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jarptica [38.1K]
3 years ago
15

Can someone answer with steps and explanation? Thanks.

Mathematics
1 answer:
Vika [28.1K]3 years ago
8 0

Answer:

A dilation by a factor of three about Point T followed by a translation of two units downwards.

Step-by-step explanation:

When transforming functions, we will reflect/dilate the figure first and then translate it. This is directly from the order of operations.

Since we are trying to determine the transformation that was performed, we can try to map ΔS'T'U' onto ΔSTU. We can start by translating the figure and then determining any reflections/dilations.

First, we can translate ΔS'T'U' up two units to map T' onto T. This is represented by the black triangle in the image below. Let the black triangle be ΔS''T''U''. (T'' and T are the same point.)

Next, notice that from Point T'' to U'', we move nine units right and six units up.

From Point T to Point U, we move three units right and two units up.

Likewise, from Point T'' to S'', we move six units left and nine units up.

From Point T to Point S, we move two units left and three units up.

Therefore, to map ΔS''T''U'' onto ΔSTU, we dilate ΔS''T''U'' about Point T by a factor of 1/3.

Hence, by reversing the transformations, to acquire ΔS'T'U', we can see that we will dilate ΔSTU by a factor of three about Point T and then a perform a translation of two units downwards.

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7. Stan is serving lemonade at the school's dance out of no
Lilit [14]

Answer: 540 cups

Step-by-step explanation:

Diameter = 20

Radius = 20/2=10

Volume of a cylinder: pi(r^2)h = 3600pi

Cone = (pi(r^2)h)/3 = (20pi)/3

3600pi/(20pi/3) = 540 cups

7 0
3 years ago
Read 2 more answers
Y/x^2 + y/x^3<br><br>perform the operation and express your answer as a single fraction​
umka2103 [35]

Answer:

\frac{y}{x^2}+\frac{y}{x^3}=\frac{yx+y}{x^3}

Step-by-step explanation:

Given the expression

\frac{y}{x^2}+\frac{y}{x^3}

Let us perform the operation and express your answer as a single fraction​

\frac{y}{x^2}+\frac{y}{x^3}=\frac{x^3y+x^2y}{x^5}

            =\:\frac{x^2\left(xy+y\right)}{x^5}

            =\frac{\left(xy+y\right)}{x^3}

Therefore, we conclude that:

\frac{y}{x^2}+\frac{y}{x^3}=\frac{yx+y}{x^3}

3 0
3 years ago
Can someone please<br> Help me?
Mila [183]

9514 1404 393

Answer:

  see below

Step-by-step explanation:

It is easiest to compare the equations when they are written in the same form.

The first set can be written in slope-intercept form.

  y = 2x +7

  y = 2x +7 . . . . add 2x

These equations are <em>identical</em>, so have infinitely many solutions.

__

The second set can be written in standard form.

  y +4x = -5

  y +4x = -10

These equations <em>differ only in their constant</em>, so have no solutions.

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The third set is already written in slope-intercept form. The equations have <em>different slopes</em>, so have exactly one solution.

5 0
3 years ago
A cup of coffee at 93 degrees Celsius is placed in a room at 23 degrees Celsius. Suppose that the coffee cools at a rate of 1 de
Alinara [238K]

Answer:

Step-by-step explanation:

The rate of change of temperature of the coffee with respect to time is expressed by the differential equation

dT/dt=k(T−A) when k = -1 and A = 23. The equation will become:

dT/dt = -(T-23)

If the coffee cools at the rate of 1°C per minute, then dT/dt = -1 and T = 70

Substituting into the equation:

-1 = k(70-23)

-1 = 47k

k = -1/47

k = -0.0213

Substituting k = -0.0213 into the original equation, the differential equation will be:

dT/dt=k(T−A)

dT/dt = -0.0213(T-23)

dT/dt = -0.0213T+0.489

To get the value of T, we will use variable separable method

dt = dT/-0.0213T+0.489

Integrate both sides

t = -0.0213ln(-0.0213T+0.489)

At t = 1 minute

1 = -0.0213ln(-0.0213T+0.489)

1/-0.0213 = ln(-0.0213T+0.489)

-46.95 = ln(-0.0213T+0.489)

Apply exp to both sides

e^-46.95 = e^ln(-0.0213T+0.489)

4.073×10^-21= -0.0213T+0.489

-0.0213T = 4.073×10^-21-0.489

-0.0213T = -0.489

T = 0.489/0.0213

T = 22.96°C

5 0
3 years ago
Help me please thank you
podryga [215]
The answer is 79 im sure of it

8 0
3 years ago
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