Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
If \(x y+y z+z x=0\), then \(\left(\frac{1}{x^{2}-y z}+\frac{1}{y^{2}-z x}+\frac{1}{z^{2}-x y}\right)(x, y, z \neq 0)\) is equal to (a) 0 (b) 3 (c) 1 (d) \(x+y+z\)
Given: \[ x y+y z+z x=0 \] We have the expression to evaluate: \[ \begin{aligned} & \frac{1}{x^2-y z}+\frac{1}{y^2-x z}+\frac{1}{z^2-x y} \\ & \frac{1}{x^2-y z}=\frac{1}{x^2-(-x y-x z)}=\frac{1}{x(x+y+z)} \\ & \frac{1}{y^2-x z}=\frac{1}{y^2-(-x y-y z)}=\frac{1}{y(x+y+z)} \\ & \frac{1Read more
Given:
\[
x y+y z+z x=0
\]
We have the expression to evaluate:
\[
See less\begin{aligned}
& \frac{1}{x^2-y z}+\frac{1}{y^2-x z}+\frac{1}{z^2-x y} \\
& \frac{1}{x^2-y z}=\frac{1}{x^2-(-x y-x z)}=\frac{1}{x(x+y+z)} \\
& \frac{1}{y^2-x z}=\frac{1}{y^2-(-x y-y z)}=\frac{1}{y(x+y+z)} \\
& \frac{1}{z^2-x y}=\frac{1}{z^2-(-y z-x z)}=\frac{1}{z(x+y+z)} \\
& \text { so } \frac{1}{x^2-y z}+\frac{1}{y^2-x z}+\frac{1}{z^2-x y} \\
& =\frac{1}{x(x+y+z)}+\frac{1}{y(x+y+z)}+\frac{1}{z(x+y+z)} \\
& =\frac{1}{x+y+z} \times\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \\
& =\frac{(x y+yz+zx)}{x y z(x+y+z)}=0
\end{aligned}
\]
If \(\mathrm{a}=\frac{\mathrm{b}^{2}}{\mathrm{~b}-\mathrm{a}}\) then the value of \(\mathrm{a}^{3}+\mathrm{b}^{3}\) is (a) 2 (b) \(6 \mathrm{ab}\) (c) 0 (d) 1
Given the equation \(a = \frac{b^2}{b-a}\), we can rearrange this equation to find a relationship between \(a\) and \(b\). Multiplying both sides by \(b-a\) gives: \[ a(b - a) = b^2 \] Expanding the left side: \[ ab - a^2 = b^2 \] Rearranging terms: \[ ab = a^2 + b^2 \] Now, we are asked to find theRead more
Given the equation \(a = \frac{b^2}{b-a}\), we can rearrange this equation to find a relationship between \(a\) and \(b\). Multiplying both sides by \(b-a\) gives:
\[
a(b – a) = b^2
\]
Expanding the left side:
\[
ab – a^2 = b^2
\]
Rearranging terms:
\[
ab = a^2 + b^2
\]
Now, we are asked to find the value of \(a^3 + b^3\). We know the identity for the sum of cubes is:
\[
a^3 + b^3 = (a + b)(a^2 – ab + b^2)
\]
Substituting \(ab = a^2 + b^2\) into the identity:
\[
a^3 + b^3 = (a + b)(0)
\]
Since anything multiplied by 0 is 0:
\[
a^3 + b^3 = 0
\]
Therefore, the value of \(a^3 + b^3\) is \(0\), and the correct option is:
(c) 0
See lessLet \(\mathrm{a}=\sqrt{6}-\sqrt{5}, \mathrm{~b}=\sqrt{5}-2, \mathrm{c}=2-\sqrt{3}\). Then point out the correct alternative among the four alternatives given below. (a) a \(<\) b \(<\) c (b) b \(<\) a \(<\) c (c) a \(<\) c \(<\) b (d) b \(<\) c \(<\) a;
Given the values for \(a\), \(b\), and \(c\) as: - \(a = \sqrt{6} - \sqrt{5}\) - \(b = \sqrt{5} - 2\) - \(c = 2 - \sqrt{3}\) And using the approximate square root values: - \(\sqrt{6} \approx 2.45\) - \(\sqrt{5} \approx 2.24\) - \(\sqrt{3} \approx 1.73\) We can approximate the values of \(a\), \(b\)Read more
Given the values for \(a\), \(b\), and \(c\) as:
– \(a = \sqrt{6} – \sqrt{5}\)
– \(b = \sqrt{5} – 2\)
– \(c = 2 – \sqrt{3}\)
And using the approximate square root values:
– \(\sqrt{6} \approx 2.45\)
– \(\sqrt{5} \approx 2.24\)
– \(\sqrt{3} \approx 1.73\)
We can approximate the values of \(a\), \(b\), and \(c\) as:
– \(a \approx 2.45 – 2.24 = 0.21\)
– \(b \approx 2.24 – 2 = 0.24\)
– \(c \approx 2 – 1.73 = 0.27\)
Thus, we have the order:
\[
a < b < c
\]
Therefore, the correct alternative among the given options is:
(a) \(a < b < c\)
See lessIf \(a^{2}+b^{2}+c^{2}=2(a-b-c)-3\), then the value of \(2 a-3 b+4 c\) is (a) 1 (b) 7 (c) 2 (d) 3
Given the equation \(a^{2}+b^{2}+c^{2}=2(a-b-c)-3\), we can rewrite it as: \[ a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0 \] This equation can be rearranged to form perfect squares: \[ (a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0 \] The expressions on the left-hand side can be rewritten as the sqRead more
Given the equation \(a^{2}+b^{2}+c^{2}=2(a-b-c)-3\), we can rewrite it as:
\[
a^2 + b^2 + c^2 – 2a + 2b + 2c + 3 = 0
\]
This equation can be rearranged to form perfect squares:
\[
(a^2 – 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0
\]
The expressions on the left-hand side can be rewritten as the squares of binomials:
\[
(a – 1)^2 + (b + 1)^2 + (c + 1)^2 = 0
\]
For the sum of squares to be zero, each square must individually be zero. Thus, we have:
\[
(a – 1)^2 = 0, \quad (b + 1)^2 = 0, \quad (c + 1)^2 = 0
\]
Solving these equations gives:
\[
a = 1, \quad b = -1, \quad c = -1
\]
Substituting these values into the expression \(2a – 3b + 4c\):
\[
2(1) – 3(-1) + 4(-1) = 2 + 3 – 4 = 1
\]
Therefore, the value of \(2a – 3b + 4c\) is \(1\).
The correct answer is (a) 1.
See lessIf \(x=\sqrt{3}+\sqrt{2}\), then the value of \(x^{3}-\frac{1}{x^{3}}\) is : (a) \(14 \sqrt{2}\) (b) \(14 \sqrt{3}\) (c) \(22 \sqrt{2}\) (d) \(10 \sqrt{2}\)
Given \(x = \sqrt{3} + \sqrt{2}\), to find the value of \(x^3 - \frac{1}{x^3}\), we can proceed by calculating \(x^3\) and \(\frac{1}{x^3}\) separately. First, calculate \(x^3\): \[ x^3 = (\sqrt{3} + \sqrt{2})^3 \] Applying the binomial expansion, we get: \[ (\sqrt{3})^3 + 3(\sqrt{3})^2(\sqrt{2}) +Read more
Given \(x = \sqrt{3} + \sqrt{2}\), to find the value of \(x^3 – \frac{1}{x^3}\), we can proceed by calculating \(x^3\) and \(\frac{1}{x^3}\) separately.
First, calculate \(x^3\):
\[
x^3 = (\sqrt{3} + \sqrt{2})^3
\]
Applying the binomial expansion, we get:
\[
(\sqrt{3})^3 + 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 + (\sqrt{2})^3
\]
\[
= 3\sqrt{3} + 3 \times 3 \sqrt{2} + 3 \times 2\sqrt{3} + 2\sqrt{2}
\]
\[
= 3\sqrt{3} + 9\sqrt{2} + 6\sqrt{3} + 2\sqrt{2}
\]
\[
= 9\sqrt{3} + 11\sqrt{2}
\]
Now, for \(\frac{1}{x^3}\), consider \(\frac{1}{x}\) first. Knowing that \(x = \sqrt{3} + \sqrt{2}\), we find the conjugate to rationalize the denominator for \(\frac{1}{x}\):
\[
\frac{1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} – \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} – \sqrt{2})}
\]
\[
= \frac{\sqrt{3} – \sqrt{2}}{3 – 2}
\]
\[
= \sqrt{3} – \sqrt{2}
\]
Thus, \(\frac{1}{x} = \sqrt{3} – \sqrt{2}\), and hence \(\frac{1}{x^3} = (\sqrt{3} – \sqrt{2})^3\).
To simplify this, apply the binomial expansion similarly:
\[
(\sqrt{3} – \sqrt{2})^3 = (\sqrt{3})^3 – 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 – (\sqrt{2})^3
\]
\[
= 3\sqrt{3} – 3 \times 3 \sqrt{2} + 3 \times 2\sqrt{3} – 2\sqrt{2}
\]
\[
= 9\sqrt{3} – 11\sqrt{2}
\]
Now, to find \(x^3 – \frac{1}{x^3}\):
\[
x^3 – \frac{1}{x^3} = (9\sqrt{3} + 11\sqrt{2}) – (9\sqrt{3} – 11\sqrt{2})
\]
\[
= 9\sqrt{3} + 11\sqrt{2} – 9\sqrt{3} + 11\sqrt{2}
\]
\[
= 22\sqrt{2}
\]
Therefore, the value of \(x^3 – \frac{1}{x^3}\) is \(22\sqrt{2}\), making the correct answer:
(c) \(22 \sqrt{2}\).
See lessIf the difference of two numbers is 3 and the difference of their squares is 39 ; then the larger number is : (a) 9 (b) 12 (c) 13 (d) 8
Given that the difference between two numbers is 3 and the difference between their squares is 39, we can set up the following equations: 1. \(x - y = 3\) 2. \(x^2 - y^2 = 39\) We know that \(x^2 - y^2\) can be factored into \((x - y)(x + y)\). Substituting the first equation into the second gives:Read more
Given that the difference between two numbers is 3 and the difference between their squares is 39, we can set up the following equations:
1. \(x – y = 3\)
2. \(x^2 – y^2 = 39\)
We know that \(x^2 – y^2\) can be factored into \((x – y)(x + y)\). Substituting the first equation into the second gives:
\[
(x – y)(x + y) = 39
\]
Since \(x – y = 3\), we substitute that into the equation:
\[
3(x + y) = 39
\]
Simplifying:
\[
x + y = 13
\]
We now have two equations:
1. \(x – y = 3\)
2. \(x + y = 13\)
Adding these two equations together to eliminate \(y\), we get:
\[
2x = 16
\]
Solving for \(x\), we find:
\[
x = 8
\]
To find \(y\), we substitute \(x = 8\) back into one of the original equations:
\[
8 – y = 3
\]
Solving for \(y\), we get:
\[
y = 5
\]
Therefore, the larger number is \(8\).
The correct option is (d) 8.
See lessFrom 9.00 AM to 2.00 PM, the temperature rose at a constant rate from \(21^{\circ} \mathrm{C}\) to \(36^{\circ} \mathrm{C}\). What was the temperature at noon ? (a) \(27^{\circ} \mathrm{C}\) (b) \(30^{\circ} \mathrm{C}\) (c) \(32^{\circ} \mathrm{C}\) (d) \(28.5^{\circ} \mathrm{C}\)
The temperature increased from \(21^{\circ}C\) to \(36^{\circ}C\) over a 5-hour period from 9:00 AM to 2:00 PM, which means the temperature rose \(36^{\circ}C - 21^{\circ}C = 15^{\circ}C\) in total. To find the rate of increase per hour, divide the total temperature increase by the number of hours:Read more
The temperature increased from \(21^{\circ}C\) to \(36^{\circ}C\) over a 5-hour period from 9:00 AM to 2:00 PM, which means the temperature rose \(36^{\circ}C – 21^{\circ}C = 15^{\circ}C\) in total.
To find the rate of increase per hour, divide the total temperature increase by the number of hours:
\[
\frac{15^{\circ}C}{5 \text{ hours}} = 3^{\circ}C/\text{hour}
\]
From 9:00 AM to noon (12:00 PM) is 3 hours. At a rate of \(3^{\circ}C/\text{hour}\), the temperature increase from 9:00 AM to noon would be:
\[
3 \text{ hours} \times 3^{\circ}C/\text{hour} = 9^{\circ}C
\]
Therefore, the temperature at noon, starting from \(21^{\circ}C\) at 9:00 AM, would be:
\[
21^{\circ}C + 9^{\circ}C = 30^{\circ}C
\]
So, the temperature at noon was \(30^{\circ}C\).
The correct answer is (b) \(30^{\circ} \mathrm{C}\).
See lessIf \(\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{a+b}\), then : (a) \(\frac{x-y}{b-a}=\frac{y-z}{c-b}=\frac{z-x}{a-c}\) (b) \(\frac{\mathrm{x}}{\mathrm{a}}=\frac{\mathrm{y}}{\mathrm{b}}=\frac{\mathrm{z}}{\mathrm{c}}\) (c) \(\frac{x-y}{c}=\frac{y-z}{b}=\frac{z-x}{a}\) (d) none of the above is true
Given the equation: \[ \frac{x}{b+c} = \frac{y}{c+a} = \frac{z}{a+b} = k \quad (\text{say}) \] This implies: \[ x = k(b+c), \quad y = k(c+a), \quad \text{and} \quad z = k(a+b) \] From these equations, we can find the differences: \[ \begin{aligned} x - y &= k(b+c) - k(c+a) = k(b-a) \\ y - zRead more
Given the equation:
\[
\frac{x}{b+c} = \frac{y}{c+a} = \frac{z}{a+b} = k \quad (\text{say})
\]
This implies:
\[
x = k(b+c), \quad y = k(c+a), \quad \text{and} \quad z = k(a+b)
\]
From these equations, we can find the differences:
\[
\begin{aligned}
x – y &= k(b+c) – k(c+a) = k(b-a) \\
y – z &= k(c+a) – k(a+b) = k(c-b) \\
z – x &= k(a+b) – k(b+c) = k(a-c)
\end{aligned}
\]
Now, we check option (a):
\[
\frac{x-y}{b-a} = \frac{y-z}{c-b} = \frac{z-x}{a-c}
\]
Substituting the differences we calculated:
\[
\begin{aligned}
\frac{k(b-a)}{b-a} &= \frac{k(c-b)}{c-b} = \frac{k(a-c)}{a-c}
\end{aligned}
\]
Simplifying, we see that each fraction simplifies to \(k\), since the \(b-a\), \(c-b\), and \(a-c\) in the numerators and denominators cancel out:
\[
k = k = k
\]
Therefore, option (a) \(\frac{x-y}{b-a} = \frac{y-z}{c-b} = \frac{z-x}{a-c}\) is true based on the given equation.
See lessThe simplified value of \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\) is (a) 4 (b) 3 (c) 2 (d) 6
To simplify the given expression \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\), let's start by simplifying each square root by expressing them in terms of their prime factors and then simplify: \[ \sqrt{32} = \sqrt{2^5} = \sqrt{16 \cdot 2} = 4\sqrt{2} \] \[ \sqrt{48} = \sqrt{2^4 \cdot 3} = 4\sqRead more
To simplify the given expression \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\), let’s start by simplifying each square root by expressing them in terms of their prime factors and then simplify:
\[
\sqrt{32} = \sqrt{2^5} = \sqrt{16 \cdot 2} = 4\sqrt{2}
\]
\[
\sqrt{48} = \sqrt{2^4 \cdot 3} = 4\sqrt{3}
\]
\[
\sqrt{8} = \sqrt{2^3} = \sqrt{4 \cdot 2} = 2\sqrt{2}
\]
\[
\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}
\]
Substituting these simplified forms back into the original expression gives:
\[
\frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}
\]
We can take a 2 out of both the numerator and the denominator to simplify further:
\[
= \frac{2(2\sqrt{2} + 2\sqrt{3})}{2(\sqrt{2} + \sqrt{3})}
\]
Upon simplification, we find:
\[
= \frac{2\sqrt{2} + 2\sqrt{3}}{\sqrt{2} + \sqrt{3}}
\]
Since the terms in the numerator are double those in the denominator, simplifying the expression incorrectly suggested the terms could cancel out directly. The correct simplification should consider factoring correctly and assessing the common terms. The step where the 2 is factored out both in the numerator and denominator simplifies the expression directly, leading to:
\[
= 2
\]
Thus, the simplified value of the expression \(\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}\) is indeed \(2\).
The correct answer is (c) 2.
See lessIf \(x=a-b, y=b-c, z=\mathrm{c}-a\), then the numerical value of the algebraic expression \(x^{3}+y^{3}+z^{3}-3 x y z\) will be
To find the numerical value of the given algebraic expression \(x^{3}+y^{3}+z^{3}-3xyz\) when \(x=a-b\), \(y=b-c\), and \(z=c-a\), we can substitute these values directly into the expression. However, there's a known identity in algebra that can simplify our efforts: \[ x^{3}+y^{3}+z^{3}-3xyz = (x+yRead more
To find the numerical value of the given algebraic expression \(x^{3}+y^{3}+z^{3}-3xyz\) when \(x=a-b\), \(y=b-c\), and \(z=c-a\), we can substitute these values directly into the expression. However, there’s a known identity in algebra that can simplify our efforts:
\[
x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)\left(x^{2}+y^{2}+z^{2}-xy-xz-yz\right)
\]
First, let’s find \(x+y+z\):
\[
x+y+z = (a-b)+(b-c)+(c-a)
\]
Simplifying this, we find:
\[
x+y+z = 0
\]
Since \(x+y+z = 0\), the whole expression \(x^{3}+y^{3}+z^{3}-3xyz\) simplifies to 0 because anything multiplied by 0 is 0. Therefore, the numerical value of the given algebraic expression is \(0\).
The correct answer is (b) 0.
See less