Quadrilateral ABCD has the following coordinates: A(-6,3) B(-4,5) C(-1,6) and D(-3,2) and
maps onto A"B"C"D". It will undergo transformations that will consist of a reflection over
the y-axis followed by a translation. Point D" is shown plotted in the plane after the
transformation.

A) Write the translation rule that maps point D' to point D".

B) Write the coordinates of point A"

Answer:

x=0 y=0

x=1 y=5

x=2 y=8

x=3 y=9

x=4 y=8

x=5 y=5

x=6 y=0

Step-by-step explanation:

Answer:

Step-by-step explanation:

-2/3 * (-1/4) =

2/3 * 1/4 =

2/12 =

1/6

Answer:

y<_0

Step-by-step explanation:

The last answer.
Hope this helps! Have a great day :)

Answer:

it is an identity that was solved

Step-by-step explanation:

cosin - the a

Answer:

-13/20

Step-by-step explanation:

-0.65=-65/100

=13/20

Answer:

area of a triangle =1/2 x b x h

which is half base x hieght

Answer:

[tex]y=-5x-8[/tex]

Step-by-step explanation:

Given the following question:

To calculate the slope intercept of a line/point we need to use the formula for slope intercept, substitute, and solve for b.

Formula for slope intercept:
[tex]y=mx+b[/tex]
[tex]m=-5[/tex]
[tex]y=2[/tex]
[tex]x=-2[/tex]
[tex]2=-5(-2)+b[/tex]

Solve for b:
[tex]2=-5(-2)+b[/tex]
[tex]-5(-2)=10[/tex]
[tex]2=10+b[/tex]
[tex]10-10=0[/tex]
[tex]2-10=-8[/tex]
[tex]-8=b[/tex]
[tex]b=-8[/tex]

Substitute for the following:
[tex]y=mx+b[/tex]
[tex]b=-8[/tex]
[tex]m=-5[/tex]
[tex]y=-5x-8[/tex]

Your answer is "y = -5x - 8."

Hope this helps.

Using it's concept, it is found that the probability that the first student chosen is a senior and the second student chosen is a sophomore is given by:

[tex]p = \frac{11}{320}[/tex]

A probability is given by the number of desired outcomes divided by the number of total outcomes.

Researching this problem on the internet, we have that there are:

31 juniors, 10 sophomores, 17 juniors, 22 seniors.

Then:

The events are independent, hence the probability of both is given by:

[tex]p = p_1p_2 = \frac{11}{40} \times \frac{1}{8} = \frac{11}{320}[/tex]

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The probability that a player will be asked a math question and then a music question is 0.03125

The size of the sections are given as:

Music = 0.5

Sport = 0.5

Others = 1

So, the total size is:

Total = 0.5 + 0.5 + 1 + 1 + 1

Evaluate

Total = 4

The probability of asking a math question is:

P(Math) = 1/4 = 0.25

The probability of asking a music question is:

P(Music) = 0.5/4 = 0.125

The required probability is:

P(Math and Music) = P(Math) * P(Music)

This gives

P(Math and Music) = 0.25 * 0.125

Evaluate

P(Math and Music) = 0.03125

Hence, the probability that a player will be asked a math question and then a music question is 0.03125

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Answer: 25

Step-by-step explanation:

[tex]\frac{23+27}{2}=\boxed{25}[/tex]

Answer:

Does not lie

Step-by-step explanation:

The equation of the circle is

[tex] {(x + 7)}^{2} + {(y - 4)}^{2} = 25[/tex]

If we substitute x=-11, and y=2, we get that:

[tex] {( - 11 + 7)}^{2} + {(2 - 4)}^{2} = 25 \\ {4}^{2} + {2}^{2} = 25 \\ 20 = 25[/tex]

Which is false, meaning it does not lie on the circle

Answer:

1/2

Step-by-step explanation:

13 total fish

remove one fish

12 total fish

6 total clown fish

6/12 simplified is 1/2

= 17.50 or 25%

Answer:

The "what" is 100

Step-by-step explanation:

Rewrite this as 37.6y = 3760 and set out to determine y:

y = 3760/37.6  = 100

The "what" is 100

71% subtracted from 100% is 29%

Add or subtract the given percent from 100%.

71%

The percentage (p) is given as:

p = 71%

When subtracted from 100%, we have:

100% - p

Substitute p = 71%

100% - 71%

71 subtracted from 100 is 29.

So, we have:

100% - 71% = 29%

Hence, 71% subtracted from 100% is 29%

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Using translation concepts, it is found that graph 3 represents the function [tex]f(x) = \sqrt{x} + 1[/tex].

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, the function is:

[tex]f(x) = \sqrt{x} + 1[/tex]

Which is a shift up of one unit of [tex]\sqrt{x}[/tex]. The square root function has y-intercept at (0,0), hence the translated function will have y-intercept at (0,1). Additionally, [tex]\sqrt{4} = 2[/tex], since f(4) = 3, which means that graph 3 is represents the function.

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Answer:

1

Step-by-step explanation:

let x be the number to be added to 5 and 3

expressing the ratios in fractional form

[tex]\frac{5+x}{3+x}[/tex] = [tex]\frac{3}{2}[/tex] ( cross- multiply )

3(3 + x) = 2(5 + x) ← distribute parenthesis on both sides

9 + 3x = 10 + 2x ( subtract 2x from both sides )

9 + x = 10 ( subtract 9 from both sides )

x = 1

then the number to be added to both is 1

To calculate the cost of the fence, we have to find the length of the semi-circle. The cost of fencing the garden would be $328.52 approximately

To find the length of the fence, we have to calculate the circumference of the semi-circle and divide it by 2

[tex]C = 2\pi r\\l = \frac{c}{2} \\l = \pi r[/tex]

Data;

Let's substitute the values and calculate the length of the fence.

[tex]l = \pi r\\l = 3.14 * 13.5\\l = 42.39ft[/tex]

The length of the fence is equal to 42.39ft

The cost of fencing would be

[tex]42.39 * 7.75 = $328.5225[/tex]

The cost of fencing the garden would be $328.52 approximately

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The range of salary is (a) the range of salaries is from $38,500 to $39,800.

The given parameters are:

Average salary = 39,800

Vary = Less than $1300

Let the salary be x.

So, we have:

x = 39,800

When the salary varies, the equation becomes

x = 39800 - 1300

Evaluate

x = 38500

This means that the salary is from 38500 to 39800

Hence, the range of salary is (a) the range of salaries is from $38,500 to $39,800.

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Answer:

43.43 degrees

Step-by-step explanation:

This is a right triangle with the opposite LEG = 11 km and the hypotenuse = 16 km

sin (angle) = opposite/hypotenuse

                 = 11/16

arc sin (11/16) = angle = 43.43 degrees

Answer:

Step-by-step explanation:

2(6²)-3(6)+2=y

(2x36)-18+2=y

72-18+2=y

y=52 BODMAS

It may also be 56 tho im not sure if thats right

Put x in Equation

Answer:

6/11*5/10 = 3/11

Step-by-step explanation:

first we can get 6/11 which is the probability.

then we minus both sides by 1 to get 5/10

then we times them

Answer:

[tex]r = \dfrac{p}{2 + \pi}[/tex]

Step-by-step explanation:

[tex]~~~~~p = 2r +\pi r\\\\\implies p = r(2 + \pi)\\\\\implies r = \dfrac{p}{2 + \pi}[/tex]

Answer:

Quadrant 4

Answer is D........

Step-by-step explanation:

A = (1, 3)

B = (3, 6)

C = (9, 2)

D = (7, -1)

the distance between 2 points is given by the Pythagoras equation based on the coordinate differences as legs of virtual right-angled triangles.

AD for example we get from

AD² = (7-1)² + (-1 - 3)² = 6² + (-4)² = 36 + 16 = 52

AD = sqrt(52) = sqrt(4×13) = 2×sqrt(13)

and AB we get from

AB² = (3-1)² + (6-3)² = 2² + 3² = 4 + 9 = 13

AB = sqrt(13)

the perimeter of the given rectangle is

2×sqrt(52) + 2×sqrt(13) = 2×2×sqrt(13) + 2×sqrt(13) =

= 6×sqrt(13) = 21.63330765...

and the area of the rectangle is

2×sqrt(13)×sqrt(13) = 2×13 = 26

Answer:.2

Step-by-step explanation:.4/(.4+.3+.6+.7)

Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).

Salt flows in at a rate of

[tex]\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}[/tex]

and flows out at a rate of

[tex]\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}[/tex]

so that the net rate of change in the amount of salt in the tank is given by the linear differential equation

[tex]\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))[/tex]

Multiply both sides by the integrating factor, [tex]e^{t/180}[/tex], and rewrite the left side as the derivative of a product.

[tex]e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))[/tex]

[tex]\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))[/tex]

Integrate both sides with respect to t (integrate the right side by parts):

[tex]\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt[/tex]

[tex]\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C[/tex]

Solve for A(t) :

[tex]\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}[/tex]

The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.

[tex]\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}[/tex]

So,

[tex]\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}[/tex]

Recall the angle-sum identity for cosine:

[tex]R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)[/tex]

so that we can condense the trigonometric terms in A(t). Solve for R and θ :

[tex]R \cos(\theta) = -\dfrac{66,096,000}{32,401}[/tex]

[tex]R \sin(\theta) = \dfrac{367,200}{32,401}[/tex]

Recall the Pythagorean identity and definition of tangent,

[tex]\cos^2(x) + \sin^2(x) = 1[/tex]

[tex]\tan(x) = \dfrac{\sin(x)}{\cos(x)}[/tex]

Then

[tex]R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}[/tex]

and

[tex]\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)[/tex]

so we can rewrite A(t) as

[tex]\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}[/tex]

As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of

[tex]24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}[/tex]

and

[tex]24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}[/tex]

which is to say, with amplitude

[tex]2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}[/tex]