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erastovalidia [21]
3 years ago
9

triangle abc is graphed on a coordinate plane, as shown below. Triangle ABC is dilated by a scale factor of 2 with a center of d

ilation at the origin to create triangle A' B' C'. What are the coordinates of the verticles of triangle A'B'C'?

Mathematics
1 answer:
Mademuasel [1]3 years ago
4 0
C is the correct answer i just took that
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Paul’s grandmother gave him a collection of 30 coins. Each month Paul adds 5 coins to his collection. Which inequality can be us
Elodia [21]

Answer:

B

Step-by-step explanation:

  • He starts out with 30
  • He adds 5 coins each month
  • x represents the number of months

>60 means greater than 60

So the inequality would be 30 + 5x > 60

3 0
3 years ago
Which of the following equations describes the line shown below? Check all<br> that apply.
scoundrel [369]

Answer:

B. y = 2x - 2

E. y + 4 = 2(x + 1)

F. y - 6 = 2(x - 4)

Step-by-step explanation:

The equation of the line can be found using either the point-slope equation or the slope-intercept equation.

✔️Equation of the line in point-slope using the slope and the coordinates of the point (4, 6):

Slope = m = change in y/change in x

Using (4, 6) and (-1, -4),

m = (-4 - 6)/(-1 - 4)

m = -10/-5

m = 2

Substitute m = 2, and (a, b) = (4, 6) into the point-slope equation form y - b = m(x - a)

Thus:

y - 6 = 2(x - 4)

✔️Equation of the line in point-slope using the slope and the coordinates of the point (-1, -4):

Slope = m = 2

Substitute m = 2, and (a, b) = (-1, -4) into the point-slope equation form y - b = m(x - a)

Thus:

y - (-4) = 2(x - (-1))

y + 4 = 2(x + 1)

✔️Equation of the line in slope-intercept form, y = mx + b

Where,

m = slope = 2

b = y-intercept = -2 (the point where the line intercepts the y-axis)

Thus, substitute m = 2 and b = -2 into y = mx + b

y = 2x + (-2)

y = 2x - 2

6 0
3 years ago
The graph of an exponential function is given. Which of the following is the correct equation of the function?
katen-ka-za [31]

Answer:

If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).

Step-by-step explanation:

4 0
3 years ago
Un automovil gasta 6 litros cada 110 kilometros ¿cuantos kilometros recorrera con 22 litros
natulia [17]

Answer:

403.33 kilometros

Step-by-step explanation:

De la pregunta anterior:

6 litros = 110 kilómetros

22 litros = x

Cruz multiplicar

6 litros × x = 22 litros × 110 kilómetros

x = 22 litros × 110 kilómetros / 6 litros

x = 403.33 kilometros

Así, 22 litros recorrerán 403.33 kilómetros

7 0
3 years ago
Is the Answer equal to ?
STatiana [176]
I believe so ,,,,,,,,,,,,,,,,,,,,
8 0
3 years ago
Read 2 more answers
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