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deff fn [24]
3 years ago
5

Help help help help help help help help help help help

Mathematics
1 answer:
oee [108]3 years ago
4 0

Answer:

A' 2,2

B' 3, -1

C' -1,0

Step-by-step explanation:

Because you're translating it by -2, 3, you're basically subtracting 2 from the x value and adding 3 to the y value.

A = 4, -1

A' = 2, 2

B = 5, -4

B' = 3, -1

C = 1, -3

C' = -1, 0

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Anuta_ua [19.1K]

They have 310 stickers in total

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3 years ago
How to find x please help
o-na [289]
You know the adjacent and you need to find the opposite, so you would use tangent. Your equation would look like this; tan 26°= x/28. You would need to multiply both sides by 28 to simplify it to this; 28*tan 26°= x
Solving this, you would get an answer of 13.7
6 0
3 years ago
Find x !!?????!!!????????????
mars1129 [50]

Answer:

x = 15

Step-by-step explanation:

Given

See attachment

Required

Find x

The figure in the attachment is a quadrilateral and the angles in a quadrilateral add up to 360.

So, we have:

90 + 6x + 5+ 10x - 40 + 4x + 5 = 360

Collect like terms

6x + 10x + 4x = 360 - 90 - 5 + 40 - 5

20x = 300

Divide both sides by 20

x = 15

Hence, the value of x is 15

4 0
3 years ago
Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.
GREYUIT [131]

Answer:

y=\dfrac{1}{1-Ke^{-t}}

Step-by-step explanation:

Given

The given equation is a differential equation

\dfrac{dy}{dt}=y-y^2

\dfrac{dy}{dt}=-(y^2-y)

By separating variable

⇒\dfrac{dy}{(y^2-y)}=-t

\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt

Now by taking integration both side

\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt

⇒\ln (y-1)-\ln y=-t+C

Where C is the constant

\ln \dfrac{y-1}{y}=-t+C

\dfrac{y-1}{y}=e^{-t+c}

\dfrac{y-1}{y}=Ke^{-t}

y=\dfrac{1}{1-Ke^{-t}}

from above equation we can say that

When t  will increases in positive direction then e^{-t} will decreases it means that {1-Ke^{-t}} will increases, so y will decreases. Similarly in the case of negative t.

4 0
3 years ago
Ilona is at a friend's home that is several miles from her home. She starts walking at a constant rate in a straight line toward
larisa [96]

Complete question is;

Ilona is at a friends home that is several miles from her home. She starts walking at a constant rate in a straight line toward her home.

The expression -2t + 10 gives the distance, in miles, that Ilona is from her home after t hours. Select True or False for each of the following.

1. The friends home is 10 miles from Ilonas home

2. The distance Ilona has walked away from her friends home after t hours is given by the absolute value of -2t

3.The negative sign in the term indicates that Ilona is walking toward her home

4. Ilona is walking at a rate of 10 miles per hour

Answer:

Statements 1 & 2 are correct.

Step-by-step explanation:

We are told that the expression -2t + 10 gives the distance, in miles, that Ilona is from her home after t hours.

Now, we know that distance = speed x time

Now, from the expression -2t + 10

It's clear that 2t is a distance value.

This means that the speed used in going home is 2 miles/hr

This means that at t hours, she has walked 2t miles from her friends home.

Thus, statement 2 is true because an absolute value of -2t will be 2t.

Since 2t is subtracted from 10 in the expression and 2t is the distance she has walked from her friends house at Time, t. It means that 10 miles is the distance from her friends house to her house.

Thus, statement 1 is also true

7 0
3 years ago
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