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

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>

4 0

2 years ago

Hello, i need help! 5/2 multiplied by what equals 1? equation: 5/2×?=1

prisoha [69]

Answer:

$\large\boxed{\dfrac{2}{5}}$

Step-by-step explanation:

$\text{The product of the number and its reciprocal gives us 1.}\\\text{Reciprocal of }\ \dfrac{5}{2}\ \text{is equal}\ \dfrac{2}{5}.\\\\\dfrac{5}{2}\times\dfrac{2}{5}=\dfrac{5\!\!\!\!\diagup}{2\!\!\!\!\diagup}\times\dfrac{2\!\!\!\!\diagup}{5\!\!\!\!\diagup}=\dfrac{1}{1}\times\dfrac{1}{1}=1\times1=1$

4 0

2 years ago