Nov 14, 2016 · I got 1.23 We can use our usual algorithm for multiplication putting the numbers in column and multiplying; at the end we have to introduce a decimal point accordingly:

Mar 9, 2018 · Rasputin ran 32 miles and walked 9 miles. Let Rasputin ran x miles at 8 mph and walked 41-x miles at 3 mph . He took total 7 hours to complete. Time taken on running is x/8 hours and time taken on walking is (41-x)/3 hours . :. x/8+(41 -x)/3=7. Multiplying by 24 on both sides we get , 3x+8(41-x)=7*24 or 3x+328-8x=168 or -5x= 168-328 or 5x=160 :. x=160/5=32 …

By changing variables, integration can be simplified by using the substitutions x=a\sin(\theta), x=a\tan(\theta), or x=a\sec(\theta). Once the substitution is made the function can be simplified using basic trigonometric identities.

Mar 27, 2016 · #5x-15+2x=41# We can combine our variables on the left to get #7x-15=41# To get our constants on one side, we can add #15# to both sides. We now have the equation #7x=56# Lastly, to completely isolate the variable, we can divide both sides by #7#. We get. #x=8# Hope this helps!

Solving concept. To solve a trig equation, transform it into one, or many, basic trig equations. Solving a trig equation, finally, results in solving various basic trig equations.

Integrating polynomial functions involves applying the reverse steps involved in differentiating polynomial functions. This typically involves using the power rule of integration.

We write #x^2+6x+9-9# If we group it this way: #(x^2+6x+9)-9# then we have a perfect square minus #9# And #(x^2+6x+9)-9=(x+3)^2-9# Solving an equation by completing the square: Solve: #x^2+6x-16=0# (by completing the square) Each of the following equations is equivalent (has exactly the same solutions) as the lines before it. #x^2+6x-16=0 # #x ...

Dec 4, 2016 · int_40^41xsqrt(x-40)color(white).dx=406/15 >I=int_40^41xsqrt(x-40)color(white).dx We should apply the substitution u=x-40. This implies that x=u+40 and that du=dx. When doing the substitution, plug the current bounds of x=40 and x=41 into u=x-40. These give bounds of x=40=>u=40-40=0 and x=41=>u=41-40=1. Thus: I=int_0^1(u+40)sqrtucolor(white).du …

Jul 20, 2015 · 7.41xx10^(-5)=0.0000741 In order to expand 7.41xx10^(-5) into standard notation, the exponent -5 indicates that the decimal should be moved to the left 5 places to get 0.0000741.

May 24, 2015 · How do you divide and reduce the expression to the lowest terms #2xy \-: \frac{2x^2}{y}#?