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alex41 [277]
1 year ago
8

The cost of 6 sandwiches and 4 drinks is $53. The cost of 4 sandwiches and 6 drinks is $47 how much does one sandwich cost

Mathematics
2 answers:
siniylev [52]1 year ago
6 0
$53-$47=$6
2 sandwiches are $6
so 1 sandwich is $3
devlian [24]1 year ago
3 0
Let sandwiches and drinks be s and d resp so according to the question,
by 1st case,
6s + 4d = 53
or, d = (53-6s)/4
again by 2nd case,
4s + 6 d = 47
4s + 6(53-6s)/4 = 47
or, (8s +159-18s)/2=47
or 159-10s = 94
or 65 =10s
or, s = $6.5
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Solve each of the following initial value problems and plot the solutions for several values of y0. Then describe in a few words
Taya2010 [7]

Answer:

Step-by-step explanation:

Answer:

a) y-8 = (y₀-8)  , b) 2y -5 = (2y₀-5)

Explanation:

To solve these equations the method of direct integration is the easiest.

a) the given equation is

          dy / dt = and -8

         dy / y-8 = dt

We change variables

          y-8 = u

         dy = du

We replace and integrate

           ∫ du / u = ∫ dt

           Ln (y-8) = t

We evaluate at the lower limits t = 0 for y = y₀

          ln (y-8) - ln (y₀-8) = t-0

Let's simplify the equation

           ln (y-8 / y₀-8) = t

           y-8 / y₀-8 =

            y-8 = (y₀-8)

b) the equation is

            dy / dt = 2y -5

            u = 2y -5

            du = 2 dy

            du / 2u = dt

We integrate

             ½ Ln (2y-5) = t

We evaluate at the limits

            ½ [ln (2y-5) - ln (2y₀-5)] = t

            Ln (2y-5 / 2y₀-5) = 2t

            2y -5 = (2y₀-5)

c) the equation is very similar to the previous one

             u = 2y -10

             du = 2 dy

             ∫ du / 2u = dt

             ln (2y-10) = 2t

We evaluate

             ln (2y-10) –ln (2y₀-10) = 2t

               2y-10 = (2y₀-10)

4 0
1 year ago
Ella completed the following work to test the equivalence of two expressions.
MrRa [10]
Answer:

The expressions are not equivalent because Ella did not know that you can’t use substitution to test for equivalence.

Step-by-step explanation:

Equivalent algebraic expressions are those expressions which on simplification give the same resulting expression.

Two algebraic expressions are said to be equivalent if their values obtained by substituting any values of the variables are same.

Two expressions 3f+2.6 and 2f+2.6 are not equivalent, because when f=1,

3f + 2.6 = 3.1 + 2.6 = 3 + 2.6 = 5.6

2f + 2.6 = 2.1 + 2.6 = 2 + 2.6 = 4.6

5.6 = 4.6

Method of substitution can only help her to decide the expresssions are not equivalent, but if she wants to prove the expressions are equivalent, she must prove it for all values of f.

3f + 2.6 = 2f + 2.6

3f = 2f

3f - 2f = 0

f = 0

This is true only when f=0.

Hence,

The expressions are not equivalent because Ella did not know that you can’t use substitution to test for equivalence.
6 0
1 year ago
Read 2 more answers
Izaak is a big baseball fan. He is comparing the cost of two options for purchasing tickets. Let g represent the number of baseb
Schach [20]

The single game ticket only costs 8g per ticket. The Season tickets costs 1.19 (I had added them). The 8 is lesser than 1.19. So the Season tickets costs more than the Single Game tickets (I hope this helped!).

6 0
2 years ago
The present age of nitin is two times the present age of his sister pooja. After 7 years their ages will add to 68 years. Find t
svlad2 [7]
N=2p

(n+7) + (p+7) = 68

2p + 7 + p + 7 = 68

3p + 14 = 68

3p = 68 - 14 = 54

p=54/3=18

n=2p=36

Check: (18+7)+(36+7)=25+43=68  good



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1 year ago
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Hunter-Best [27]
It’s called a pi fight
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1 year ago
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