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3 years ago

Fill in the blank. (5x)2 a0x2

Mathematics

1 answer:

Yuki888 [10]3 years ago

5 0

Answer: $a_0=25$

Step-by-step explanation:

It is given that, $(5x)^2=a_0x^2$

We need to find the value of $a_0$

As $(5x)^2=a_0x^2$

Solving LHS

$25x^2=a_0x^2$

Now comparing the coefficients of $x^2$

In LHS the coefficient of $x^2$ is 25

In RHS the coefficient of $x^2$ is $a_0$

It implies that, $a_0=25$

So, the value of $a_0$ is 25.

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<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

$\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}$

<h3>Hence the standard form of inequality is y²+3y +2 ≤ 0 and the Solution set of the inequality is [ -2 , -1 ] .</h3>

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