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ollegr [7]
3 years ago
7

What is the domain of the function f(x) = 2x + 5?

Mathematics
1 answer:
alukav5142 [94]3 years ago
8 0

There is no value that x cannot be for linear equations.

Thus, the domain is ALL REAL NUMBERS.

CHOICE B.

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2 cars are 700 miles apart. They travel toward each other. One of the cars travels at 75mph, the other travels at 100 mph. How l
olga2289 [7]

Answer: C 4

Step-by-step explanation:

Use the formula: distance = rate x time

We can say that train 1 travels a distance of x, and train 2 travels a distance of 700 - x

The rate of train 1 is 75 mph, and the rate of train 2 is 100 mph

The time traveled for the two trains will be the same. We can represent that with the variable t.

We have the following equation for train 1:

x = 75t

For train 2, we have this equation:

700 - x = 100t

Use the Substitution Method by replacing x in the equation for train 2 with the value 75t.

700 - 75t = 100t

700 = 175t

700/175 = 4 hours.

It will take 4 hours for the two trains to meet.

6 0
3 years ago
Multiplying a certain even number by 4 then subtracting 16 gives you the next consecutive even number. Find the original number
anyanavicka [17]

Let 2k represent the original even number (such that k is an integer)

then 2(k + 1) is the next consecutive even number.

4(2k) - 16 = 2(k + 1)

8k - 16 = 2k + 2

6k = 18

k = 3

since 2k represents the original number, then 2(3) = 6 is the oroginal number

Answer: 6

3 0
3 years ago
Which of the following reveals the minimum value for the equation 2x^2 − 4x − 2 = 0?
12345 [234]

Answer:

2(x-1)^{2}=4

Step-by-step explanation:

<u><em>The options of the question are</em></u>

2(x − 1)2 = 4

2(x − 1)2 = −4

2(x − 2)2 = 4

2(x − 2)2 = −4

we have

2x^{2} -4x-2=0

This is a vertical parabola open upward

The vertex represent the minimum value

The quadratic equation in vertex form is

y=a(x-h)^2+k

where

a is a coefficient

(h,k) is the vertex

so

Convert the quadratic equation in vertex form

Factor 2 leading coefficient

2(x^{2} -2x)-2=0

Complete the squares

2(x^{2} -2x+1)-2-2=0

2(x^{2} -2x+1)-4=0

Rewrite as perfect squares

2(x-1)^{2}-4=0

The vertex is the point (1,-4)

Move the constant to the right side

2(x-1)^{2}=4

3 0
3 years ago
Use the method of reduction of order to find a second independent solution of the given differential equation. t2y'' + 3ty' + y
Charra [1.4K]

I assume you mean y_1=t^{-1}, and not y_1=t-1, since this doesn't satisfy the ODE.

Assume a second solution of the form y_2=vy_1, where v is a function of t. Then

{y_2}'=v'y_1+v{y_1}'

{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''

Substituting into the ODE gives

t^2\left(\dfrac{v''}t-\dfrac{2v'}{t^2}+\dfrac{2v}{t^3}\right)+3t\left(\dfrac{v'}t-\dfrac v{t^2}\right)+\dfrac vt=0

\implies tv''+v'=0

\implies(tv')'=0

\implies tv'=C

\implies v'=\dfrac Ct

\implies v=C_1\ln|t|+C_2

\implies y_2=\dfrac{\ln t}t

where we omit the second term because it's already accounted for by y_1.

3 0
3 years ago
!!!!!?!??!<br> Pls help me plsss
irakobra [83]

Answer:

A is only ur answer

Hope it helps you.

8 0
2 years ago
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