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Original price of a truck 25,000.00

Discount 38%

Tax5%

Find the sell price

I don't get that. Could someone re-explain it differently?

how to solve ((2/15 + 1 7/12)*30/103-2 / 25*0.28125)/((0.375+0.35)+2.75/8.8), please help

((2/15 + 1 + 7/12)×30/103 - 2/25×0.28125)/((0.375 + 0.35) + 2.75/8.8)=191 / 415

y=1x\ 0+1/4

what was stated in the question??? 

It's area is 40square feets

Thanks, that's right!

a=listfor(n, 1, 11, (11+n) nCr 4);print a, "=", sum a

(495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315) = 32857 - total number of solutions.

17=30/2+2

19=22-3+0

20=22-3+0!

21=23-2+0

22=22+(3*0)

23=23+(2*0)

25=23+2+0

26=20+(3*2)

27=(2*2)!+3-0

28=(2*2)!+3+0!

29=32-2-0!

30=32-2-0

31=32-2^0

33=32+2^0

34=32+2+0

35=32+2+0!

37=3(2*2)+0!

40=30+2/.2

17=30/2+2

19=22-3+0

20=22-3+0!

21=23-2+0

22=22+(3*0)

23=23+(2*0)

25=23+2+0

26=20+(3*2)

27=(2*2)!+3-0

28=(2*2)!+3+0!

29=32-2-0!

30=32-2-0

31=32-2^0

33=32+2^0

34=32+2+0

35=32+2+0!

37=3(2*2)+0!

40=30+2/.2

Can u just give me the answer?

Here is the answer to part (f);

Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

Here is the answer to part (f);

Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

Here is the answer to part (f);

Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

Let the 3 pieces by   x,     6 - y   and  y-x     long

Note that when you add these up you get 6

You know that each of these must be between 0 and 6 units long.

GRAPH the sample space.  Find Area of sample space

Now overlay that with the desired area where all the pieces  are less than 3 units long.  Find the area of this event

Probability = area of event/ area of sample space

This is called probability contour mapping.

Guest, please stop posting random answers. 

We know PQ is 18, and by pythagorean theorem, we can conclude that RS (the points of tangency of circles P and Q respectively) has length sqrt(18^2 - 2^2). sqrt(320) simplifies to 8sqrt(5). (unnessecary but you can do it this way)

Let QR = x. By similar triangles, x/8 = (x + PQ)/10

10x = 8x + 8*18

2x = 144

x = 72 = QR

There is no diagram below, please repost your question with "the diagram below" please, thank you.