1
Simplify \frac{1}{2}\imath n(x+3)21ın(x+3) to \frac{\imath n(x+3)}{2}2ın(x+3)
\frac{\imath n(x+3)}{2}-\imath nx=02ın(x+3)−ınx=0
2
Add \imath nxınx to both sides
\frac{\imath n(x+3)}{2}=\imath nx2ın(x+3)=ınx
3
Multiply both sides by 22
\imath n(x+3)=\imath nx\times 2ın(x+3)=ınx×2
4
Regroup terms
\imath n(x+3)=nx\times 2\imathın(x+3)=nx×2ı
5
Cancel \imathı on both sides
n(x+3)=nx\times 2n(x+3)=nx×2
6
Divide both sides by nn
x+3=\frac{nx\times 2}{n}x+3=nnx×2
7
Subtract 33 from both sides
x=\frac{nx\times 2}{n}-3x=nnx×2−3
The answer is in my attachment
Surface area= Area of all the shapes added together
Area of square =s^2
Area of triangle =bh/2
S=12
B=12 B is s
H=8
12^2 =144 square
12×8= 96/2=48 triangle
There are 4 triangles and 1 square
144+48+48+48+48=336 Answer is f hope that helps
Answer:
4 roots: 2 , -3, 1+√2, 1 - √2
Step-by-step explanation:
x⁴ − x³ − 9x² + 11x + 6 = 0 leading: 1 (x⁴) Trailing: 6 (1,2,3,6)
possible roots: trailing/leading (±1 , ±2, ±3, ±6)
Test: x = 2 (2)⁴ - (2)³ - 9*(2)² + 11*(2) + 6 = 0
x = -3 (-3)⁴ - (-3)³ - 9*(-3)² + 11*(-3) + 6 = 0
By rational roots test: possible roots (x-2)(x+3) are factors
Long division: (x⁴ − x³ − 9x² + 11x + 6) / (x-2)(x+3) = x²-2x-1
roots of x²-2x-1: x= (-b±√b²-4ac) / 2a
x = (2 ± √4+4) / 2 = (2 ± 2√2) / 2 = 1 ± √2
In order to compare two fractions, they must both have the same denominator. In this case, we'll begin by converting 1/4 to twelfths. Multiply the denominator of 1/4 by 3:
4 × 3 = 12
This gives us 1/12. But to avoid changing the overall value of the fraction, we must also multiply the numerator of 1/4 by 3:
1 × 3 = 3
This leaves us with a final value of 3/12. We have now proven that 1/4 is equal to 3/12. This is the answer to your question:
A) 1/4 = 3/12
I hope this helps!
