Answer:
Solution
p = {-3, 1}
Step-by-step explanation:
Simplifying
p2 + 2p + -3 = 0
Reorder the terms:
-3 + 2p + p2 = 0
Solving
-3 + 2p + p2 = 0
Solving for variable 'p'.
Factor a trinomial.
(-3 + -1p)(1 + -1p) = 0
Subproblem 1
Set the factor '(-3 + -1p)' equal to zero and attempt to solve:
Simplifying
-3 + -1p = 0
Solving
-3 + -1p = 0
Move all terms containing p to the left, all other terms to the right.
Add '3' to each side of the equation.
-3 + 3 + -1p = 0 + 3
Combine like terms: -3 + 3 = 0
0 + -1p = 0 + 3
-1p = 0 + 3
Combine like terms: 0 + 3 = 3
-1p = 3
Divide each side by '-1'.
p = -3
Simplifying
p = -3
Subproblem 2
Set the factor '(1 + -1p)' equal to zero and attempt to solve:
Simplifying
1 + -1p = 0
Solving
1 + -1p = 0
Move all terms containing p to the left, all other terms to the right.
Add '-1' to each side of the equation.
1 + -1 + -1p = 0 + -1
Combine like terms: 1 + -1 = 0
0 + -1p = 0 + -1
-1p = 0 + -1
Combine like terms: 0 + -1 = -1
-1p = -1
Divide each side by '-1'.
p = 1
Simplifying
p = 1
Solution
p = {-3, 1}
Answer: 90 is a number which has 9 as the sum of its digits, 9+0=9 and if the digits are reversed, 09 is 81 less than 90.
Step-by-step explanation: Since the number must be a 2-digit number, it seems that 90 is the only number that qualifies.
If you need some equations, here is a way to set up a system to solve:
x+y= 9 rewrite to get a value for y to substitute:
y = 9-x
10x +y = 10y + x + 81 rewrite by moving (subtracting variable terms from both sides)
10x -x + y-10y = 81 combine like terms
9x - 9y = 81
Substitute for y and solve for x
9x - 9(9-x)= 81 distribute.
9x - 81 + 9x = 81
18x = 162. 162/18 = 9
x = 9
Substitute in the original equation to solve for y
9 + y = 9
y = 0
10(9) + 0 = 10(0) + 9 +81
90 - 81 = 09
Answer:
9 mi.
Step-by-step explanation:
sorry if i get it wrong ):
I'm going to separate this into sections so it makes more sense for you to read. For the problems with π where you have to round, ask your teacher where to round, unless your textbook specifies it:
A – 100 cm^2
To calculate area of squares, you multiply l • w. It's a square, so all sides are equal, and since we know that one side = 10 cm, the area is 10 • 10 = 100
B – πr^2 (not sure if the r shows up very well, so I'm retyping it in words - pi • radius squared)
C – 25π cm^2 or an approximate round like 78.54 cm^2 (ask your teacher about this – it could be to the nearest tenth, hundredth, etc.)
To find the area of a circle, you must follow the formula πr^2. In this case, the diameter is 10. The radius is half the diameter, so to substitute the values you must find 10 ÷ 2 = 5. So the radius is 5 cm. From there you can substitute r for 5, ending up with π • 5^2. 5^2 = 25, so the area is 25π, or about 78.54, depending on where the question wants you to round.
D – An approximate round (to the nearest hundredth it is 21.46 cm^2)
To find the area of the shaded region, just subtract the circle's area from the square's area, or 100 – 25π ≈ 21.46. Again, though, ask your teacher about where to round, unless your textbook specifies it.
E – dπ (diameter • pi)
F – 10π cm^2 or an approximate round like 31.42 cm^2
The diameter is 10. 10π ≈ 31.42
Hope this helps!
The answer for this question would be 156
