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Nikitich [7]
3 years ago
13

A new graph is formed from y = 5x by changing the slope to 9 and y-intercept to 8. Which statement about the new relationship is

true?
The new graph has a steeper slope and a y-intercept of 13.

The new graph has a steeper slope and y-intercept of 8.

The new graph has a less steep slope and a y-intercept of 13

The new graph has a less steep slope and a y-intercept of 8
Mathematics
1 answer:
shusha [124]3 years ago
8 0

Answer:

The new graph has a steeper slope and a y-intercept of 8.

Step-by-step explanation:

From Analytical Geometry, we define a straight line by following first order polynomial (linear function):

y = m\cdot x + b (1)

Where:

x - Independent variable.

y - Dependent variable.

b - y-Intercept.

m - Slope.

And the slope represents the change in dependent variable (\Delta y) divided by the change in independent variable (\Delta x). That is:

m = \frac{\Delta y}{\Delta x}

If m_{1} > 0 and m_{2} > m_{1}, then the new line is steeper with respect to the original line.

Let y = 5\cdot x, its slope and y-intercept are 5 and 0, respectively. If slope is changed into 9 and y-intercept becomes 8, then the new graph has a steeper slope and a y-intercept of 8.

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Pls help! Pls do this step by step . <br>factorise: <br>m^3-27​
Verdich [7]

Answer:

m-3

Step-by-step explanation:

cube root the two to get m-3

7 0
3 years ago
In 2013, the moose population in a park was measured to be 5,100. By 2018, the population was measured again to be 5,200. If the
natka813 [3]

Answer:

P(t) = 5100e^{0.0039t}

Step-by-step explanation:

The exponential model for the population in t years after 2013 is given by:

P(t) = P(0)e^{rt}

In which P(0) is the population in 2013 and r is the growth rate.

In 2013, the moose population in a park was measured to be 5,100

This means that P(0) = 5100

So

P(t) = 5100e^{rt}

By 2018, the population was measured again to be 5,200.

2018 is 2018-2013 = 5 years after 2013.

So this means that P(5) = 5200.

We use this to find r.

P(t) = 5100e^{rt}

5200 = 5100e^{5r}

e^{5r} = \frac{52}{51}

\ln{e^{5r}} = \ln{\frac{52}{51}}

5r = \ln{\frac{52}{51}}

r = \frac{\ln{\frac{52}{51}}}{5}

r = 0.0039

So the equation for the moose population is:

P(t) = 5100e^{0.0039t}

5 0
3 years ago
Which choice is equivalent to the quotient shown here for acceptable
Gennadij [26K]
The answer to this question is A
7 0
3 years ago
A provider order has been received for a dobutamine (dobutrex®) drip at 7 mcg/kg/min. the patient's weight is 220 lbs. how many
Liula [17]
To figure this out you first need to determine how many kg there are in 220 lbs. 1 lb equals 0.454 kg, so 220 lbs= 99.88 kg.

It states that a patient receives 7mcg per kg per minute, so the patient will receive 7 x 99.88=699.16 mcg per minute.

You need to convert this to a number of mcg per hour, so multiply the amount by 60 to get 41949.6 mcg per hour.
4 0
3 years ago
The quadratic function f (x) = - 45x2 + 350x + 1,590 models the population of a city where x represents the number of years sinc
kramer

Answer: 2,215,000

Step-by-step explanation:

Given: The quadratic function f (x) = - 45x^2 + 350x + 1,590 models the population of a city where x represents the number of years since 2005.

To Find:  population of the city in 2010

We need to put x= 2010-2005 = 5

f (5) = - 45(5)^2 + 350(5) + 1,590\\\\=-1125+1750+1590 =2215

Hence, the estimated population of the city in 201 = 2215 thousands or 2,215,000 .

3 0
3 years ago
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