Answer:
that is scientifically impossible
Step-by-step explanation:
one you can't add, subtract, round up, or use decimals write that down.
Answer:
80π
Step-by-step explanation:
Circle O with radius IO:
large radius = IO = 12 = R
Circle O with diameter JK
IJ = JK = KL = 2 × IO = 24
IJ = JK = KL = 24/3 = 8
small radius = 8/2 = 4 = r
The shaded area is a semicircle of radius 12 minus a semicircle of radius 4 plus two semicircles of radius 4.
That is the same as
1 semicircle of radius 12 plus 1 semicircle of radius 4
A = πR²/2 + πr²/2
A = π(12² + 4²)/2
A = 160π/2
A = 80π
<h3>
Answer: B) Only the first equation is an identity</h3>
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I'm using x in place of theta. For each equation, I'm only altering the left hand side.
Part 1
cos(270+x) = sin(x)
cos(270)cos(x) - sin(270)sin(x) = sin(x)
0*cos(x) - (-1)*sin(x) = sin(x)
0 + sin(x) = sin(x)
sin(x) = sin(x) ... equation is true
Identity is confirmed
---------------------------------
Part 2
sin(270+x) = -sin(x)
sin(270)cos(x) + cos(270)sin(x) = -sin(x)
-1*cos(x) + 0*sin(x) = -sin(x)
-cos(x) = -sin(x)
We don't have an identity. If the right hand side was -cos(x), instead of -sin(x), then we would have an identity.
One red ribbon, 6 green ribbons and 3 blue ribbons total together would be 10 ribbons
Answer: 5x - 1 = -6
Step-by-Step Explanation:
We have this diagram here showing numbers and symbols. Just a remainder, "X" is a variable which can be found through and equation of Algebra. Alright, so we have 5 boxes with the symbol "X". Alright, that's a start, so we have <u>5x</u> as a start of the equation. We have <u>-1</u> on the same side as well.
Alright, so on the other side- we have Six symbols of "-1".
That concludes to... <u>-6</u>. Alright, notice how all of the symbols are on a Scale, so we can heavily imply that the "=" symbol will take place between the two sets of symbols. Alright, combining them we get.. <u>5x - 1 = -6.</u>
<em>If you need further instructions/help, don't hesitate to give me a shout out. Bye, and I hope this helps.</em>
