Question

$\huge{\green{\underline{\underline{\tt{Solution:}}}}}$

Given,

☞ x = -1

Let, P(x) = 5x-4x²+3

$→\;{\sf{P(-1) = 5(-1)-4(-1)²+3}}$

$→\;{\sf{P(-1) = -5-4(+1)+3}}$

$→\;{\sf{P(-1) = -5-4+3}}$

$→\;{\sf{P(-1) = -9+3}}$

$➝\;\red{\bf{P(-1) = -6}}$

Hope It Helps You ✌️
$\huge\underbrace{\green{\tt{\diamond\; Solution:}}}$

Let

Total Pocket Money be 'x'

Now, Savings = 30% of x

$→\;\Large{\frac{30}{100}×x}$

☞ $→\;\Large{\frac{30x}{100}}$

Given,

Amount Spend = Rs. 280.

$→\;\Large{\sf{x - \frac{30x}{100} = 280}}$

LCM is '100'

$→\;\Large{\sf{\frac{100x-30x}{100} = 280}}$

$→\;{\sf{70x = 280×100}}$

$→\;\Large{\sf{x = \cancel{\frac{28000}{70}}}}$

$→\;\boxed{\sf{x = 400}}$

.:. Total Pocket Money is $\fbox\red{Rs.400}$

Hope It Helps You ✌️
Refer The Attachment ⬆️

$\huge\underbrace{\green{\tt{\diamond\; SoluTion:}}}$

$→\;{\sf{2x² = 128}}$

$→\;{\sf{x² = \Large\frac{128}{2}}}$

$→\;{\sf{x = \sqrt{64}}}$

$→\;{\sf{x = \sqrt{8²}}}$

$☞\;\Large{\red{\bf{x = 8}}}$

$\purple{\bf{MrMonarque}}$

Hope It Helps You ✌️
Refer The Attachment ⬆️

$\red{\sf{MrMonarque}}$

Hope It Helps You ✌️$\huge{\green{\underline{\underline{\tt{Solution:}}}}}$

Given,

☞ x = -1

Let, P(x) = 5x-4x²+3

$→\;{\sf{P(-1) = 5(-1)-4(-1)²+3}}$

$→\;{\sf{P(-1) = -5-4(+1)+3}}$

$→\;{\sf{P(-1) = -5-4+3}}$

$→\;{\sf{P(-1) = -9+3}}$

$➝\;\red{\bf{P(-1) = -6}}$

Hope It Helps You ✌️
$\huge\underbrace{\green{\tt{\diamond\; Solution:}}}$

Let

Total Pocket Money be 'x'

Now, Savings = 30% of x

$→\;\Large{\frac{30}{100}×x}$

☞ $→\;\Large{\frac{30x}{100}}$

Given,

Amount Spend = Rs. 280.

$→\;\Large{\sf{x - \frac{30x}{100} = 280}}$

LCM is '100'

$→\;\Large{\sf{\frac{100x-30x}{100} = 280}}$

$→\;{\sf{70x = 280×100}}$

$→\;\Large{\sf{x = \cancel{\frac{28000}{70}}}}$

$→\;\boxed{\sf{x = 400}}$

.:. Total Pocket Money is $\fbox\red{Rs.400}$

Hope It Helps You ✌️
Refer The Attachment ⬆️

$\huge\underbrace{\green{\tt{\diamond\; SoluTion:}}}$

$→\;{\sf{2x² = 128}}$

$→\;{\sf{x² = \Large\frac{128}{2}}}$

$→\;{\sf{x = \sqrt{64}}}$

$→\;{\sf{x = \sqrt{8²}}}$

$☞\;\Large{\red{\bf{x = 8}}}$

$\purple{\bf{MrMonarque}}$

Hope It Helps You ✌️
Refer The Attachment ⬆️

$\red{\sf{MrMonarque}}$

Hope It Helps You ✌️
Let $\sf p \geq 5$ be a prime number.... Prove that there exists an integer $\sf a$ with $\sf 1 \leq a \leq p - 2$ such that neither $\sf {a}^{(p - 1)} - 1$ nor [tex
$\huge\star\underline\mathtt\green{Question:-}$(◍•ᴗ•◍)❤ $\tt\pink{✰\:Define}$ $\sf 1. \:Acid$ $\sf 2.... \:Base$ $\sf 3. \:Salt$ No Spam‼️ ​

Answer 1

Explanation:

$\huge{\green{\underline{\underline{\tt{Solution:}}}}}$

Given,

☞ x = -1

Let, P(x) = 5x-4x²+3

$→\;{\sf{P(-1) = 5(-1)-4(-1)²+3}}$

$→\;{\sf{P(-1) = -5-4(+1)+3}}$

$→\;{\sf{P(-1) = -5-4+3}}$

$→\;{\sf{P(-1) = -9+3}}$

$➝\;\red{\bf{P(-1) = -6}}$

Hope It Helps You ✌️

$\huge\underbrace{\green{\tt{\diamond\; Solution:}}}$

Let

Total Pocket Money be 'x'

Now, Savings = 30% of x

$→\;\Large{\frac{30}{100}×x}$

☞ $→\;\Large{\frac{30x}{100}}$

Given,

Amount Spend = Rs. 280.

$→\;\Large{\sf{x - \frac{30x}{100} = 280}}$

LCM is '100'

$→\;\Large{\sf{\frac{100x-30x}{100} = 280}}$

$→\;{\sf{70x = 280×100}}$

$→\;\Large{\sf{x = \cancel{\frac{28000}{70}}}}$

$→\;\boxed{\sf{x = 400}}$

.:. Total Pocket Money is $\fbox\red{Rs.400}$

Hope It Helps You ✌️

Refer The Attachment ⬆️

$\huge\underbrace{\green{\tt{\diamond\; SoluTion:}}}$

$→\;{\sf{2x² = 128}}$

$→\;{\sf{x² = \Large\frac{128}{2}}}$

$→\;{\sf{x = \sqrt{64}}}$

$→\;{\sf{x = \sqrt{8²}}}$

$☞\;\Large{\red{\bf{x = 8}}}$

$\purple{\bf{MrMonarque}}$

Hope It Helps You ✌️

Refer The Attachment ⬆️

$\red{\sf{MrMonarque}}$

Hope It Helps You ✌️$\huge{\green{\underline{\underline{\tt{Solution:}}}}}$

Given,

☞ x = -1

Let, P(x) = 5x-4x²+3

$→\;{\sf{P(-1) = 5(-1)-4(-1)²+3}}$

$→\;{\sf{P(-1) = -5-4(+1)+3}}$

$→\;{\sf{P(-1) = -5-4+3}}$

$→\;{\sf{P(-1) = -9+3}}$

$➝\;\red{\bf{P(-1) = -6}}$

Hope It Helps You ✌️

$\huge\underbrace{\green{\tt{\diamond\; Solution:}}}$

Let

Total Pocket Money be 'x'

Now, Savings = 30% of x

$→\;\Large{\frac{30}{100}×x}$

☞ $→\;\Large{\frac{30x}{100}}$

Given,

Amount Spend = Rs. 280.

$→\;\Large{\sf{x - \frac{30x}{100} = 280}}$

LCM is '100'

$→\;\Large{\sf{\frac{100x-30x}{100} = 280}}$

$→\;{\sf{70x = 280×100}}$

$→\;\Large{\sf{x = \cancel{\frac{28000}{70}}}}$

$→\;\boxed{\sf{x = 400}}$

.:. Total Pocket Money is $\fbox\red{Rs.400}$

Hope It Helps You ✌️

Refer The Attachment ⬆️

$\huge\underbrace{\green{\tt{\diamond\; SoluTion:}}}$

$→\;{\sf{2x² = 128}}$

$→\;{\sf{x² = \Large\frac{128}{2}}}$

$→\;{\sf{x = \sqrt{64}}}$

$→\;{\sf{x = \sqrt{8²}}}$

$☞\;\Large{\red{\bf{x = 8}}}$

$\purple{\bf{MrMonarque}}$

Hope It Helps You ✌️

Refer The Attachment ⬆️

$\red{\sf{MrMonarque}}$

Hope It Helps You ✌️

Let $\sf p \geq 5$ be a prime number.... Prove that there exists an integer $\sf a$ with $\sf 1 \leq a \leq p - 2$ such that neither $\sf {a}^{(p - 1)} - 1$ nor [tex

$\huge\star\underline\mathtt\green{Question:-}$(◍•ᴗ•◍)❤ $\tt\pink{✰\:Define}$ $\sf 1. \:Acid$ $\sf 2.... \:Base$ $\sf 3. \:Salt$ No Spam‼️

Answer 2

Answer:

Let, P(x) = 5x-4x²+3

→\;{\sf{P(-1) = 5(-1)-4(-1)²+3}}→P(−1)=5(−1)−4(−1)²+3

→\;{\sf{P(-1) = -5-4(+1)+3}}→P(−1)=−5−4(+1)+3

→\;{\sf{P(-1) = -5-4+3}}→P(−1)=−5−4+3

→\;{\sf{P(-1) = -9+3}}→P(−1)=−9+3

➝\;\red{\bf{P(-1) = -6}}➝P(−1)=−6