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fiasKO [112]
3 years ago
7

What is the solution set of the quadratic inequality x^2+x-2>0?

Mathematics
2 answers:
vekshin13 years ago
8 0

Answer is A.

The values of x must be greater than or equal to 1 and the value of x must be less than or equal to -2.

x >= 1.....x <= -2

Please see attachment!


Svetradugi [14.3K]3 years ago
8 0

Answer: The solution set is (-\infty, -2)\cup (1, \infty).

Step-by-step explanation:  We are given to find the solution set of the following quadratic inequality:

x^2+x-2>0.

The solution is as follows:

x^2+x-2>0\\\\\Rightarrow x^2+2x-x-2>0\\\\\Rightarrow x(x+2)-1(x+2)>0\\\\\Rightarrow (x+2)(x-1)>0.

We know that if 'a' and 'b' are two numbers such that a × b > 0, then either both 'a' and 'b' are greater than 0 or both of them are less than 0.

Therefore, we have

either

x+2>0,~~~x-1>0\\\\\Rightarrow x>-2,~~~~x>1\\\\\Rightarrow x>1,

or

x+2

Thus, the required solution set is x < -2 and x > 1. We can also write the solution set in terms of intervals as (-\infty, -2)\cup (1, \infty).

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Step-by-step explanation:

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4 0
3 years ago
Part 3: Write the equation of and graph an ellipse.
tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

The center is halfway between vertices, at (4, -6).

It is also halfway between foci.

:::::

The vertices are vertically aligned, so the parabola is vertical.

General equation for a vertical ellipse:

 (y-k)²/a² + (x-h)²/b² = 1

with

 center (h,k)

 vertices (h,k±a)

 co-vertices (h±b,k)

 foci (h,k±c), c² = a²-b²

Apply your data and solve for h, k, a, and b.

center (h,k) = (4, -6)

h = 4

k = -6

vertices (4,-6±a) = (4,-6±9)

a = 9

foci (4,-6±c) = (4,-6±5√2)

b² = a² - c² = 9² - (5√2)² = 31

b = √31

The equation becomes

 (y+6)²/81 + (x-4)²/31 = 1

:::::

length of major axis = 2a = 18

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8 0
3 years ago
in 25 minutes Li can run 10 laps around the track. consider the number of laps she can run per minute
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She can run 2.5 miles a minute
8 0
3 years ago
Read 2 more answers
The figures shown are similar
Harlamova29_29 [7]
Hi there!

Based on your options, I think we're going from the big figure to the small figure. 12/8 = 1 1/2

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D - 1.5

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8 0
3 years ago
In​ 2000, a company had 1147 stores nationwide. By​ 2002, this total had grown to 1542. If the number of stores continues to gro
mr_godi [17]

12226 stores will be there in 2016

<em><u>Solution:</u></em>

<em><u>The growth function is given as:</u></em>

y = a(1+r)^t

Where,

y is the future value

a is the initial value

r is the growth rate

t is the number of years

From given,

In​ 2000, a company had 1147 stores nationwide

By​ 2002, this total had grown to 1542

Therefore,

y = 1542

a = 1147

t = 2000 to 2002 = 2 years

r = ?

<em><u>Substituting we get,</u></em>

1542 = 1147(1+r)^2\\\\(1+r)^2 = \frac{1542}{1147}\\\\(1+r)^2 = 1.34437

Taking square root of both sides

1+ r = 1.1594\\\\r = 1.1594 - 1\\\\r = 0.1594

<em><u>If the number of stores continues to grow exponentially at the same​ rate, how many stores will there be in​ 2016?</u></em>

Therefore,

y = ?

a = 1147

r = 0.1594

t = 2000 to 2016 = 16 years

<em><u>Substituting we get,</u></em>

y = 1147(1 + 0.1594)^{16}\\\\y = 1147(1.1594)^{16}\\\\y = 1147 \times 10.6593\\\\y = 12226.3311 \approx 12226

Thus 12226 stores will be there in 2016

4 0
3 years ago
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