All Answers 355980 Answers

Assuming when you say a music disc in state a is 5 dollars, you mean without sales tax, then here is the solution:

The total is $1.12(5)+1.09(8)=5.60+8.72=\boxed{14.32}$ dollars

Nice, Riemann!!!!

4 (12)  +  5(10)   =  48  + 50   =   98  books total

Sorry, that is wrong because the point B is supposed to travel on a curved path, not just including the start and endpoints.

We  can use  the  Law  of Cosines

AC^2  =  AB^2  + BC^2  -  2(AB* BC) cos (ABC)

40^2  =  25^2  + 25^2  - 2( 25*25) cos (ABC)

cos ABC  =   [  ( 40^2    - 25^2  - 25^2 )   / ( - 2 * 25 *25)  ]

cos ABC  =  -  7/25

sin ABC  =  sqrt   (  1  -(-7/25)^2  )   =  sqrt  [ 625  - 49] / 25  =  sqrt  [ 576] / 25   =  24 / 25

1)

Per week, let's say she goes to school 5 times a week, which is reasonable. So she spends 5 miles to school*5 days=25 miles for school. But, she needs to drive back, so 2*25 miles for school=50 miles for school. Also, her soccer practice is 2 miles, so it's 2 miles*3 days=6 miles. So the total per week is 50+6=56. But wait! She drives 20 miles on the weekend. So the total per week is 56+20=76 miles. Let's estimate 52 weeks per year, so there would be 39 weeks of school(school is about 3/4 of the year), so there would be 76*39

$\approx$80*40=3200 miles per year. But wait! She still drives in the summer 100 miles per week for about 13 weeks, so the total is 3200+13*100=$\boxed{4500}$ miles per year.

2)

Because 3 miles is 1 kilowatt hour, we have that April spends $\frac{4500}{3}=1500$ kilowatt hours. Because it is 0.12 dollars per kilowatt hour, we have that the electricity cost is $0.12(1500)=\boxed{180}$ dollars on electricity (feel free to round this yourself)

3)

Assuming that Megan lives on a highway and not city, so she gets 25 miles per gallon. So this means Megan spends $\frac{4500}{25}=180$ gallons of gas. Because she spends 2.85 dollars per gallon, her total cost is $2.85(180)=\boxed{513}$ dollars on gas costs (feel free to round this yourself)

3/ 5    +   3/ 8   = ?

LCM = 40

24/40    + 15/40   =  39/40 of  a  bowl

M = 1/6 J    or    J = 6M

and

M + 1/2J = 1/2J + 9

     M = 9           then J = 6M = 54

Ok I posted another one of it correctly typed

1483 lb/life   1 life / 75 yr   *  1 yr/52 week =    1483 / (75 * 52) =    ..........   lb/week

1)

Per week, let's say she goes to school 5 times a week, which is reasonable. So she spends 5 miles to school*5 days=25 miles for school. But, she needs to drive back, so 2*25 miles for school=50 miles for school. Also, her soccer practice is 2 miles, so it's 2 miles*3 days=6 miles. So the total per week is 50+6=56. But wait! She drives 20 miles on the weekend. So the total per week is 56+20=76 miles. Let's estimate 52 weeks per year, so there would be 39 weeks of school(school is about 3/4 of the year), so there would be 76*39$\approx$80*40=3200 miles per year. But wait! She still drives in the summer 100 miles per week for about 13 weeks, so the total is 3200+13*100=$\boxed{4500}$ miles per year.

2)

I can't read your question and the statement before it because you messed up the LaTeX, please retype it

3)

I also can't read your question and the statement before it because you messed up the LaTeX, please retype it

4/21 * 77 =  14 2/3 ~~ 15 dogs chosen

$k^6\equiv 4\pmod{10}$ has solutions $k\equiv \pm 2\pmod{10}$.

So $2,8,12,18,22,28, ...$ all work.

10 of the levels is 1/3 of the total.

30 levels in total.

3 out 5 voted for Tommy. This is equal to 9 out of 15 (copy the ratio three times). So 15 voted in total.

It was correct! Thank you so much! The explanation cleared it up. 

I think the link has malware.

It says larger than 3456 @itachiuchihakujo, so exactly 3456 would not work

ivy$=5k$ for some part k

ken$=4k$ for some part k

ron$=2k$ for some part k

$4k-c=0.75(4k)=3k$

$2k+c=1.5(2k)=3k$

$4k-c=3k$

$2k+c=3k$

Setting these equal, ignoring the 3k, we see $4k-c=2k+c$, so:

$4k-c=2k+c$

$2k=2c$

$k=c$.

So Ivy's sweets increased by $k$, so her percentage increase was $\frac{6}{5}=120=\boxed{20}$ percent increase

Let the side length of the base be $s$. Then $\frac{1}{3}s^2\cdot 30=63960$. Solve for $\sqrt{s^2+30^2}$.

Yep, believe so :)

If $a-b=1$ there are $10, 21, 32, ..., 98=9$ integers.

If $a-b=2$ then we have $20, 42, 64, 86=4$ integers.

If $a-b=3$, then we have $30, 63, 96=3$ integers.

If $a-b=4$, then we have $40, 84=2$ integers.

If $a-b=5$, then we have $50=1$ integer.

If $a-b=6$, then we have $60$.

If $a-b=7$, we have $70$, if $a-b=8$ we have $80$, if $a-b=9$ we have $90$.

So it's $9+4+3+2+1+1+1+1+1=\boxed{23}$. 

So 17 then?