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Hatshy [7]
3 years ago
15

Jessica wants to go to Mexico for spring break.The total cost includes round trip airfare of $150 and $40 per night for a hotel.

She can spend a maximum of $300.Which inequality can be used to find the maximum number of nights Jessica can afford?
Mathematics
1 answer:
mars1129 [50]3 years ago
6 0

Answer:

150 + 40x ≤ 300

Step-by-step explanation:

The flat rate is $150.  "Per" represents multiplication.  $40 per night = 40x

Since she can only spend a MAXIMUM, the sign would be ≤ (less than or equal to) because she can't go over $300.

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Step-by-step explanation:

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Please do step by steps for each!
Schach [20]

Answer:

  • x² - 8x + 12
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Step-by-step explanation:

#1) Find the polynomial with roots at 2 and 6

  • (x -2)(x - 6) = x² - 8x + 12

#2) Find the polynomial with a double root at -3 and another root at 4

  • (x+3)(x+3)(x-4) = (x²+6x+9)(x-4) = x³ + 2x² - 15x - 36

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  • (x -0)(x+3)(x-5) = x(x²-2x - 15) = x³ -2x² - 15x
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Solve the inequality then graph. (you can graph on paper)<br> d+4&lt;-5
soldi70 [24.7K]

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Step-by-step explanation:

4 0
3 years ago
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One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides
Elina [12.6K]
<h2>Answer:</h2>

<em> The side of the triangle is either 38.63ft or 10.35ft</em>

<h2>Step-by-step explanation:</h2>

This problem can be translated as an image as shown in the Figure below. We know that:

  • The side of the square is 10 ft.
  • One of the vertices of an equilateral triangle is on the vertex of a square.
  • Two other vertices are on the not adjacent sides of the same square.

Let's call:

Since the given triangle is equilateral, each side measures the same length. So:

x: The side of the equilateral triangle (Triangle 1)

y: A side of another triangle called Triangle 2.

That length is the hypotenuse of other triangle called Triangle 2. Therefore, by Pythagorean theorem:

\mathbf{(1)} \ x^2=100+y^2

We have another triangle, called Triangle 3, and given that the side of the square is 10ft, then it is true that:

y+(10-y)=10

Therefore, for Triangle 3, we have that by Pythagorean theorem:

(10-y)^2+(10-y)^2=x^2 \\ \\ 2(10-y)^2=x^2 \\ \\ \\ \mathbf{(2)} \ x^2=2(10-y)^2

Matching equations (1) and (2):

2(10-y)^2=100+y^2 \\ \\ 2(100-20y+y^2)=100+y^2 \\ \\ 200-40y+2y^2=100+y^2 \\ \\ (2y^2-y^2)-40y+(200-100)=0 \\ \\ y^2-40y+100=0

Using quadratic formula:

y_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ y_{1,2}=\frac{-(-40) \pm \sqrt{(-40)^2-4(1)(100)}}{2(1)} \\ \\ \\ y_{1}=37.32 \\ \\ y_{2}=2.68

Finding x from (1):

x^2=100+y^2 \\ \\ x_{1}=\sqrt{100+37.32^2} \\ \\ x_{1}=38.63ft \\ \\ \\ x_{2}=\sqrt{100+2.68^2} \\ \\ x_{2}=10.35ft

<em>Finally, the side of the triangle is either 38.63ft or 10.35ft</em>

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