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Zina [86]
3 years ago
15

PLEASE HELP!! Dee bought 6 dolls and 2 toy trains for $55. At the same prices, Joy bought 4 dolls and 7 toy trains for $65. What

is the price of a doll?
Mathematics
1 answer:
GuDViN [60]3 years ago
5 0

Answer:

The cost of 1 doll   =  $ 7.5

and The  Cost of 1 toy train =   $ 5

Step-by-step explanation:

Let the price of 1 doll  = $ x

The price of  1 toy train  = $ y

Now, according to the question:

6 x + 2 y  = $ 55

and  4 x   + 7 y = $ 65

Solving the given system of equation,

From (1), we get

x= \frac{55 - 2y}{6}

4 ({\frac{55 - 2y}{6}} )  + 7y = 65

or, solving for y , we get

110y - 4y + 21 y = 195

or, 17 y = 85

⇒ y= 85/17 = 5

This implies :  6 x + 2 (5)   =55

or, 6x = 55 - 10 = 45

or, x =  45/ 6 = 15/2

Hence, the cost of 1 doll  = x = $ 7.5

and the Cost of 1 toy train =  y = $ 5

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Katena32 [7]

Answer:

y= -240750/11

Step-by-step explanation:

44y + 321. 30000 = 0

44y = - 963000

y= -240750/11

6 0
3 years ago
Please write:
liubo4ka [24]

Answer:

1. No vertical compression or stretching but will open up downward.

2. 1/4 vertical stretch

3. 4 vertical compression

Step-by-step explanation:

Any time your dealing with vertical stretching or compression it will always be the number before the parentheses.

So in 1. it is - or basically -1 which means the parabola will be open up downward. And in this problem there is no stretching or compression

2. Is the 1/4 is a vertical stretch

3. 4 is a vertical compression.

Or what also could help is when the number before the parentheses is bigger such as number 4 the closer together the parabola will be. And the smaller the number like 1/4 the wider the parabola will be.

8 0
3 years ago
What is 65 less than j
lisabon 2012 [21]

j-65

hope this helps :)


6 0
3 years ago
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2 x ( __ x 7 ) - ( __ x 6 )
Maslowich
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7 0
3 years ago
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Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One common ch
enot [183]

Answer:

y=\frac{-7t^2+22t-7}{7t-22}

Step-by-step explanation:

We are given that

Initial value problem

y'=(t+y)^2-1, y(3)=4

Substitute the value z=t+y

When t=3 and y=4 then

z=3+4=7

y'=z^2-1

Differentiate z w.r.t t

Then, we get

\frac{dz}{dt}=1+y'

z'=1+z^2-1=z^2

z^{-2}dz=dt

Integrate on both sides

-\frac{1}{z}dz=t+C

z=-\frac{1}{t+C}

Substitute t=3 and z=7

Then, we get

7=-\frac{1}{3+C}

21+7C=-1

7C=-1-21=-22

C=-\frac{22}{7}

Substitute the value of C then we get

z=-\frac{1}{t-\frac{22}{7}}

z=\frac{-7}{7t-22}

y=z-t

y=\frac{-7}{7t-22}-t

y=\frac{-7-7t^2+22t}{7t-22}

y=\frac{-7t^2+22t-7}{7t-22}

8 0
3 years ago
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