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Darya [45]
3 years ago
6

Ben sells homemade cards at a craft fair. He wants to earn more than $50 at the fair. He sells his cards for $2 and he has alrea

dy earned $36. How many cards foes he need to sell to reach his goal? Write and solve an inequality.​
Mathematics
1 answer:
Kryger [21]3 years ago
7 0

Answer:

He must sell 8 cards to reach the minimum goal.

Step-by-step explanation:

Giving the following information:

He wants to earn more than $50 at the fair.

He sells his cards for $2 and he has already earned $36.

<u>First, we need to calculate the money required to reach the minimum goal:</u>

51 - 36= $15

<u>Now, we write the inequality:</u>

2*x >15

x= number of cards sold.

x>15/2

x> 7.5

He must sell 8 cards to reach the minimum goal.

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lina2011 [118]

Answer:

(x - 3)^2 + (y - 4)^2 = 3^2

Step-by-step explanation:

Notice that both endpoints are on the vertical line x = 3.  Thus, to determine the diameter of this circle, we need only subtract 1 from 7, obtaining 6.  The diameter is 6, so the radius is 3.

The center is on the vertical line x = 3 and is halfway between the endpoints of the diameter, that is, at y = 4.  (3, 4) describes this point.

Then we insert the known quantities into the general equation for a circle with center at (h, k) and radius r:  3 for h, 4 for k and 3 for r:

(x - 3)^2 + (y - 4)^2 = 3^2 is the desired equation.

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3 years ago
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damaskus [11]

<u>ANSWER:  </u>

x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

<u>SOLUTION:</u>

Given, f(x)=x^{2}+12 x+24 -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here, for (1) a = 1, b= 12 and c = 24

X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}

\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}

Hence the x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

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Answer:

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Step-by-step explanation:

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