Answer:
This is khan academy. I am pretty sure they have a video for this but whatever.
Step-by-step explanation:
So we are using sin, cosine, and tan. 3.4 is the hypotenuse. the side is X. The angle is 20 degrees. So, this is adjacent to hypotenuse. This means it is cosine.
3.4cos(20)= around 3.19
And thats that! let me know if this is wrong and I will try to fix it.
Answer:
- x = 0 or 1
- x = ±i/4
- x = -5 (twice)
Step-by-step explanation:
Factoring is aided by having the equations in standard form. The first step in each case is to put the equations in that form. The zero product property tells you that a product is zero when a factor is zero. The solutions are the values of x that make the factors zero.
1. x^2 -x = 0
x(x -1) = 0 . . . . . x = 0 or 1
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2. 16x^2 +1 = 0
This is the "difference of squares" ...
(4x)^2 - (i)^2 = 0
(4x -i)(4x +i) = 0 . . . . . x = -i/4 or i/4 (zeros are complex)
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3. x^2 +10x +25 = 0
(x +5)(x +5) = 0 . . . . . x = -5 with multiplicity 2
Since the radius is 12, the equation is
.. x^2 +y^2 = 144
40s+2bs+200=number of books needed
if we want 12 books and 8 stalls
b=12
s=8
solve
40(8)+2(12)(8)+200=
320+192+200=
712
needs 712 books
Strange question, as normally we would not calculate the "area of the tire." A tire has a cross-sectional area, true, but we don't know the outside radius of the tire when it's mounted on the wheel.
We could certainly calculate the area of a circle with radius 8 inches; it's
A = πr^2, or (here) A = π (8 in)^2 = 64π in^2.
The circumference of the wheel (of radius 8 in) is C = 2π*r, or 16π in.
The numerical difference between 64π and 16π is 48π; this makes no sense because we cannot compare area (in^2) to length (in).
If possible, discuss this situatio with your teacher.
