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Alisiya [41]
4 years ago
13

1. Keisha needs to make at least 28 costumes for the school play. Since she can make 4 costumes each week, Keisha plans to work

on the costumes for at least 7 weeks.
2. If Keisha has to have the costumes complete in 10 weeks or fewer, how will our solution change?

For each question, write an inequality. Then, graph your solution.

Mathematics
1 answer:
snow_lady [41]4 years ago
7 0

Answer:

1. Let n represents the number of weeks she needs to make at least 28 costumes,

Since, she can make 4 costumes each week,

\implies n \geq \frac{28}{4}

\implies n \geq 7-----(1)

i.e. In number line closed circle on 7 and shaded right from 7.

2. If the maximum time taken by her is 10 weeks,

⇒ n ≤ 10 ----(2),

From equation (1) and (2),

7\leq n\leq 10

I.e. closed circle on both 7 and 10 and shaded right from 7 and shaded left from 10.

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Solve 5x^2 − 3x + 17 = 9.
Vedmedyk [2.9K]

Answer:

The roots of the equation are x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

and there are no real roots of the equation given above

Step-by-step explanation:

To solve:

5x² − 3x + 17 = 9

or

⇒ 5x² − 3x + 17 - 9 = 0

or

⇒ 5x² − 3x + 8 = 0

Now,

the roots of the equation in the form ax² + bx + c = 0 is given as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

in the above given equation

a = 5

b = -3

c = 8

therefore,

x = \frac{-(-3)\pm\sqrt{(-3)^2-4\times5\times8}}{2\times5}

or

x = \frac{3\pm\sqrt{9-160}}{10}

or

x = \frac{3+\sqrt{-151}}{10} and x = \frac{3-\sqrt{-151}}{10}

or

x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

here i = √(-1)

Hence,

The roots of the equation are x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

and there are no real roots of the equation given above

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