He can run:
40/5.79 metres per second
that is: 6.9 metres per second (which is how to measure how fast can a body move)
Given that 2 large puzzle costs $16 then
2l=16
or
l=8
That means cost of 1 large puzzle = $8
Given that 3 small puzzle costs $15 then
3s=15
or
s=5
That means cost of 1 small puzzle = $5
Now we will use this information to find the correct choice.
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First choice is <u>wrong </u>as coefficient 8 is with 8l so that gives cost of 1 large not small puzzle.
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Second choice is <u>correct </u>as coefficient 8 is with 8l so that gives cost of 1 large puzzle.
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Third choice is <u>correct </u>as coefficient 5 is with 5s so that gives cost of 1 small puzzle.
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Fourth choice is <u>Wrong </u>as coefficient 5 is with 5s so that gives cost of 1 small not large puzzle.
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Fifth choice is <u>wrong </u>as constant 128 gives total cost of not the total number of puzzles.
L=Lim tan(x)^2/x x->0
Since both numerator and denominator evaluate to zero, we could apply l'Hôpital rule by taking derivatives.
d(tan^2(x))/dx=2tan(x).d(tan(x))/dx = 2tan(x)sec^2(x)
d(x)/dx = 1
=>
L=2tan(x)sec^2(x)/1 x->0
= (2(0)/1^2)/1
=0/1
=0
Another way using series,
We know that tan(x) = x+x^3/3+2x^5/15+.....
then tan^2(x), using binomial expansion gives
x^2+2*x^4/3+.... (we only need two terms)
and again apply l'Hôpital's rule, we have
L=d(x^2+2x^4/3+...)/d(x) = (2x+8x^3/3+...)/1
=0 as x->0
The chance of a king and a ace being drawn out of a deck of cards would be 8 out of 52
Answer:
4^5x-1=16
4^5x-1=4²eliminate 4
5x-1=2
5x=2+1
5x=3
x=3/5
log4(10x+2)=1
log4(10x+2)=log4 ⁴
eliminate log4
10x+2=4
10x=4-2
10x=2
x=2/10
x=1/5
difference between the is we use law of indices and logarithm to the questions respectively
similarities is we eliminate there base