Answer:
SEE THE IMAGE ABOVE FOR ANSWER.
Answer:
The irrational conjugate theorem states that if a polynomial equation has a root (a + √b), then we can say that the conjugate of (a + √b), i.e. (a - √b) will also be another root of the polynomial.
Step-by-step explanation:
The irrational conjugate theorem states that if a polynomial equation has a root (a + √b), then we can say that the conjugate of (a + √b), i.e. (a - √b) will also be another root of the polynomial.
For example, if we consider a quadratic equation x² + 6x + 1 = 0, then two of its roots are - 3 + √8 and - 3 - √8 and they are conjugate of each other. (Answer)
Answer:
you can find a radius through its volume and height. Multiply the volume by 3. For example, the volume is 20. Multiplying 20 by 3 equals 60
Step-by-step explanat
I hope to help you
Answer:
103= p small
plarge = 123
Step-by-step explanation:
We know that the the ratio of the areas is the scale factor squared/
Larger triangle over smaller triangle
72
----- = scale factor ^2
50
simplify by dividing by 2 on top and bottom
36
----- = scale factor ^2
25
Take the square root of each side
sqrt(36)
-------------- = sqrt(scale factor ^2)
sqrt(25)
6
-------------- = scale factor
5
The ratio of the perimeters is the scale factor
p large 6
-------------- = ----------------
p small 5
Using cross products
6 p large = 5 p small
We know the sum is 226
p large + p small = 226
p large = 226 - p small
We have 2 equation and 2 unknowns
6 p large = 5 p small
Substitute for p large
6 (226 - p small) = 5 p small
1356 - 6 p small = 5 p small
Add 6 p small to each side
1356 = 11 p small
divide by 11
1356/11 = p small
P large = 226-1356/11
p large = 2486/11-1356/11
plarge = 1130/11
The solution requires whole number answers so
1356/11 = p small
123.27 which rounds to 123
plarge = 1130/11
plarge = 102.7272 = 103
Answer: 
Step-by-step explanation:
Remember the logarithms properties:

Then,simplifying:

Apply base 8 to boths sides and then solve for "x":
