MathProblemSolver101

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Last digit of $196^{213}:$

We only need to care about the last digit. That is $6.$

We notice a repetition in the last digit of every $6^{n}$: it is $6.$

Now, last digit of $213^{196}:$

Similarily, we notice a repetition between $3^{1,2,3,4}:$ it is $3, 9, 7, 1.$

$196$ is a multiple of $4$ so the last digit is $1.$

$6 \cdot 1 = \boxed{6}.$

3 of the same colour

3*2 = 6 of (2 colour,1 colour)

1 of all colour

3+6+1=10

(a) $\frac{1+11+3+5+1}{5}=\frac{21}{5}=\boxed{4.2}.$

(b) All other values except for $1$ appear exactly once. Thus the mode is $\boxed{1}.$

You are correct on both of them.

 - Jimmy

$2^5 = \boxed{32}$

Let $X$ be the position of the be.

Let $A$ be the highest point of the building.

Let $B$ be the lowest point of the building.

Let $C$ be the point that has the same altitude to $X$ (aka the shortest distance, which is what we are trying to find.)

We are given $AB = 400.$

Since we are given two angles and $AB$, we will try to relate $AB$ to $XC.$

Note $AB = AC + BC.$

We have that $AC = XC \tan 4^{\circ} \cong 0.070 XC.$

We also have that $BC = XC \tan 2^{\circ} \cong 0.035 XC.$

Thus we have $AB \cong 0.105 XC.$

And since we also have $AB = 400,$ we have that $0.105 XC \cong 400.$

$XC \cong \frac{200}{21} \cdot 400 \cong \boxed{3800 \text{ feet.}}$

Paul has 2/3 as many postcards as Shawn . The number of postcards Shawn has is 3/5 of the number of postcards Tim has. If the three boys have 280 postcards altogether,how many more postcards does Tim have than Paul ?

$p = \frac{2}{3}s$

$s = \frac{3}{5}t$

$p+s+t=280$

$p=\frac{2}{5}t$

$2t=280$

$t=140$

$(1-\frac{2}{5})t = \frac{3}{5} \cdot 140 = 3 \cdot 28 = \boxed{84}$

I don't think anyone did! 

$5-2(4a+1)+3a=13$

$5-(8a+2)+3a=13$

$(-8a+3a)+(5-2)=13$

$(-8+3)a+3=13$

$-5a=10$

$a=\frac{10}{-5}=-\frac{10}{5}=\boxed{-2}$

I think you meant 54, but good job!