How would you solve this?
(5 * pi) /12 = (5 * 180)/12 = (5*15) = 75 cos ((5*pi)/12) = cos 75 = 0.92175127
How would you solve this? z^2+12z=48?
z^2 +12z - 48 = 0 z = -12 ± sqrt(144 -(4 x 1 x(-48))/2 z = (-12 ± 18.3)/2 z1 = -15.15 z2 = 3.15
How would you solve this? (1+cot^2x)/(tan^2x+1)?
(1+cot²x)/(tan²x+1) = (1+[1/tan²x])/(tan²x+1) = (1+ [1/ [sin²x/cos²x] ]) / ([sin²x/cos²x]+1) = ([sin²x/sin²x]+ [cos²x/sin²x])/([sin²x/cos²x]+[cos²x/cos... = ( [sin²x+cos²x] /sin²x) / ( [sin²x+cos²x] /cos²x) = [sin²x+cos²x][cos²x] / [sin²x+cos²x][sin²x] = [cos²x] / [sin²x] = 1/tan²x = cot²x
How would you solve this problem?
Volume of a sphere: V=(D^3)(pi)/6, where D is the volume of the ice and iron ball. Surface area of a sphere: A=(pi)(D^2) Take derivative of V wrt time: dV/dt = (D^2)2 (dD/dt) (using the chain rule.) From the problem statement, the ice is melting at a rate of 0.008 L/min, which is dV/dt (change in volume wrt time). So you can solve for dD/dt, which will yield the change in diameter wrt time: dD/dt = 2(0.008 L/min)/(D^2(pi)) Since we're given the diameter in cm, we need to convert 0.008 L to cm^3 so our units will match (0.008L = 8 cm^3): dD/dt = 2(8 cm^3/min)/(D^2(pi)) We need to find the change in surface area wrt time, so take the derivative of A wrt time: dA/dt = 2(D)(pi) dD/dt Since we want to find dA/dt when D = 20 cm, we can substitute 20 cm for D in dD/dt and find dD/dt = 0.0127 cm/min Finally, with dD/dt = 0.0127 cm/min, and D=20cm, we can substitute to find dA/dt: dA/dt = 1.6 cm^2/min Hope this helps!
How do I solve this math problem 10/5 +8-10?
There is a rule which guides us how to solve such equations called Order of Operations rule (Operations are /, +, - etc). It is BODMAS Order of Operations - BODMAS. It stands for:Brackets Of Division Multiplication Addition SubtractionThese are the rules you should follow while solving any equation. So lets apply this rule in your equation:There are no Brackets and Ofs, so we do not need to do anything for those. Next is Division so lets performs the division operation wherever needed10/5 + 8 - 10 becomes 2 + 8 - 10Next is Multiplication, and there is no multiplication. So no changesNow comes Addition, lets do that2 + 8 - 10 becomes 10 - 10Last step is Subtraction, lets so that also10 - 10 = 0This is your answer, Zero (0)
How would you solve this GCSE math question?
Umm, maybe if you specified what your mathematical problem is?
How would you solve this? PLease help.?
24500 x 1.05 = 25725 $25725
How would you solve this equation?--Grade 9?
So you are solving for w; you have to get w by itself on one side of the equation, saying w = something. Remember that to do anything to one side of an equation, you have to do the same to the other side in order to keep the equation balanced. So first we would subtract 4w on both sides of the equation, to keep it equal. 6w - 4w + 8 = 4w - 4w + 18 So the 4w and -4w cancel, while the 6w - 4w = 2w 2w + 8 = 18 So lets get rid of the 8; subtract 8 from both sides 2w = 10 Now we have to get rid of the 2 being multiplied by w; to get rid of 2 being multiplied, we divide by 2 2w/2 = 10/2 That leaves w = 10/2 = 5 So, w = 5
How would you solve this series problem?
I’m definitely not a mathematician, but my mind goes something like this.Newton’s approximate is:[math](x^n + y)^{\frac{1}{n}} \approx x + \frac{y}{nx^{n-1}}[/math]We could set n = 2, y = the remainder of the series. Therefore, the series equals approximately:[math]2+\frac{y}{2\times2^{2–1}} = 2 + \frac{y}{2\times2}[/math]Of course, the hard part is y. We can rewrite the remaining series as:[math]y = (4+(4+(4+…)^{\frac{1}{n+3}})^{\frac{1}{n+2}})^{\frac{1}{n+1}}[/math]Interestingly, we find that [math]\lim_{n \to \infty} 4^{\frac{1}{n}} = 1 [/math].So, what is [math]\lim_{n \to \infty} (4+(4+(4+…)^{\frac{1}{n+3}})^{\frac{1}{n+2}})^{\frac{1}{n+1}}[/math] ?Well, I am led to believe that it is [math]\infty[/math] because, you will ultimately have [math]y \approx (1 + 1 + 1 … )^{\frac {1}{2}}= \infty[/math]. If we plug that in for y, in our Newtonian approximate, we get[math]2 + \frac{\infty}{4}[/math]And, two plus infinity divided by four is still infinity.The challenge is that the series converges toward one. If it converged toward zero, we could take a limit and yield a finite result. But, ultimately, the square root of infinity is still infinity.However, that may not be right (me thinking out loud here) because[math]\lim_{y \to \infty} x^{\frac{1}{y}} = 1[/math]Ohhhh, but wait, the problem is the order of the series. If the series was[math](4+(4+(4+…)^{\frac{1}{n-2}})^{\frac{1}{n-1}})^{\frac{1}{n}}[/math]then the limit as n approached infinity would be 1!Pretty sure (about 80% sure) the answer is “infinity.” Hey, if nothing else, I got to practice my LaTeX skills!