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Veronika [31]
3 years ago
14

Given f(x) = 2x-3, what is the value of f(3)?

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

3

Step-by-step explanation:

f(3)=2(3)-3

You just plug in the number for the variable.

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the area of a rectangular field is x^2-x-72m^2. the length of the field is x+8m. what is the width of the field in meters?
olga nikolaevna [1]
So first off, remember the formula for the area of a rectangle. it's length time width right? or, you could think that A= 2 things multiplied together. how can we simplify x^2 - x -72 to equal two things multiplied together? well, we can factor it! We know that one of the factors is x+8 right? so what's the other factor? it would be x-9. so, we know width is x-9. I'm not sure if you need an actual value for width. if so, I could probably go further.
7 0
3 years ago
Read 2 more answers
MATH HELP!!! 100PTS!!!
Lostsunrise [7]

Answer:

x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}

Step-by-step explanation:

<u>Given functions</u>:

  p(x)=2x-4

  r(x)=\dfrac{6x-1}{9x+1}

Solve for p(x) = r(x):

\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

<u>Quadratic Formula</u>

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

Therefore:

a=18, \quad b=-40, \quad c=-3

Substitute the values of a, b and c into the <u>quadratic formula</u> and solve for x:

\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}

Therefore, the solutions are:

x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}

Learn more about quadratic equations here:

brainly.com/question/27750885

brainly.com/question/27739892

6 0
2 years ago
The goal of a nutritional study was to compare the caloric intake of adolescents living in rural areas of the US with the calori
stiks02 [169]

Answer:

To give more clarity to the question, lets examine the attached back-to-back stem plot.

A)

Having examined the stem plot, we can using quick calculations, summarize that:

The mean (40.45 cal/kg) and median (41 cal/kg) daily caloric intake of ninth-grade students in the rural school is higher than the corresponding measures of center, mean (32.6 cal/kg) and median (32 cal/kg), for ninth-graders in the urban school.

The median and the mean for the students in the 9th grade in the urban school is lower than that of those of their contemporaries in the rural school. The respective medians and means are stated below:

Urban 9th Grade Students

Median = 32 cal/kg

Mean = 36 cal/kg

Rural 9th Grade Students

Median = 41 cal/kg

Mean = 41 cal/kg

Please note that all figures above have been approximated to the nearest whole number.

B)

It is unreasonable to generalize the findings of this study to all rural and urban 9th-grade students in the United States because the sample is too small compared to the target population size.

To allow for generalization, they would have to collect and analyze more samples say from every state within America.

Cheers!

5 0
3 years ago
Find the distance between each pair of points
dolphi86 [110]

Answer: the correct answer is 20

Step-by-step explanation:

The formula for determining the distance between two points on a straight line is expressed as

Distance = √(x2 - x1)² + (y2 - y1)²

Where

x2 represents final value of x on the horizontal axis

x1 represents initial value of x on the horizontal axis.

y2 represents final value of y on the vertical axis.

y1 represents initial value of y on the vertical axis.

From the graph given,

x2 = - 7

x1 = 5

y2 = - 7

y1 = 9

Therefore,

Distance = √(- 7 - 5)² + (- 7 - 9)²

Distance = √(- 12²) + (- 16)²

= √(144 + 256) = √400

Distance = 20

5 0
3 years ago
Can someone please help me with this math problem.
mr Goodwill [35]
PART A:
Find the rate of change between 1980 and 1989
d for P₁ = 80 - 60
d for P₁ = 20

d for P₂ = 76 - 82
d for P₂ = -6

The rate of change in P₁ is 20 hundred per year. The rate of change in P₂ is -6 hundred per year.

PART B:
Find the rate of change between 1989 and 1996
d for P₁ = 100 - 80
d for P₁ = 20

d for P₂ = 70 - 76
d for P₂ = -6

The rate of change in P₁ is 20 hundred per year. The rate of change in P₂ is -6 hundred per year.

PART C:
Find the rate of change between 1980 and 1996
d for P₁ = 100 - 60
d for P₁ = 40

d for P₂ = 70 - 82
d for P₂ = -12

The rate of change in P₁ is 40 hundred per year. The rate of change in P₂ is -12 hundred per year.
4 0
3 years ago
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