Answer:
x=4
Step-by-step explanation:
Let's solve the equation:
3x+12=4x+8
12=4x-3x+8
12=x+8
12-8=x
4=x
x=4
The way i got this answer was by solving the equation using the following steps. Since you're solving for one side and have two different equations, put an equal sign in between the two equations to get the equation set up above. Then you need to have the x variable on one side, instead of both sides, so you take 3x and subtract it from both sides, leaving x on one side, because 4x-3x is equal to 1x, or just x. Then we need to have what is not attached to a variable on the other side to make it easier to solve, so you would need to subtract 8 from both sides to get rid of the 8 on the side with the variable, because if you subtract 12 from both sides, it will just make it more confusing to solve, and then 12-8 is equal to 4, so you get x is equal to 4.
Answer:
-cos^4(x)
Step-by-step explanation:
Step 1: Use the Pythagorean identity : 1=cos^2(x) + sin^2(x)
1-sin^2(x) = cos^2(x)
-1+sin^2(x) = -cos^2(x)
cos^2(x) (-cos^2(x))
Step 2: Factor out common terms cos^2(x)
cos^2(x) (sin^2(x)-1)
Ans: -cos^4(x)
12, because 4 rows and 3 sections in each
4x3=12
Answer:
Step-by-step explanation:
Given that a rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

the parabola is open down with vertex at (0,2)
We can find that the rectangle also will be symmetrical about y axis.
Let the vertices on x axis by (p,0) and (-p,0)
Then other two vertices would be (p,2-p^2) (-p,2-p^2) because the vertices lie on the parabola and satisfy the parabola equation
Now width = 
Area = l*w = 
Use derivative test
I derivative = 
II derivative = 
Equate I derivative to 0 and consider positive value only since we want maximum
p = 
Thus width= 
Length =
Width = 
<u>The interpreter attended </u><u>22 parent-teacher conference</u><u> and </u><u>5 school related assembly</u>
To solve this problem, we would write out two set of linear equations.
The data given on this problem are
- parent-teacher conference = $35.50
- school related assembly = $42
- The total number of engagements = 27
Let x represent the number of parent-teacher conference
let y represent the number of school related conference
<h3>Equations</h3><h3>

</h3>
From equation (i)

Put equation(iii) into equation (ii)

Put y = 5 into equation 1

From the above calculations, he attended 22 parent-teacher conference and 5 school related assembly.
Learn more on linear equations here;
brainly.com/question/4074386