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serious [3.7K]
3 years ago
6

You are considering buying two different vehicles. • The first is a van at $20,523, and an estimated annual operating cost of $3

200. • The second is a fuel-efficient car at $23,540 and an estimated annual operating cost of $2450. Use these equations to represent the cost (y) of the cars over x years: Y = 20,523 + 3200 X Y = 23,540 + 2450 X
Mathematics
1 answer:
torisob [31]3 years ago
4 0

Answer:

Fill in the following tables, using the equations.

Cost of owning and operating the vehicles

Vehicle    0              1              2             3             4               5             6    

Van     $20,523   $23,723  $26,923  $30,123  $33,323 $36,523  $39,723

Car      $23,540  $25,990  $28,440  $30,890 $33,340 $35,790  $38,240

How much does each vehicle cost initially?

  • Van costs $20,523
  • Car costs $23,540

How much does each cost to operate for one year?

  • Van costs $3.200
  • Car costs $2,450

Which one initially costs less?

  • Van

Which one will cost less over the six years?

  • Car

When do they cost about the same?

  • year 4: van accumulated costs $33,323 and car accumulated costs = $33,340

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Answer:

Q13. y = sin(2x – π/2); y = - 2cos2x  

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Step-by-step explanation:

Question 13

(A) Sine function

y = a sin[b(x - h)] + k

y = a sin(bx - bh) + k; bh = phase shift

(1) Amp = 1; a = 1

(2) The graph is symmetrical about the x-axis. k = 0.

(3) Per = π. b = 2

(4) Phase shift = π/2.  

2h =π/2

h = π/4

The equation is

y = sin[2(x – π/4)} or

y = sin(2x – π/2)

B. Cosine function

y = a cos[b(x - h)] + k

y = a cos(bx - bh) + k; bh = phase shift

(1) Amp = 1; a = 1

(2) The graph is symmetrical about the x-axis. k = 0.

(3) Per = π. b = 2

(4) Reflected across x-axis, y ⟶ -y

The equation is y = - 2cos2x  

Question 14

(A) Sine function

(1) Amp = 2; a = 2

(2) Shifted down 1; k = -1

(3) Per = π; b = 2

(4) Phase shift = 0; h = 0

The equation is y = 2sin2x -1

(B) Cosine function

a = 2, b = -1; b = 2

Phase shift = π/2; h = π/4

The equation is

y = -2cos[2(x – π/4)] – 1 or

y = -2cos(2x – π/2) - 1

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Step-by-step explanation:

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A quadratic equation is of the format:

ax² + bx + c = 0,  where a ≠ 0.

The first option is not quadratic, it is a cubic, it has highest power of 3.

The second option f (x) = 3/4 x ^2 + 2x − 5, is quadratic, that is the answer.

The third f(x) = 4/x^2 - 2/x + 1, doesn't qualify because of the 2/x.

The last f(x) = 0x^2 − 9x + 7 , doesn't qualify because of the a = 0, and the rule is that a ≠ 0.

Therefore  f (x) = 3/4 x ^2 + 2x − 5 represents a quadratic function.

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If eggs in the basket are removed two at a time, one egg will remain. If the eggs are removed three at a time, 2 eggs will remai
Mamont248 [21]
<h2>Answer:</h2>

<em><u>Smallest number of eggs in the basket = 119</u></em>

<h2>Step-by-step explanation:</h2>

In the question,

We know that the number of eggs in the basket should be a a multiple of 7.

But it can not be a multiple of 2, 3, 4, 5 and 6 because every time we pick up the eggs 2, 3, 4, 5 or 6 at a time we are left with some eggs with us.

Therefore, the number of eggs can not be a multiple of these numbers.

Now,

Let us say the number of eggs in the basket be 7x.

So,

Let us take the LCM of 2, 3, 4, 5 and 6.

So,

LCM = 60

Now, the number would be greater than 60 and the multiple of 7.

So, checking on all the multiples of 7 above 60 and checking the condition that, the remainder left on dividing by,

2 is 1. (2x + 1)

by 3 is 2. (3x + 2)

by 4 is 3. (4x + 3)

by 5 is 4. (5x + 4)

by 6 is 5. (6x + 5)

So,

On Checking the multiples of 60 which are divisible by 7 are ,

60 + 1 , 120 + 1, 60 - 1, 120 - 1.

So,

For 61 it is not satisfying all the conditions.

121 is also not satisfying all the conditions.

59 is also not satisfying all the conditions.

But,

The number 119 on checking its divisibility by 2. Leaves remainder 1 as, (2(59) + 1).

Divisibility by 3 leaves remainder 2 as, {3(39) + 2}.

Divisibility by 4 leaves remainder 3 as, {4(29) + 3}.

Divisibility by 5 leaves remainder 4 as, {5(23) + 4}.

Divisibility by 6 leaves remainder 5 as, {6(19) + 5}.

Divisibility by 7 leaves remainder 0 as, {7(17) + 0}.

<em><u>Therefore, the minimum number of eggs in the basket are 119.</u></em>

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3 years ago
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