24$
2/6 of 24$
divide 24 by 6
24/6 = 4
each one sixth will equal 4, since there is two of them it will equal 8$
1/3 of 24$ is equal to 2/6 but simplified
so 8$ + 8$ =16$
then 24$ - 16$ = 8$
8$ left over
he waits until night then goes through the first door.
he kills the dragon
You need to supply the table!!
.
Try using the same approach as indicated here: https://web2.0calc.com/questions/math_76239#r1
Testing:
N/600 = 18/48
N = 600*18/48
N = 225
The formulary (see very top of the page) can be useful here.
https://web2.0calc.com/formulary/math/geometry/cylinder
Like so:
Compute the definite integral: integral_0^π sin^2(x) cos^2(x) dx Write cos^2(x) as 1 - sin^2(x): = integral_0^π sin^2(x) (1 - sin^2(x)) dx Expanding the integrand sin^2(x) (1 - sin^2(x)) gives sin^2(x) - sin^4(x): = integral_0^π (sin^2(x) - sin^4(x)) dx Integrate the sum term by term and factor out constants: = - integral_0^π sin^4(x) dx + integral_0^π sin^2(x) dx Use the reduction formula, integral sin^m(x) dx = -(cos(x) sin^(m - 1)(x))/m + (m - 1)/m integral sin^(-2 + m)(x) dx, where m = 4: = 1/4 sin^3(x) cos(x) right bracketing bar _0^π + 1/4 integral_0^π sin^2(x) dx Evaluate the antiderivative at the limits and subtract. 1/4 sin^3(x) cos(x) right bracketing bar _0^π = (1/4 sin^3(π) cos(π)) - 1/4 sin^3(0) cos(0) = 0: = 1/4 integral_0^π sin^2(x) dx Write sin^2(x) as 1/2 - 1/2 cos(2 x): = 1/4 integral_0^π (1/2 - 1/2 cos(2 x)) dx Integrate the sum term by term and factor out constants: = -1/8 integral_0^π cos(2 x) dx + 1/8 integral_0^π 1 dx For the integrand cos(2 x), substitute u = 2 x and du = 2 dx. This gives a new lower bound u = 2 0 = 0 and upper bound u = 2 π: = -1/16 integral_0^(2 π) cos(u) du + 1/8 integral_0^π 1 dx Apply the fundamental theorem of calculus. The antiderivative of cos(u) is sin(u): = (-(sin(u))/16) right bracketing bar _0^(2 π) + 1/8 integral_0^π 1 dx Evaluate the antiderivative at the limits and subtract. (-(sin(u))/16) right bracketing bar _0^(2 π) = (-1/16 sin(2 π)) - (-(sin(0))/16) = 0: = 1/8 integral_0^π 1 dx Apply the fundamental theorem of calculus. The antiderivative of 1 is x: = x/8 right bracketing bar _0^π Evaluate the antiderivative at the limits and subtract. x/8 right bracketing bar _0^π = π/8 - 0/8 = π/8: Answer: | = π/8
A school survey produced the following results. If about 300 students were surveyed, about how many have 4 or more pets at home?
The average price for gasoline rose from $2.85 to $3.24 in one day. Find the percent increase.
379 x 25% =379 x 25/100 =94.75
Up to last Friday, Apr. 7, 2017, you would have worked for EXACTLY 31 weeks. So, we have:
31 x $450 per week =$13,950 - amount you would have received to Apr. 7, 2017.
Using synthetic divsion, we have.:
-1 [ 1 1 m - 7 ] 1 [ 1 1 m -7 ]
-1 0 -m 1 2 m + 2
___________ ___ ____________
1 0 m ( -7 - m) 1 2 (m + 2) ( m - 5 )
So
-7 - m = m - 5
-2 = 2m
-1 = m
So..... the remainder is (-1 - 5) = -6
811397 : 787 = 1031
I don't have any great insights to offer! I agree the total number of 3s is 12*10^11.
There is a distinction to be made between number of individual digits that are 3s and number of integers that contain 3s. For example there are 20 individual digits that are 3s between 1 and 100, but only 19 different integers that contain 3s (because 33 counts two towards individual digits, but just one towards individual integers). The figure of 12*10^11 refers to individual digits, not individual integers.
4 + 8 + x = 17
x = 5
3^(x+1)=2^(x+2)
\(\begin{array}{|rcll|} \hline 3^{x+1}& = & 2^{x+2} \\ 3^{x+1}& = & 2^{x+1}\cdot 2^1 \quad & | \quad : 2^{x+1} \\ \frac{ 3^{x+1} } {2^{x+1}} & = & 2 \\ (\frac32)^{x+1} & = & 2 \\ (1.5)^{{x+1}} & = & 2 \quad & | \quad \ln \text{both sides} \\ \ln\Big((1.5)^{x+1} \Big) & = & \ln(2) \quad & | \quad \ln(a^b) = b\cdot \ln(a) \\ (x+1)\cdot \ln(1.5) & = & \ln(2) \quad & | \quad : \ln(1.5) \\ x+1& = & \frac{\ln(2)} { \ln(1.5) } \quad & | \quad -1 \\ x & = & \frac{\ln(2)} { \ln(1.5) } -1 \\ x & = & 0.70951129135 \\ \hline \end{array}\)
But heureka answered correctly.
My answer: = z
\(3^{x+1}=2^{x+2}\\ (x+1)\ln 3=(x+2)\ln2\\ x\ln 3 - x\ln 2 = 2\ln 2 - \ln 3\\ x = \dfrac{2\ln 2 - \ln 3}{\ln 3 - \ln 2}\)
Use the calculator and you get the answer.
Your solution and mine posted as "Guest #6", which begins with:"The way to approach this problem is as follows:" are exactly the same!! So, I do agree with your solution! My beef is also with the people who are dividing by 2. I asked Alan and heureka to take a look at it and see if they have anything NEW to add. Got it?.
What do you NOT understand from my solution? tell me and ill explain it further
