In Algebra a term is either a single number or variable, or numbers and variables multiplied together. Terms are separated by + or − signs, or sometimes by divide.

The above operations involve BODMAS theorem( which states the order of preference of operations which are “Brackets Open,Division,Multiplication, Addition, Subtraction” in the decreasing order of preference. So the answer is 3+3*3+3=15 in which 3*3 operation is done first followed the additions.

The answer is 12. Originally Answered: What is 3+3×3-3+3=?

How to solve for exponents

1. xn=y. Take the log of both sides:
2. logxn=logy. By identity we get:
3. n⋅logx=logy. Dividing both sides by log x: n=logylogx. Find the exponent of a number.
4. 3n=81. Take the log of both sides:
5. log3n=log81. By identity we get:
6. n⋅log3=log81. Dividing both sides by log 3: n=log81log3.

Here is the algorithm for finding power of a number….Multiply(res[], x)

1. Initialize carry as 0.
2. Do following for i=0 to res_size-1. …. a. Find prod = res[i]*x+carry. …. b. Store last digit of prod in res[i] and remaining digits in carry.
3. Store all digits of carry in res[] and increase res_size by number of digits.

How to calculate the modulus of a big number on a calculator?

1. I start by dividing the number by the modulus. 41/5=8.2.
2. Remove the integer part of the answer. 0.2.
3. Multiply by the modulus.

How to compute mod of a big number? Given a big number ‘num’ represented as string and an integer x, find value of “num % x” or “num mod x”. Output is expected as an integer. The idea is to process all digits one by one and use the property that xy (mod a) ≡ ((x (mod a) * y) (mod a)).

Pressure is force per unit area: P = F/A in P ascal = N/m2. Work is the force exerted on an object times the distance over which the force is exerted: W = Fx, where W is the work in Nm = Joule. Work done to lift an object of mass m a height h: W = Fgh = mgh. Work done against friction: W = Ff x.

At terminal velocity, the rate of energy transfer for those two processes (GPE ==> KE, and KE ==> air) are equal; GPE is transforming into KE, and KE is transforming into energy of the air at the same rate, which is why speed stops increasing.

The loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy. This can be written in equation form as −ΔPEg = ΔKE. Using the equations for PEg and KE, we can solve for the final speed v, which is the desired quantity.

Examples of Gravitational Potential Energy

• A raised weight.
• Water that is behind a dam.
• A car that is parked at the top of a hill.
• A yoyo before it is released.
• River water at the top of a waterfall.
• A book on a table before it falls.
• A child at the top of a slide.
• Ripe fruit before it falls.

The formula v2=u2+2GS(where v is final velocity, u is initial velocity, G is acceleration due to gravity and S is the linear displacement) assumes that gravity remains constant throughout.