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2 years ago

Write a function rule for "The output is half of the input x."

Mathematics

1 answer:

yKpoI14uk [10]2 years ago

5 0

Answer:

$y = \frac{x}{2}$

Step-by-step explanation:

Functions:

Functions have the following format:

y = f(x)

In which y is the output, x is the input, and f is the relation between them.

In this question:

The output is half of the input x. This means that:

$y = \frac{x}{2}$

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Which equation represents a line that passes through (2,-1/2) and has a slope of 3 ?

Anna11 [10]

Answer:

C. $y + \frac{1}{2} = 3(x - 2)$

Step-by-step explanation:

Given

$Point: (2,\frac{-1}{2})$

$Slope: 3$

Required

Equation of line

Let m represents the slope of the line;

m is calculated as thus

$m = \frac{y - y_1}{x - x_1}$

$where (x_1,y_1) = (2,\frac{-1}{2})$

$So; x_1 = 2; y_1 = \frac{-1}{2}$

$m = 3$

By substituting the right values in the formula above;

$m = \frac{y - y_1}{x - x_1}$ becomes

$3 = \frac{y - \frac{-1}{2}}{x - 2}$

Multiply both sides by $(x - 2)$

$3 *(x - 2) = \frac{y - \frac{-1}{2}}{x - 2} *(x - 2)$

$3 *(x - 2) = (y - \frac{-1}{2})$

$3 *(x - 2) = (y + \frac{1}{2})$

$3(x - 2) = (y + \frac{1}{2})$

$3(x - 2) = y + \frac{1}{2}$

Reorder

$y + \frac{1}{2} = 3(x - 2)$

Hence, the equation that represents the line is $y + \frac{1}{2} = 3(x - 2)$

4 0

3 years ago

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

3 years ago

What is the value of x?

Kazeer [188]

Answer:

Step-by-step explanation:

The volume of a sphere is

$V=\frac{4}{3}\pi r^3$

We are told our volume is 500pi/3, so we fill that in and solve for r:

$\frac{500\pi}{3}=\frac{4}{3}\pi r^3$

Start by multiplying both sides by 3 over 4 pi:

$(\frac{3}{4\pi})\frac{500\pi}{3}=(\frac{3}{4\pi})\frac{4\pi}{3}r^3$

Simplifying gives us

$125=r^3$

Take the cubed root of both sides to get that

r = 5

6 0

2 years ago

What is the explicit rule for this sequence?<br> -5, 15, – 45, 135, ...

hoa [83]

Answer:

Step-by-step explanation:

In a geometric sequence, consecutive terms differ by a common ratio. The formula for determining the nth term of a geometric progression is expressed as

an = a1r^(n - 1)

Where

a represents the first term of the sequence.

r represents the common ratio.

n represents the number of terms.

From the given sequence,

a1 = - 5

r = 15/- 5 = - 3

Therefore, the explicit rule for this sequence is

an = - 5(- 3)^n - 1

4 0

3 years ago

How much money will be in a bank account after 4 years if $8 is deposited at an interest rate of 5%? a. $12.00 b. $10.41 c.$16.0

vichka [17]

C: 16.00 because when I multipled it was 16

4 0

2 years ago

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