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olga nikolaevna [1]
3 years ago
6

FOR 20 points

Mathematics
1 answer:
DochEvi [55]3 years ago
3 0

Answer:

one tin of cheese $23

one tin of caramel $27

Step-by-step explanation:

let 'x' = cost of cheese tin

let 'y' = cost of caramel tin

10x + 8y = 446

22x + 11y = 803

i multiplied the first equation by -11 and the second by 8 to eliminate the 'y' terms

  -110x - 88y = -4906

+  <u>176x + 88y = 6424</u>

    66x = 1518

     x = 1518 / 66

     x = 23

find 'y':  10(23) + 8y = 446

             230 + 8y = 446

              8y = 216

               y = 27

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6 0
3 years ago
Which of the following has a solution set of {x | x = 0}?
notka56 [123]

Answer:

(b)  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

Step-by-step explanation:

Here, the given expression is : {x | x = 0}

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Now, take each option and solve the given expression:

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Adding -1 BOTH sides, we get:

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Adding -1 BOTH sides, we get:

x + 1 -1 < 1  -1

or, x <0 ⇒ x = { -∞ , .... , -4,-3,-2,-1}

So, (x + 1 < -1) ∩ (x + 1 < 1) =  { -∞ , .... , -4,-3}∩ { -∞ , .... , -4,-3,-2,-1}  

=  { -∞ , .... , -4,-3}

⇒ (x + 1 < -1) ∩ (x + 1 < 1) ≠ {0}

Similarly, solving

(b) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1)

(x + 1 ≤ 1) =  x≤ 0 =   { -∞ , .... , -4,-3,-2,-1, 0}

(x + 1 ≥ 1) =  x ≥ 0 =  {0,1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {0}

(b)(x + 1 < 1) ∩ (x + 1 > 1)

(x + 1 < 1) =  x <  0 =   { -∞ , .... , -4,-3,-2,-1}

(x + 1 > 1) =  x  > 0 =  { 1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = Ф

Hence,  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

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