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luda_lava [24]
2 years ago
6

Set of integers x such that x - 5 ​

Mathematics
1 answer:
My name is Ann [436]2 years ago
3 0

The set of integers that satisfy the inequality is x ∈ (-∞, 30) where x is an integer.

<h3>What is inequality?</h3>

It is defined as the expression in mathematics in which both sides are not equal they have mathematical signs either less than or greater than known as inequality.

The question is incomplete.

The complete question is:

Set of integers x such that x - 5 is less than 25​.

We have an inequality:

x - 5 < 25

Adding 5 both sides:

x - 5 + 5 < 25 + 5

x < 30

x ∈ (-∞, 30) where x is an integer.

Thus, the set of integers that satisfy the inequality is x ∈ (-∞, 30) where x is an integer.

Learn more about the inequality here:

brainly.com/question/19491153

#SPJ1

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3 years ago
Solve for q q+5/4=-2
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4 0
3 years ago
Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum
lara [203]

Answer:

Hello,

Step-by-step explanation:

We divide the interval [a b] in n equal parts.

\Delta x=\dfrac{b-a}{n} \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^{th} \ rectangle=R(x_i)=\Delta x * y_i\\

\displaystyle \sum_{i=1}^{n} R(x_i)=\dfrac{b-a}{n}*\sum_{i=1}^{n}\  (a^2 +(\dfrac{b-a}{n})^2*i^2+2*a*\dfrac{b-a}{n}*i)\\

=(b-a)^2*a^2+(\dfrac{b-a}{n})^3*\dfrac{n(n+1)(2n+1)}{6} +2*a*(\dfrac{b-a}{n})^2*\dfrac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} R(x_i)\\\\=(b-a)*a^2+\dfrac{(b-a)^3 }{3} +a(b-a)^2\\\\=a^2b-a^3+\dfrac{1}{3} (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=\dfrac{b^3}{3}-ab^2+ab^2+a^2b+a^2b-2a^2b-\dfrac{a^3}{3}  \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =\dfrac{b^3}{3} -\dfrac{a^3}{3}}\\

4 0
2 years ago
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