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3:5 means that there are 8 equal parts to the suppplementary angle....and the smaller one is 3/8 of this....so we have

(3/8)(180)    =  67.5°

Here is a crude rendition of a 3 scale ammeter

M is the meter    Rm is the resistance of the meter

S is a switch which places various resistors (R1 and R2) in PARALLEL with the meter resistance

Thanks!

Yes lim is 3.     Use L'Hopital's theorem.  That is   lim  f(x)/ g(x)      =  lim f'(x) /g'x)   and you will get

lim  =ln'(1+3x)  /(x')   =    [ {1/(1+3x)} (3)  ]  /1   = 3/(1 +3x)  which tends to 3/(1  + 0)    = 3

Second one, lim  xcot(x)    = lim x { cos(x)/sin(x)}   = lim  { x cos (x)}  /sin (x)

divide numerator and denomimator by x   , lim  = cos(x) /  { sin(x) /x }

Now,{sin(x)}/x  tends to 1 as x tends to zero    so lim  = cos(x) /1  as x tends to zero,which is   1   since cos zero =1

2+2=4

The fox says talk more about math and less of what he says

Here is another take:

16 is to 5% as x is to 100%    then set it up as a ratio:

16/5% = x/100%                    Multiply both sides by 100%

16/5%  x 100% = x

16/5*100 = 320 = x

16 = 5/100   times   x        where x is the number you want

16= 5x  /100

multiply both sides by  100

1600 = 5x

x= 1600/5

x= 320

18=3|4x-10|         x = ?

\(\begin{array}{|rcll|} \hline 3|4x-10| &=& 18 \quad & | \quad :3 \\ |4x-10| &=& 6 \quad &|\quad \text{square both sides} \\ (4x-10)^2 &=& 6^2 \\ (4x-10)^2 &=& 36 \\ 16x^2-80x+100 &=& 36 \quad & | \quad -36 \\ 16x^2-80x+64 &=& 0 \quad & | \quad :16 \\ x^2-5x+ 4 &=& 0\\ (x-4)(x-1) &=& 0\\ x-4 = 0 & \text{or} & x-1 = 0 \\ x = 4 & \text{or} & x = 1 \\ \hline \end{array}\)

i tried so hard and got so far and the end it doesnt even matter

XD

3) -

Solve for a:
a/(x + 1) + 3/(2 x - 1) = (5 x + 2)/((x + 1) (2 x - 1))

Subtract 3/(2 x - 1) from both sides:
a/(x + 1) = 1/(x + 1)

Multiply both sides by x + 1:
Answer: |a = 1

Thanks for the help guys 

2) - Solve for a:
a/(x - 1) - 2/(x + 1) = 4/((x - 1) (x + 1))

Add 2/(x + 1) to both sides:
a/(x - 1) = 2/(x - 1)

Multiply both sides by x - 1:
Answer: |a = 2

nothing much

nothing much

I think the other two are easier because you just need to multiply both sides by the lowest common denominator.

With the first on, just multiply both sides by  (x-1)(x+1)

that will get rid of all the fractions and make it much more friendly  :)

See if you can do it :)

Hi James,

\(\frac{x}{2(3-x)}-\frac{x^2}{7(x-3)^2}\\ \text{you have to get a common denominator}\\ =\frac{x}{-2(x-3)}-\frac{x^2}{7(x-3)^2}\\ =\frac{-x}{2(x-3)}-\frac{x^2}{7(x-3)^2}\\ =\frac{-x}{2(x-3)}\times \frac{7(x-3)}{7(x-3)}-\frac{x^2}{7(x-3)^2}\times \frac{2}{2}\\ =\frac{-7x(x-3)}{14(x-3)^2}-\frac{2x^2}{14(x-3)^2}\\ =\frac{-7x^2+21x-2x^2}{14(x-3)^2}\\ =\frac{-9x^2+21x}{14(x-3)^2}\\ \)

*

The 2 figures are congruent.

Rotation and reflection does not affect the size of the figures.

\(E^2 - 1 = 3\\ E^2 =4\\ E = \pm 2\)

(8t - 32)(t + 1)

= 8t² - 32t + 8t - 32

= 8t² - 24t - 32

Are you trying to find the value of X?

Hi Rauhan

A cylindrical can of internal radius 20 cm stands upright on a flat surface. It contains water to a depth of 20 cm. Calculate the rise in the level of the water when a brick of brick of volume 1500cm3  is immersed in the water.

A really important thing to remember is that  1cm^3 holds 1mL of water.

so

a 1500cm^3 brick will displace 1500ml of water.

The original height of the water is irrelavant.  Unless the water overflows.

How much height is this?

\(\pi r^2 h=Volume\\ \pi *20^2*h=1500\\ h=\frac{1500}{400\pi}\\ h=\frac{15}{4\pi}\;cm\)

h=1.193662073189215 cm

The water will rise approx 1.2cm