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Georgia [21]
3 years ago
10

The cash price of a bike is $3000. The bike can be bought on hire purchase by making a 25% down-payment and monthly payments of

125 for 2 years. The hire purchase is
A) $3000
B) $750
C) $ 3750
D) $125​
Mathematics
2 answers:
horrorfan [7]3 years ago
8 0
25% downpayment of $750 + $125 x 24 months

= $3750
ASHA 777 [7]3 years ago
3 0

Answer:

C) $3750

Step-by-step explanation:

We can start by calculating the down payment. To do this we can just simply calculate 25% of 3000, which equals 0.25*3000=7500. Now we can find out how much money it will cost to pay not including the down payment. $125 monthly for 2 years= $125 monthly for 24 months= 125*24= $3000. Finally we add the downpayment and the total cost of monthly payment to get $7500+$3000=$3750.

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2 Here are two equations:
MatroZZZ [7]

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Step-by-step explanation:

a. Decide whether each neither of the equations,

i (3,4)

ii. (4,2.5)

ill. (5,5)

iv. (3,2)

To decide whether each point is solution to equations or not we will put the point in the equations

Equations are:

Equation 1: 6x + 4y = 34

Equation 2: 5x – 2y = 15

<u>i (3,4) </u>

Putting in Equation 1:

6(3) + 4(4) = 34\\18+16=34\\34=34\\

Putting in Equation 2:

5(3) - 2(4) = 15\\15-8 = 15\\7\neq 15

<u>ii. (4,2.5)</u>

Putting in Equation 1:

6(4) + 4(2.5) = 34\\24+10=34\\34=34\\

Putting in Equation 2:

5(4) - 2(2.5) = 15\\20-5 = 15\\15=15

<u>ill. (5,5)</u>

6(5) + 4(5) = 34\\30+20=34\\50\neq 34

Putting in Equation 2:

5(5) - 2(5) = 15\\25-10 = 15\\15=15

<u>iv. (3,2)</u>

6(3) + 4(2) = 34\\18+8=34\\26\neq 34

Putting in Equation 2:

5(3) - 2(2) = 15\\15-4 = 15\\11\neq 15

Hence,

(3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. Is it possible to have more than one (x, y) pair that is a solution to both

equations?

The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution

So,

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Keywords: Linear equations, Ordered pairs

Learn more about linear equations at:

  • brainly.com/question/10534381
  • brainly.com/question/10538663

#LearnwithBrainly

8 0
3 years ago
Given the formula K = LMN, what is the formula for M?
Andre45 [30]
To start off devide both sides by M i.e
K/M=LMN/M
-> K/M=LN
Then devide by K to remove it on the left , this will end you up with M=LN/K
So the answer is (c).
Hope this helps :).
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3 years ago
Express the following expression in the form of a + bi: (39 + 34i) + (76 + 89i) - (47 +26i).
andre [41]

Answer:

68+97i

Step-by-step explanation:

work shown and pictured

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What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of
notsponge [240]

Answer:

A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})

Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

A(0, 0), B(2, 0) and C(0, 2).

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?

=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, (x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)

<u>So, The coordinates of the triangle A'B'C' are:</u>

A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})

6 0
3 years ago
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