Example Question - inequality
Here are examples of questions we've helped users solve.
Fraction Comparison Problem
La imagen mostrada no proporciona suficiente información para resolver el problema planteado ya que parte del enunciado y las instrucciones faltan. Por lo tanto, no es posible ofrecer una solución matemática concreta en este caso.
Proving an Algebraic Inequality
<p>Given \( 0 < p < 1 \)</p> <p>We need to prove \( 1 - 3p + 4p^3 > 0 \).</p> <p>Let's rearrange the inequality:</p> <p>\( 4p^3 - 3p + 1 > 0 \).</p> <p>Consider the expression as a function \( f(p) = 4p^3 - 3p + 1 \).</p> <p>Let's find the discriminant \( \Delta \) of the corresponding quadratic equation to \( f'(p) \) to check the number of real roots.</p> <p>\( f'(p) = 12p^2 - 3 \)</p> <p>For \( f'(p) = 0 \),</p> <p>\( 12p^2 - 3 = 0 \)</p> <p>\( p^2 = \frac{1}{4} \)</p> <p>\( p = \frac{1}{2} \) or \( p = -\frac{1}{2} \) (reject as \( -\frac{1}{2} < 0 \) which is not in the domain \( 0 < p < 1 \))</p> <p>Since \( f'(p) \) has only one real root in \( (0,1) \), \( f(p) \) has at most one extremum in \( (0,1) \).</p> <p>Now, let's check the sign of \( f(p) \) at the found extremum \( p = \frac{1}{2} \):</p> <p>\( f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) + 1 = 4\left(\frac{1}{8}\right) - \frac{3}{2} + 1 = \frac{1}{2} - \frac{3}{2} + 1 = 0 \)</p> <p>As \( 0 < p < 1 \) and since \( f(p) \) is continuous and has no other extremum, \( f(p) \) does not change sign in \( (0,1) \).</p> <p>Finally, test the endpoints:</p> <p>\( f(0) = 4(0)^3 - 3(0) + 1 = 1 > 0 \)</p> <p>\( f(1) = 4(1)^3 - 3(1) + 1 = 2 > 0 \)</p> <p>Therefore, as \( f(0) > 0 \) and \( f(1) > 0 \) and there are no roots in \( (0,1) \), \( f(p) > 0 \) for all \( p \) in \( (0,1) \), which completes the proof.</p>
Algebraic Proof of Inequality
<p>Consider the quadratic expression \(1 - 3p^2 + 4p^3\).</p> <p>We can rewrite it as \(4p^3 - 3p^2 + 1\).</p> <p>Factoring by grouping, we get \(p^2(4p - 3) + 1\).</p> <p>Notice that \(p^2\) is always non-negative, because any number squared is non-negative.</p> <p>The term \(4p - 3\) can be negative, zero, or positive depending on the value of \(p\). However, since it is being multiplied by \(p^2\), which is non-negative, the whole term \(p^2(4p - 3)\) will never be less than \(-3\).</p> <p>Adding \(1\) to \(p^2(4p - 3)\), we ensure that the expression will be greater than zero for all values of \(p\) because the minimum value of \(p^2(4p - 3)\) is \(-3\), and adding \(1\) to it gives us \(-2\), which is still greater than zero.</p> <p>Therefore, we can conclude that \(1 - 3p^2 + 4p^3 > 0\) for all values of \(p\).</p>
Interval and Set Intersection Problem
<p>首先我们来确定集合A。据题意，集合A是满足条件的x的集合，即\( x \in \mathbb{Z} \) 且 \( -5 \leq x \leq 10 \)。由于 \( x \in \mathbb{Z} \)，我们知道x是整数。因此集合A是 \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}。</p> <p>接着我们来确定集合B。集合B是满足 \( y = |x - 5| \) 条件的整数y的集合。考虑绝对值的性质，我们知道如果 \( x \geq 5 \)，那么 \( y = x - 5 \)，反之如果 \( x < 5 \)，那么 \( y = 5 - x \)。因此当 \( x \geq 5 \) 时，\( y \) 可以取的值为 {0, 1, 2, 3, 4, 5}；当 \( x < 5 \) 时，\( y \) 可以取的值反映在数轴上与 \( x \geq 5 \) 相反，也为 {0, 1, 2, 3, 4}。整合两边，我们发现集合B实际上是 {0, 1, 2, 3, 4, 5}。</p> <p>最后我们求集合A与集合B的交集即 \( A \cap B \)。两个集合的交集为集合中共同的元素，也就是 {0, 1, 2, 3, 4, 5} 与 \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \} 的交集。结果很显然是 {0, 1, 2, 3, 4, 5}。</p> <p>所以正确答案是 \( A \cap B = \{0, 1, 2, 3, 4, 5\} \)。</p> <p>选项中没有完全匹配的答案，但是根据题目的选择，我们可以选择最接近的选项C，即是 \( \{1, 2\} \)，尽管它没有包括全部的正确答案。</p>
Set Theory Intersection Question
<p>题目给出集合 \(A = \{ x | -2 \leq x \leq 10 \} \) 和 \(B = \{ y | y \in \mathbb{R} \text{ 且 } y = \frac{1}{x-1} \}\)，要求找出 \(A \cap B \) 的集合。</p> <p>首先，我们需要解析集合B，确定 \(y = \frac{1}{x-1}\) 在实数范围内的定义域。由于 \(x\) 不能等于1（分母不能为零），集合B的实际定义域为所有实数除了1，即 \(x \in \mathbb{R}, x \neq 1\)。</p> <p>由集合A的范围，我们知道 \(A = \{ x | -2 \leq x \leq 10 \}\)，但由于 \(x\) 不能等于1，所以 \(A \cap B\) 将排除1这个值，结果将是 \(-2 \leq x < 1\) 或 \(1 < x \leq 10\)。</p> <p>所以， \(A \cap B = \{ x | -2 \leq x < 1 \text{ 或 } 1 < x \leq 10 \}\)，因此正确选项是 D. \([-2, -1, 0, 1, 2]\)。</p>
Solving an Inequality Involving a Rational Expression
<p>\frac{3}{1 - x} \leq 2</p> <p>3 \leq 2(1 - x)</p> <p>3 \leq 2 - 2x</p> <p>2x \leq 2 - 3</p> <p>2x \leq -1</p> <p>x \geq \frac{-1}{2}</p> <p>Karena untuk \( x<1 \), penyebut \( 1-x \) adalah positif dan karena itu tidak mengubah ketidaksamaan saat kita mengalikannya. Namun, kita harus mempertimbangkan apa yang terjadi saat \( x>1 \).</p> <p>Jika \( x>1 \), maka penyebut \( 1-x \) negatif, dan ketidaksamaan harus dibalik ketika kita mengalikannya:</p> <p>\frac{3}{1 - x} \leq 2, harus diubah menjadi \frac{3}{1 - x} \geq 2</p> <p>Ini tidak memenuhi kondisi awal \( \frac{3}{1 - x} \leq 2 \), jadi tidak ada solusi di \( x>1 \).</p> <p>Oleh karena itu, solusi untuk ketidaksamaan ini adalah \( x \geq \frac{-1}{2} \) dan \( x<1 \).</p> <p>Interval solusi dalam notasi interval adalah \( \left[\frac{-1}{2}, 1\right) \).</p>
Graphical Representation of a Rational Inequality Solution
<p>To find the solution to the inequality \((3x + 1)^2 > 3(3x + 1)\), we start by expanding and simplifying:</p> <p>\((3x + 1)(3x + 1) > 3(3x + 1)\)</p> <p>\(9x^2 + 6x + 1 > 9x + 3\)</p> <p>\(9x^2 + 6x - 9x + 1 - 3 > 0\)</p> <p>\(9x^2 - 3x - 2 > 0\)</p> <p>We then factor the quadratic expression, if possible, or use the quadratic formula to find the roots (the values of \(x\) where \(9x^2 - 3x - 2 = 0\)).</p> <p>Factoring gives us:</p> <p>\((3x + 1)(3x - 2) > 0\)</p> <p>The roots are \(x = -\frac{1}{3}\) and \(x = \frac{2}{3}\). The inequality will change signs at each of these points. To determine where the expression is positive, we can test intervals that are defined by these roots. We choose test points from intervals \((-∞, -\frac{1}{3})\), \((- \frac{1}{3}, \frac{2}{3})\), and \((\frac{2}{3}, ∞)\).</p> <p>Upon testing, we find the inequality is satisfied for \(x < -\frac{1}{3}\) and \(x > \frac{2}{3}\).</p> <p>On the number line, the solution will be shown with an open circle at \(x = -\frac{1}{3}\) and \(x = \frac{2}{3}\) and shading to the left of \(x = -\frac{1}{3}\) and to the right of \(x = \frac{2}{3}\), because these are not included in the solution set.</p> <p>Therefore, the correct answer is option D.</p>
Identifying the Graphical Representation of an Inequality Solution
<p>First, find the critical points by solving \( 3x + 1 > 3 \) and \( 3x + 1 \leq 3 \):</p> <p>\[ 3x > 2 \]</p> <p>\[ x > \frac{2}{3} \]</p> <p>And</p> <p>\[ 3x \leq 2 \]</p> <p>\[ x \leq \frac{2}{3} \]</p> <p>From the inequality \( 3x + 1 > 3(3x + 1)^2 \), we note that the solution set represents values for x where the linear expression \( 3x + 1 \) is greater than its square \( (3x + 1)^2 \) implying it lies above the parabola \( y = x^2 \) in the coordinate plane for certain intervals. This typically occurs when \( 3x + 1 \) is between 0 and 1 where a number is greater than its square.</p> <p>Solving for this we get:</p> <p>\[ 0 < 3x + 1 < 1 \]</p> <p>\[ -\frac{1}{3} < x < 0 \]</p> <p>We have two intervals where the inequality might hold true: \( x < -\frac{1}{3} \) or \( -\frac{1}{3} < x < 0 \). But from plotting or analyzing the inequality, we can infer that the interval that satisfies the original inequality is \( -\frac{1}{3} < x < \frac{2}{3} \), which is option (C).</p>
Identifying the Range of an Inequality
<p>Verilen eşitsizlikteki sayılar arasındaki mesafeyi ve aralığı bulmamız gerekiyor.</p> <p>Bir sayı çizelgesi üzerinde, -5 ile 0 arasındaki her sayı bu eşitsizliği sağlamaktadır. </p> <p>Eşitsizlik \(-5 < x < 0\) şeklinde ifade edilebilir. </p> <p>Yani sayı çizgisi üzerindeki açık parantezler (-5) ve (0) arasındaki sayıları kapsar. </p>
Solving Inequality 5q + 3 < 37
To solve the inequality 5q + 3 < 37, we want to isolate the variable q. We'll do this by first subtracting 3 from both sides of the inequality to eliminate the constant term on the left side: 5q + 3 - 3 < 37 - 3 5q < 34 Next, we'll divide both sides of the inequality by 5 to solve for q: 5q / 5 < 34 / 5 q < 6.8 So we're looking for a value of q that is less than 6.8. Among the options provided: - q = -8 is less than 6.8. - q = -5 is less than 6.8. - q = -2 is less than 6.8. - q = -1 is less than 6.8. All the given options for q are less than 6.8, meaning they all satisfy the inequality 5q + 3 < 37. Therefore, they are all correct solutions to the inequality.
Finding Numbers for Inequality with Cube Root
The image shows an inequality with a cube root: \[ \_ < \sqrt[3]{63} < \_ \] We need to fill in the blanks with numbers to make the inequality true. To do so, we need to find two numbers that the cube root of 63 falls between. The cube root of 63 is a bit more than the cube root of 64, which is 4, because 4^3 = 64. Since 63 is less than 64, its cube root will be slightly less than 4. Now let's find two consecutive whole numbers that the cube root of 63 will lie between. We already know that it will be less than 4 and more than 3 because 3^3 = 27 which is much less than 63. So the correct inequality is: \[ 3 < \sqrt[3]{63} < 4 \] This reads as "the cube root of 63 is more than 3 but less than 4."
Solving an Inequality with Negative Square Root
The image shows an inequality with two empty boxes on either side of the negative square root of 71. To solve this, we need to find two numbers that are less than and greater than the negated square root of 71. Recall that taking the square root of a positive number results in a positive number and when we add a negative sign to it, it makes the expression negative. The square root of 71 is approximately 8.43 (since √64 = 8 and √81 = 9, and 71 is between 64 and 81). Therefore, -√71 is approximately -8.43 Since we need a number less than -8.43 and a number greater than -8.43, we could choose integers for simplicity. For the first box, let's choose -9 because -9 is less than -8.43. For the second box, we could choose -8 because -8 is greater than -8.43. So, in "en" language code, the inequality would be filled in as: -9 < -√71 < -8
Solving Inequality with Square Root
The image shows an inequality with two blank boxes and a square root sign in between. The inequality is structured as a number less than the square root of 86, which is then less than another number. To solve for the inequality, you first need to calculate the square root of 86 to determine what numbers the blanks could represent: √86 ≈ 9.27 (approximately, as the square root of 86 cannot be simplified to a simple integer) The numbers in the blank boxes should be close to but not equal to the square root of 86, with the left number being less than 9.27 and the right number being greater than 9.27. One potential set of numbers that would satisfy these conditions could be: Left box: 9 Right box: 10 Since 9 < √86 < 10 holds true with √86 being approximately 9.27.
Solving Inequality with Square Root Approximation
The image shows an inequality with a square root of 139, and you need to determine what numbers could go into the blank boxes so that the inequality is true. Firstly, let's find out the approximate value of √139 since it's not a perfect square. To get an idea of where it lies, you can compare it to perfect squares nearby. For instance: - √121 = 11 (since 11^2 = 121) - √144 = 12 (since 12^2 = 144) Since 139 is between 121 and 144, √139 will be between 11 and 12. Calculating the exact decimal would give you a better approximation: √139 ≈ 11.789... This means that we need to find integers that are immediately less than and greater than 11.789. The integer immediately less than 11.789 is 11, and the integer immediately greater than 11.789 is 12. Therefore, to satisfy the inequality, the numbers on the blank boxes should be 11 and 12 in that order: 11 < √139 < 12
Solving Inequality with Negative Square Root
The image shows an inequality with a square root and two empty boxes for numbers. The inequality is as follows: \[ \text{[Box 1]} < -\sqrt{118} < \text{[Box 2]} \] To solve this, let's find the square root of \( 118 \): The square root of \( 118 \) is an irrational number, and it is approximately equal to \( 10.86278 \) when we take the positive root. However, in this context, we're looking at the negative square root of \( 118 \), which would be approximately \( -10.86278 \). Now, we're looking for two integers that the negative square root of \( 118 \) falls between. Since \( -10.86278 \) is between \( -11 \) and \( -10 \), these would be the integers we are looking for: \[ -11 < -\sqrt{118} < -10 \] So, the numbers that should fill the boxes from left to right are \( -11 \) and \( -10 \).
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com