Iampanda

since it could land in a-z order first try it could take only 1 1/2 seconds but if it was on the very last try it would (26^26)x1.5/amount of seconds in a year. 

we are trying to find the real numbers that satisfy 1/(64a^3)-7=0

first we add 7 to both sides 1/(64a^3)=7

then we multiply  to get     1=488a^3+49

subtract 49 to get -48=488a^3

-48/488=a^3

-3/28=a^3

so we have cuberoot(-3/28)=a

hope this helps

we can write 2 equations 40 mph is d/40 and 55 mph is d/55 since we saved 25 min we can write d/40-d/55=5/12 the lcm of 40 and 55 is 440

so we have 11d/440-8d/440=5/12

3d/440=5/12

36d=2200

18d=1100

9d=550

d=550/9 miles

which is around 61.1 miles

To find the asnwer we can plug in (-21-sqrt201)/100 as r

we square that number since we have to find x^2 to get (642+42sqrt201)/100 mutiply that 5 since that is the coeffcient to get (3210+210sqrt201)/100 and simplify to 32.1+2.1sqrt201

Then we do 21 times our number to get (-441-21sqrt201)/10 to get -44.1-2.1sqrt201

so we have

32.1+2.1sqrt201-44.1-2.1sqrt201+v=0

(32.1-44.1)+(2.1sqrt201-2.1sqrt201)+v=0 canceling gives us

-12+v=0

v=12

soorry for the misspelled words

We can first expand the left side to get 2x^2-x-28=-31+jx

Then we shift everything to one side to get 2x^2-x-jx+3=0 and then to get 2x^2-(1+j)+3=0

We know that if the it only has one real solution then we can use the discriminant condition from teh quadratic equation to get

sqrt(b^2-4ac)=0

we blu everything in to get sqrt((1+j)^2-24)=0

(1+j)^2-24=0 by squaring both sides

(1+j)^2=24 by adding 24 to both sides

1+j=sqrt24 by square rooting both sides

j=-1±sqrt24

j=-1±2sqrt6

Using Apollonius's Theorem we know that the median from the right angle is half the hypotenuse so the answer is 2x1 which is 2

We want to find the sum of f(0)+f(1)+f(2)+f(3)+f(4) where f(n) is the remainder of n^8+n^16 divided by 5

We start with 0 and get 0 so the remainder is 0

Then we plug in 1 to get 1+1 is 2 so the remainder is 2

For f(2) we know 2^8 is 256. 256 ends in a 6 so 2^16 which is (2^8)^2 ends in 6. 6+6=12 so 12/5 the remainder is 2

For f(3) we know 3^4 is 81 so the last diit of 3^8 is 1 then square 3^8 to get 3^16 and the last digit is also 1 so 1+1=2 so the remainder is also 2

For f(4) is the same as f(2) since it is bascically square all of f(2) so the remainder is still 2

Adding 0+2+2+2+2=8

To find the leading term you multiply the leading term of the functions. (-7x^4)(2x^8)=(-7)(2)(x^4)(x^8)=-14x^12

Leading term:-14x^12

The leading coefficient is -14 since that is the coefficient of the leading term

Leading coefficient: -14

Degree:12

To get the constant term you multiply -5 and 2 because they are the 2 constants

Constant term: -10

To find the coefficient of x^6 we multiply

(-7x^4)(18x^2)=-126x^6

(-3x^2)(-11x^4)=33x^6

(2)(5x^6)=10x^6

-126+33+10=-83

Coefficient of x^6: -83

To find the coefficient of x^2 you have to multiply

(x^2)x1=x^2

and

(x)(8x+7x)=(x)(15x)=15x^2

you add x^2+15x^2 to get 16x^2 so the answer is 16