The inequality that represents the range of k so that inequality 1 has real solution is -3 < k < 5
<h3>How to determine the range?</h3>
<u>Inequality 1</u>
x^2 + (k + 1)x + k + 4 > 0
The discriminant is calculated using
D = b^2 - 4ac
So, we have:
D = (k + 1)^2 - 4 * 1 * (k + 4)
Evaluate
D = k^2 + 2k + 1 - 4k - 16
This gives
D = k^2- 2k - 15
Factorize the equation
D = (k+ 3)(k -5)
Solve for k
k = -3 and k = 5
Express the k values as inequalities
-3 < k < 5
<u>Inequality 2</u>
The inequality cannot be read, because of the image quality
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Addition and Subtraction in Scientific Notation. A number written in scientific notation is written as the product of a number between 1 and 10 and a number that is a power of 10. That is, it is written as a quantity whose coefficient is between 1 and 10 and whose base is 10.
Answer:
9 - m = 12
Step-by-step explanation:
9 - m = 12
mark as brailnlest:)
Answer:
(3, 2), (2, 3)
Step-by-step explanation:
x + y = 5
xy = 6
Solve the first equation for x.
x = 5 - y
Substitute 5 - y for x in the second equation.
xy = 6
(5 - y)y = 6
5y - y² = 6
y² - 5y + 6 = 0
Factor.
(y - 2)(y - 3) = 0
y - 2 = 0 or y - 3 = 0
y = 2 or y = 3
Now substitute 2 for y in the first equation and solve for x.
x + y = 5
x + 2 = 5
x = 3
One solution is x = 3; y = 2, or (3, 2).
Now substitute 3 for y in the first equation and solve for x.
x + y = 5
x + 3 = 5
x = 2
Another solution is x = 2; y = 3, or (3, 2).
Answer: (3, 2), (2, 3)
Answer:
2b^2
Step-by-step explanation:
according to the question equation is
(a^2 + b^2 ) - (a^2 - b^2)
a^2 + b^2 - a^2 + b^2
plus a square and minus a square gets cancel
b^2 + b^2
since they are like terms they can be added
2b^2
