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Lesechka [4]
3 years ago
8

A collection of dimes and quarters is worth $6.50. There are 35 coins in all. How many dimes are there?

Mathematics
1 answer:
DIA [1.3K]3 years ago
8 0
Lets do this like this:
X = number of quarters and y=number of dimes.
Now, we know that the total number is 35 and that the change adds up to 6.50.
One equation we can form is <span>x+y=35</span>Remember the total number of coins is 35.
The other equation is <span>25x+.1y=6.5</span>Because the number of dimes times their value plus the number of quarters times their value gives us a total of 6.5. Now we can solve for x or y easily in the first equation.
Solving for y like this:<span>y=35−x</span>We can substitute this into the other equation to obtain<span>.25x+.1(35−x)=6.5</span>This simplifies to<span>.15x+3.5=6.5</span><span>.15x=3</span><span>x=<span>20
I hope this can help you indeed</span></span>
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vertex = (0, -4)

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Step-by-step explanation:

Given:

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  • parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola:  y=a(x-h)^2+k

(where (h, k) is the vertex and a is some constant)

Substitute point (0, -4) into the equation:

\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}

Substitute point (-2, 8) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}

Substitute point (1, -1) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}

Equate to find h:

\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}

Substitute found value of h into one of the equations to find a:

\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}

Substitute found values of h and a to find k:

\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}

Therefore, the equation of the parabola in vertex form is:

\implies y=3(x-0)^2-4=3x^2-4

So the vertex of the parabola is (0, -4)

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