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Nookie1986 [14]

3 years ago

5

A group of 20 adults and x students will visit the Museum of Science and History. Tickets for adults cost \$14 each , and ticket

s for students cost $12 each. Which inequality can be used to determme any number of student tickets that can be purchased with a maximum $300 when 20 tickets for adults are purchased ?

1 answer:

dimaraw [331]3 years ago

4 0

Answer:

280 + 12x ≤ 300

Step-by-step explanation:

A group of 20 adults and x students will visit the Museum of Science and History.

Each adult ticket costs $14 and each student ticket costs $12.

If there are maximum $300 to spend on ticket purchasing, then the inequality that gives the number of student tickets can be purchased will be

$14 \times 20 + 12 \times x \leq 300$

⇒ 280 + 12x ≤ 300 (Answer)

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Answer:

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Step-by-step explanation:

Given

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Required

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Take square roots of both sides

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The options are not clear enough; however the roots of the equation are $b = 5 \sqrt{3} \ or\ b = -5 \sqrt{3}$

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3 years ago

. Use the quadratic formula to solve each quadratic real equation. Round

Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

$\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$ , with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}$

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}$

$\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5$

So B has two solutions of 5 and -1.5.

Now to C!

$\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}$

$\frac{44}{8} = 5.5$

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

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2 years ago