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3 years ago

Did I get this right? And please explain why or why not please

Mathematics

1 answer:

schepotkina [342]3 years ago

3 0

$\bf 7x^2-10\implies 7(\stackrel{x}{6})^2-10\implies 7(36)-10\implies 252-10\implies 242
\\\\\\
43\ \textless \ y-30\implies 43+30\ \textless \ y\implies 73\ \textless \ y\qquad y=
\begin{cases}
74\\
75\\
76\\
77\\
...+\infty
\end{cases}$

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3 years ago

What is the solution of the system of equations shown in the graph?

Rudiy27

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3 years ago

A man has 32 coins in his pocket, all of which are dimes and quarters. If the total value of his change is 545 cents, how many d

inn [45]

A dime is worth 10 cents and a quarter is worth 25 cents.

Let D be the number of dimes and Q be the number of quarters.

Since the total amount of coins in the pocket is 32, then:

$D+Q=32$

On the other hand, the total value of D dimes is 10D, while the total value of Q dimes is 25Q. Then, the total value of D dimes and Q cuarters is 10D+25Q, which must be equal to 545. Then:

$10D+25Q=545$

Notice that we have found a 2x2 system of equations:

$\begin{gathered} D+Q=32 \\ 10D+25Q=545 \end{gathered}$

Solve the system using the substitution method. To do so, isolate D from the first equation and replace the expression for D into the second equation to obtain a single equation in terms of Q:

$\begin{gathered} D+Q=32 \\ \Rightarrow D=32-Q \\ \\ 10D+25Q=545 \\ \Rightarrow10(32-Q)+25Q=545 \\ \Rightarrow320-10Q+25Q=545 \\ \Rightarrow25Q-10Q=545-320 \\ \Rightarrow15Q=225 \\ \Rightarrow Q=\frac{225}{15} \\ \\ \therefore Q=15 \end{gathered}$

Replace back Q=15 into the expression for D to find the amount of dimes:

$\begin{gathered} D=32-Q \\ =32-15 \\ =17 \end{gathered}$

Therefore, the amount of dimes that the man has is 17.

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1 year ago