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diamong [38]
3 years ago
9

Write the linear function if f(0)=5 and f(2)=35

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
3 0

Answer: y=15x+5

Step-by-step explanation:

Looking at what we are given, we can tell that the y-intercept is 5. To ge tthe slope, you use \frac{y_{2}-y_{1}  }{x_{2} -x_{1} }. With the slope , you get 15.

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19. Find x:<br> € (17x-7)*<br> G<br> 19<br> HELP W ALL 3 !!
blsea [12.9K]

Answer:

24

(17x-7)

x = 7 + 17

x = 24

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7 0
3 years ago
May I have some help with these, please?
Nookie1986 [14]

Minimum = 18, Q₁ = 27.5, median = 39.5, Q₃ = 43, maximum = 49.

The five-number summary is a descriptive statistic that provides information on a series of observations. It consists of the following statistics:

1. Minimum: the smallest observation

2. First quartile  Q₁: the average of the values ​​below the median.

3. Medium  M: the average term.

4. third quartile  Q₃: the average of values ​​above the median.

5. Maximum: The largest observation.

The data represents the numbers of runs allowed by 8 college pitchers.

{18, 49, 38, 41, 33, 44, 42, 22}

The five-number summary is:

First, we have to sort the data from least to greatest.

{18, 22, 33, 38, 41, 42, 44, 49}

From the ordered data we can see that the minimum value is 18 and the maximum is 49.

The median is the middle term of the data set. In this case of an even number of terms, the median is the average of the terms located in the middle. So, the terms located in the middle if the data are in bold:

{18, 22, 33, 38, 41, 42, 44, 49}

Median= (38 + 41)/2 = 79/2 = 39.5

To calculate the first quartile Q₁, the values ​​below the median are {18, 22, 33, 38}. So, the median of this values is the first quartile Q₁:

{18, 22, 33, 38}

Q₁ = (22 + 33)/2 = 55/2 = 27.5

To calculate the third quartile  Q₃, the values ​​above the median are {41, 42, 44, 49}. So, the median of this values is the third quartile Q₃:

{41, 42, 44, 49}

Q₃ = (42 + 44)/2 = 86/2 = 43

8 0
3 years ago
a model car is one tenth of the size of the real car the models measures 42cm longs what the length of length of the real car
trapecia [35]
Just multiply 42 by 10 and that gives you 420 , so the real car has a length of 420cm or 4.2 metres :)
8 0
3 years ago
Read 2 more answers
For questions 8 – 14, use the following functions:
pogonyaev

we are given

f(x)=5

g(x)=\sqrt{x+2}

(8)

(f+g)(x)=f(x)+g(x)

we can plug it

(f+g)(x)=5+\sqrt{x+2}

(9)

(f-g)(x)=f(x)-g(x)

we can plug it

(f-g)(x)=5-\sqrt{x+2}

(10)

(f*g)(x)=f(x)*g(x)

we can plug it

(f*g)(x)=5\sqrt{x+2}

(11)

(\frac{f}{g} )(x)=\frac{f(x)}{g(x)}

we can plug it

(\frac{f}{g} )(x)=\frac{5}{\sqrt{x+2}}

(12)

(\frac{g}{f} )(x)=\frac{g(x)}{f(x)}

we can plug it

(\frac{g}{f} )(x)=\frac{\sqrt{x+2}}{5}

(13)

(fog)(x)=f(g(x))

f(x)=5

we can replace g(x)

we get

(fog)(x)=5

(14)

(gof)(x)=g(f(x))

f(x)=5

we can replace f(x)

(gof)(x)=\sqrt{f(x)+2}

we get

(gof)(x)=\sqrt{5+2}

(gof)(x)=\sqrt{7}


8 0
3 years ago
The square below has area 100. Find the area of the shaded region.
Degger [83]

Answer:

21.46

Step-by-step explanation:

Start with the square

A = a^{2} \\100 = a^{2}\\10 = a

Each side of the square is 10

This also means the diameter of the circle is 10, with a radius of 5

A = \pi r^{2} \\A = \pi 5^{2} \\A = 78.54

Now substract the area of the circle from the square

x = 100 - 78.54

6 0
3 years ago
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