For a single 20 sided dice, a chance to get a certain number would be 1/20. For 2 numbers it will be a 2/20 chance, which can be simplified to a 1/10 chance that you will roll a 15 or a 17.
Answer:
3 <u>> </u>x
Step-by-step explanation:
(Sorry if im wrong)
Divide both sides by 6
18/6 <u>></u> 6x/6
3 <u>></u> x
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ
Use the Sum/Difference Identities:
sin(α + β) = sinα · cosβ + cosα · sinβ
cos(α - β) = cosα · cosβ + sinα · sinβ
Use the Unit circle to evaluate: sin45 = cos45 = √2/2
Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ
Use the Pythagorean Identity: cos²Ф + sin²Ф = 1
<u />
<u>Proof LHS → RHS</u>
LHS: 2sin(45 + 2A) · cos(45 - 2A)
Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)
Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]
Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]
Distribute: cos²2A + 2cos2A·sin2A + sin²2A
Pythagorean Identity: 1 + 2cos2A·sin2A
Double Angle: 1 + sin4A
LHS = RHS: 1 + sin4A = 1 + sin4A 
Question 1:
"Match" the letters
DE are the last two letters of BCDE
The last two letters of OPQR is QR
DE is congruent to QR
Question 2:
Blank 3: Reflexive property (shared side)
Blank 4: SSS congruence of triangles (We have 3 sets of congruent sides)
Question 3:
I'm guessing those two numbers are 7.
Since both are 7, AB and AE are congruent.
We know that all the other sides are congruent because it is given.
We also know that there is a congruent angle in each triangle.
Thus, the two triangles are congruent by SAS or SSS.
(Note: I couldn't prove this without the two "7"s because there is no such thing as SSA congruence)
Have an awesome day! :)
The expanded form of 0.368 is
.3 + .06 + .008
Hope this helps. :)