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3 years ago

Please help!!!!!!!!!

Mathematics

2 answers:

Anna11 [10]3 years ago

4 0

Distributive as you are distributing the 3 to the x and 5 (to get (3x+15))

Blababa [14]3 years ago

4 0

Distributive because you’re distributing the 3 into (x + 5)

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Find the solution of this system of equations<br> 4x - 5y = -12<br> -X + 5y = 18

Natalka [10]

4x - 5y = -12

-x + 5y = 18

Add the equations together:

4x + -x = 3x

-5y + 5y = 0

-12 + 18 = 6

So now you have:

3x = 6

Divide both sides by 3:

X = 2

Now you have the value of x, replace x with 2 in one of the equations and solve for y:

4(2) -5y = -12

8 -5y = -12

Subtract 8 from both sides:

-5y = -20

Divide both sides by -5:

Y = 4

The solution is x = 2 and y = 4 which is written as (2,4)

7 0

3 years ago

Write the equation of the line that passes through the point (3,5) and is perpendicular to the line y=3.

3241004551 [841]

Answer:

x = 3

Step-by-step explanation:

y = 3 is the equation of a horizontal line parallel to the x- axis.

A line perpendicular to it must therefore be a vertical line parallel to the y- axis with equation

x = c

where c is the value of the x- coordinates the line passes through.

The line passes through (3, 5) with x- coordinate 3 , thus

x = 3 ← equation of perpendicular line

6 0

4 years ago

I am in need of help :) please and thank you :):

nika2105 [10]

It’s the second answer

8 0

3 years ago

Read 2 more answers

Mike owes his friend $6. Each day, Mike earns $3 per day.

raketka [301]

Answer:

the x intercept would mean the number of days and y would mean the amount theyre making

Step-by-step explanation:

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2 years ago

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

3 years ago