Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

at Slide Arrange Tools Extensions Help Accessibility Background Layout Theme Transition Let x represent any number in the set of even integers greater than 1 . Which inequality is true for all values of x ?

at Slide Arrange Tools Extensions Help Accessibility\newlineBackground\newlineLayout\newlineTheme\newlineTransition\newlineLet x x represent any number in the set of even integers greater than 11 .\newlineWhich inequality is true for all values of x x ?

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Precalculusright chevron icon
Solve higher-degree inequalitiesright chevron icon

Full solution

Q. at Slide Arrange Tools Extensions Help Accessibility\newlineBackground\newlineLayout\newlineTheme\newlineTransition\newlineLet x x represent any number in the set of even integers greater than 11 .\newlineWhich inequality is true for all values of x x ?
  1. Define Even Integers Set: Let's define the set of even integers greater than 11. An even integer can be represented as 2n2n, where nn is an integer. Since xx is greater than 11, we start with n=1n = 1, which gives us the smallest even integer greater than 11 as 2(1)=22(1) = 2. Therefore, xx can be represented as 2n2n for n≥1n \geq 1.
  2. Establish Inequality x≥2x \geq 2: We need to find an inequality that is true for all values of xx in the form of 2n2n, where n≥1n \geq 1. Since xx is even and greater than 11, xx will always be greater than or equal to 22. This can be written as x≥2x \geq 2.
  3. Test Inequality for All Values: To check if our inequality x≥2x \geq 2 is true for all even integers greater than 11, we can test a few values. Let's test for n=1n = 1, which gives us x=2(1)=2x = 2(1) = 2. The inequality holds because 2≥22 \geq 2. Now let's test for n=2n = 2, which gives us x=2(2)=4x = 2(2) = 4. The inequality holds because 4≥24 \geq 2. We can see that for any n≥1n \geq 1, the inequality will hold because multiplying by 22 will always give us an even number greater than or equal to 22.

More problems from Solve higher-degree inequalities

Question
Solve for xx\newline3(x+2)(x+5)>03(x+2)(x+5) > 0\newlineWrite the solution using interval notation. Use the union symbol ∪\cup to express the solution as a union of disjoint intervals. Finite endpoints of all intervals should be integers. If there are no solutions, use the symbol ∅\varnothing for the empty set. Use the set notation {a}\{a\} to represent an isolated solution aa. \newline□\square
Get tutor helpright-arrow

Posted 3 months ago

Question
Solve the equation. Check your solution.\newline14(−16k+52)=8+10k\frac{1}{4}(-16 k+52)=8+10 k\newlineThe solution set is _____\_\_\_\_\_ (Type an integer or a simplified fraction).
Get tutor helpright-arrow

Posted 6 days ago

Question
Solve:\newline4x−2x−1≤3 \frac{4 x-2}{x-1} \leq 3 \newlinex∈ x \in (Enter your answer in INTERVAL notation, using U U to indicate a union of intervals; or enter DNE if no solution exists)
Get tutor helpright-arrow

Posted 2 months ago

Question
The Hodge conjecture is that for projective algebraic varieties, Hodge cycles are rational linear combinations of algebraic cycles.\newlineHdg(k)(X)=H(2k)(X,Q)∩H(k,k)(X)\text{Hdg}^{(k)}(X) = \text{H}^{(2k)}(X,\mathbb{Q}) \cap \text{H}^{(k,k)}(X).\newlineWe call this the group of Hodge classes of degree 2k2k on XX.\newlineThe modern statement of the Hodge conjecture is:\newlineLet XX be a non-singular complex projective manifold. Then every Hodge class on XX is a linear combination with rational coefficients of the cohomology classes of complex subvarieties of XX.
Get tutor helpright-arrow

Posted 2 months ago

Question
Given:\newlineUniversal Set U=2,4,5,6 \mathrm{U}=2,4,5,6 \newlineSubset B = 44\newlineWhat is the complement of Set B in Set U?\newline{4} \{4\} \newline{2,4,5,6} \{2,4,5,6\} \newline{2,5,6} \{2,5,6\} \newline{2,4,5} \{2,4,5\}
Get tutor helpright-arrow

Posted 2 months ago

Question
Given:\newlineUniversal Set U=5,6,10,11,12 \mathrm{U}=5,6,10,11,12 \newlineSubset B =10 =10 \newlineWhat is the complement of Set B in Set U?\newline{5,6,11,12} \{5,6,11,12\} \newline{5,6,10,11} \{5,6,10,11\} \newline{10} \{10\} \newline{5,6,10,11,12} \{5,6,10,11,12\}
Get tutor helpright-arrow

Posted 2 months ago

Question
Given:\newlineUniversal Set U=4,5,6,7,11,12 \mathrm{U}=4,5,6,7,11,12 \newlineSubset B = 44\newlineWhat is the complement of Set B in Set U?\newline{4,5,6,7,11,12} \{4,5,6,7,11,12\} \newline{5,6,7,11,12} \{5,6,7,11,12\} \newline{4,5,6,11} \{4,5,6,11\} \newline{4} \{4\}
Get tutor helpright-arrow

Posted 2 months ago

Question
Given:\newlineUniversal Set U=4,5,7,8,10 \mathrm{U}=4,5,7,8,10 \newlineSubset B =8 =8 \newlineWhat is the complement of Set B in Set U?\newline{4,5,7,8,10} \{4,5,7,8,10\} \newline{8} \{8\} \newline{4,5,8} \{4,5,8\} \newline{4,5,7,10} \{4,5,7,10\}
Get tutor helpright-arrow

Posted 2 months ago

Question
Given:\newlineUniversal Set U=2,5,7,9,10 \mathrm{U}=2,5,7,9,10 \newlineSubset B = 22\newlineWhat is the complement of Set B in Set U?\newline{2,5,7,9,10} \{2,5,7,9,10\} \newline{5,7,9,10} \{5,7,9,10\} \newline{2,5,7,10} \{2,5,7,10\} \newline{2} \{2\}
Get tutor helpright-arrow

Posted 2 months ago

Question
In which of the following does the shaded region represent the solution set to y≤32x+2 y \leq \frac{3}{2} x+2 ?\newlineChoose 11 answer:\newline(A)\newline(B)\newline(c)\newline(D)
Get tutor helpright-arrow

Posted 18 hours ago

View all (23)