0.000290372)/log(0.50) ≈ -3.537045/-0.301030

  t ≈ 11.75

It will take about 11.75 years to achieve Six Sigma quality (0.99 ppb error rate).

_____

<em>Comment on Six Sigma</em>

A 3.4 ppm error rate is customarily associated with "Six Sigma" quality. It assumes that the process may have an offset from the mean of up to 1.5 sigma, so the "six sigma" error rate is P(z ≤ (1.5 -6)) = P(z ≤ -4.5) ≈ 3.4·10^-6.

Using that same criteria for the "4.5-Sigma" quality level, we find that error rate to be P(z ≤ (1.5 -4.5)) = P(z ≤ -3) ≈ 1.35·10^-3.

Then the improvement ratio needs to be only 0.00251699, and it will take only about ...

  t ≈ log(0.00251699)/log(0.5) ≈ 8.6 . . . . years

5 0
2 years ago
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