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2 years ago

Which equation represents a linear function?

Mathematics

2 answers:

bogdanovich [222]2 years ago

5 0

Answer:

Step-by-step explanation:

The equation of a linear function:

y = mx + b

m - slope

b - y-intercept

x = 3 - it's a vertical line. It's not equation of a linear function

y = 16 - it's a horizontal line. It's an equation of a linear function

where m = 0, b = 16

y = -3x + 10 - it's an equation of a linear function

where m = -3, b = 10

y = 3x² + 1 - it's a quadratic function ( x² ).

katen-ka-za [31]2 years ago

3 0

Answer:

y = 16 and y = -3x + 10

Step-by-step explanation:

hello

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A circle is divided into 9 uneven "pie" shaped pieces. Each section is labeled 1, 2, ... , 9. The circle is then spun and the nu

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There are three outcomes of 4 out of eighteen outcomes, so the fraction of angle of spinner numbered 4 is $\frac{3}{8}$

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3 years ago

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PLEASE ANSWER THIS QUESTION Show that 4x – 7 is equivalent to 4(x - 1) – 3 when x = 3.

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Answer:

4(3) - 7 = 4(3 - 1) - 3

12 - 7 = 4(2) - 3

5 = 8 - 3

= 5

They are equivalent because when they are both simplified they have the same answer 5

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3 years ago

A marina is in the shape of a coordinate grid. Boat A is docked at (3.1, −1) and Boat B is docked at (−4.1, −1). The boats are _

kirill115 [55]

Answer:

d = 7.2

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra II

Distance Formula: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Step-by-step explanation:

Step 1: Define

Boat A(3.1, -1)

Boat B(-4.1, -1)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d.

Substitute [DF]: $d = \sqrt{(-4.1-3.1)^2+(-1+1)^2}$
Subtract/Add: $d = \sqrt{(-7.2)^2+(0)^2}$
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2 years ago

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Human body temperatures have a mean of 98.20° F and a standard deviation of 0.62° F. Sally's temperature can be described by z =

irina1246 [14]

Answer:

Sally's temperature is 97.27 °F.

Step-by-step explanation:

All the information given in the question tells us that the human body temperatures are normally distributed with a population's mean = 98.20°F and a standard deviation = 0.62°F.

The question gives us Sally's temperature in a z-score. We have to remember that the standard normal distribution is a particular case of a normal distribution where the mean = 0 and the standard deviation = 1.

Using the standard normal distribution, we can determine every probability associated with a normal distribution "transforming" the raw scores, coming from normally distributed data, into z-scores.

A z-score gives us the distance from the population's mean and is in standard deviation units. So, a z = 1.5 tells us that the value is 1.5 standard deviations above the mean. Conversely, a z = -1.5 tells us that the raw score is also 1.5 standard deviation from the mean, but in the opposite direction, that is, below the mean.

The formula for a z-score is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ (1)

Where

$\\ x\;is\;the\;raw\;score$ .

$\\ \mu\;is\;the\;population\;mean$ .

$\\ \sigma\;is\;the\;population\;standard\;deviation$ .

Then to find x (or the raw score, that is, Sally's temperature), we need to solve the formula (1) for it to finally solve the question.

Then

$\\ \mu = 98.20^\circF$ °F

$\\ \sigma = 0.62^\circF$ °F

$\\ z = -1.5$

Thus (with no units)

$\\ -1.5 = \frac{x - 98.20}{0.62}$

$\\ (-1.5*0.62) = x - 98.20$

$\\ (-1.5*0.62) + 98.20 = x$

$\\ x = (-1.5*0.62) + 98.20$

$\\ -0.93 + 98.20$

$\\ x = 97.27$ °F

Thus, Sally's temperature is $\\ x = 97.27$ °F (rounding the answer to the nearest hundredth).

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3 years ago

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Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

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$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago