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sp2606 [1]
3 years ago
5

In the fraction 15⁄7, which of the following is 15?

Mathematics
1 answer:
Ket [755]3 years ago
3 0
The numerator. the number on top is always the numerator of a fraction.                                                                                                                                  
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Kate bought 3 used CDs and 1 used DVD at the bookstore. Her friend Joel bought 2 used CDs and 2 used DVDs at the same store. If
natka813 [3]

Answer:

Cost of 1 DVD is $6.5

Cost of one CD is $4.5

Combine cost of one DVD and CD is $11

Step-by-step explanation:

Given :

Kate bought 3 used CDs and 1 used DVD at the bookstore.

Her friend Joel bought 2 used CDs and 2 used DVDs at the same store.

Kate spent $20

Joel spent $22

To Find :  the cost of a used CD and a used DVD

Solution :

Let cost of 1 used CDs be $x

Let cost of 1 used DVDs be $y

Cost of 3 used CDs = $3x

Cost of 2 used CDs = $2x

Cost of 2 used DVDs = $2y

Kate bought 3 used CDs and 1 used DVD at the bookstore.

She spent $20

So, equation becomes :

⇒3x+y=20 ---(a)

Joel bought 2 used CDs and 2 used DVDs. she spent $22.

So, equation becomes:

⇒2x+2y=22 --(b)

Now solve (a) and (b) to determine the value of x and y

We will use substitution method

Substitute the value of x from (b) in (a)

3(\frac{22-2y}{2} )+y=20

33-3y+y=20

33-2y=20

33-20=2y

13=2y

\frac{13}{2} =y

6.5 =y

Thus cost of 1 DVD is $6.5

Now substitute this value of y in b to get value of x

⇒2x+2(6.5)=22

⇒2x+13=22

⇒2x=22-13

⇒2x=9

⇒x=\frac{9}{2}

⇒x=4.5

Thus the cost of one CD is $4.5

Hence the combine cost of one DVD and CD =  $6.5+ $4.5=$11


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3 years ago
How much pure acid do you mix with 2L of 40% acid to get 70% acid
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Express 3 as power of 3​
Alex17521 [72]

Answer:

3 can be expressed as 3^1.

5 0
3 years ago
Fill in the blank: A piecewise function uses different rules for different intervals of the _________.
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<span>A piecewise function uses different rules for different intervals of the DOMAIN</span>
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List the ordered pairs in the equivalence relations produced by these partitions of {a, b, c, d, e, f, g}: a) {a,b},{c,d},{e,f,g
finlep [7]

Answer:

a)  {a,b} , {c,d} , {e,f,g},{a , b } = {(a,a) (b,b) (a,b) (b,a)},{c , d } = {(c,c) (d,d) (c,d) (d , c) } ,{e,f,g} = {(e,e) (e,f)  (f,e) (e,g) (f,f) (f,g) ( g,e) (g,f)(g,g)}  

b) { (a,a)(b,b)(c,c)(c,d)(d,c)(d,d)(e,e)(e,f)(f,e)(f,f)(g,g)

c)  {(a,a)(a,b)(b,b)(b,a)(a,c)(c,a)(b,c)(c,b)(c,c)(d,a)(a,d)(b,d)(d,b)(d,c)(c,d)(d,d)(c,c)(c,f)(f,c)(f,f)(c,g)(g,c)(f,g)(g,f)(g,g)

Step-by-step explanation:

a) {a,b} , {c,d} , {e,f,g}

  {a , b } = {(a,a) (b,b) (a,b) (b,a)}

  {c , d } = {(c,c) (d,d) (c,d) (d , c) }

  {e,f,g} = {(e,e) (e,f)  (f,e)  (e,g) (f,f) (f,g) ( g,e) (g,f) (g,g)}  

b) {a} ,{b},{c,d} ,{e,f},{g}

     { (a,a)(b,b)(c,c)(c,d)(d,c)(d,d)(e,e)(e,f)(f,e)(f,f)(g,g)}

c)  {a,b,c,d},{e,f,g}

   {(a,a)(a,b)(b,b)(b,a)(a,c)(c,a)(b,c)(c,b)(c,c)(d,a)(a,d)(b,d)(d,b)(d,c)(c,d)(d,d)(c,c)(c,f)(f,c)(f,f)(c,g)(g,c)(f,g)(g,f)(g,g)

4 0
3 years ago
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