Subtract sqrt 3 to isolate 2sin x
Divide by 2 to isolate sin x
Take the arcsin of each side to isolate x
Thinking of a 30-60-90 triangle, we know that x sqrt 3 is opposite 60° and 2x acts as the hypotenuse, thus it will be some variation of 60°.
In order to produce a negative result, x must be -60° (-π/3)
It is worth noting that this is a restricted value. Any addition/subtraction of 2π from this answer will produce the same result, thus your full answer is the above plus or minus 2kπ where k ∈ Z (k is an interger)
Answer:
Polynom degree: 5
Y intercept point: (0, 80)
Step-by-step explanation:
P(x)=(x+5)(x+4)²(x+1)²
When you expand, the highest power of x is 1 for first term (x+5), 2 for second term (x+4)² and again 2 for (x+1)². Overall, x⁵ will be the x term with highest power. So the degree of the polynom is 5.
The y intercept, i.e. intersection with OY axis, happens for x=0. Thus, P(0)=5×4²×1²=5×16=80. The y intercept point is (0, 80)
Let's find out.
David needs to score an average of 145 across all four games, so first we do:
145 × 4 = 580
Then we add all the scores he currently has:
139 + 143 + 144 = 426
Then we subtract 426 from 580 to find out how much he needs to score for his fourth:
580 - 426 = 154
David needs to score 154 on his 4th game to have an average score of 145 for all 4 games.
Answer:
2/5
Step-by-step explanation:
3+2 = 5
2 kitkats, 5 total chocolates, 2/5
hope this helps.
Answer:
1 and −1 are the only solutions to the equation x^2 = 1.
Step-by-step explanation:
We shall proceed as he suggests
![x=a+bi](https://tex.z-dn.net/?f=x%3Da%2Bbi)
Given ![x^{2}=1](https://tex.z-dn.net/?f=x%5E%7B2%7D%3D1)
substitute a+bi in x, we get
![(a+bi)^{2}=1](https://tex.z-dn.net/?f=%28a%2Bbi%29%5E%7B2%7D%3D1)
Rewriting the both sides in standard form for a complex number
![(a^{2}-b^{2})+2abi=1+0i](https://tex.z-dn.net/?f=%28a%5E%7B2%7D-b%5E%7B2%7D%29%2B2abi%3D1%2B0i)
Equating the real parts on each side of the equation, and equating the imaginary parts on each side of the equation.
and ![2ab=0](https://tex.z-dn.net/?f=2ab%3D0)
So either a=0 or b=0. If a=0 then
![0^{2}-b^{2}=1](https://tex.z-dn.net/?f=0%5E%7B2%7D-b%5E%7B2%7D%3D1)
. has no real solution.
If b=0 then
![a^{2}-0^{2}=1](https://tex.z-dn.net/?f=a%5E%7B2%7D-0%5E%7B2%7D%3D1)
![a^{2}=1](https://tex.z-dn.net/?f=a%5E%7B2%7D%3D1)
![a^{2}-1^{2}=0](https://tex.z-dn.net/?f=a%5E%7B2%7D-1%5E%7B2%7D%3D0)
![(a-1)(a+1)=0](https://tex.z-dn.net/?f=%28a-1%29%28a%2B1%29%3D0)
. or ![a=-1](https://tex.z-dn.net/?f=a%3D-1)
Hence proved.