Given the quadrilateral ABCD, what is the length of diagonal AC?

Fittoniya [83]

<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>

4 0

2 years ago

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2−x2. What are the dimensions of su

Paul [167]

Answer:

Step-by-step explanation:

Given that a rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

$y=2-x^2$

the parabola is open down with vertex at (0,2)

We can find that the rectangle also will be symmetrical about y axis.

Let the vertices on x axis by (p,0) and (-p,0)

Then other two vertices would be (p,2-p^2) (-p,2-p^2) because the vertices lie on the parabola and satisfy the parabola equation

Now width = $2-p^2$

Area = l*w = $2(2p-p^3)$

Use derivative test

I derivative = $2(2-3p^2)$

II derivative = $-12p$

Equate I derivative to 0 and consider positive value only since we want maximum

p = $\sqrt{\frac{2}{3} }$

Thus width= $\sqrt{\frac{2}{3} }$

Length = $2\sqrt{\frac{2}{3} }$

Width = $2-2/3 = 4/3$

3 0

2 years ago

Find the equation of a line parallel to the line x=-10 and contains the point (-4,4)

earnstyle [38]

Hello!

The line x = -10, is an example of an undefined line. It does not pass the vertical line test and each element of the domain is not paired with a unique (one and only one) element of the range. Remember that no two ordered pairs in a function have the same first element.

To find the line parallel to the line given line, we need to use the x-value in the given ordered pair, which will create the parallel line.

Therefore, the line parallel to the line x = -10 is x = -4.

6 0

3 years ago

What number is its own oppoite

SashulF [63]

The number that is its own opposite is zero. Zero is both a positive and a negative number at the same time.

4 0

2 years ago

Read 2 more answers

Fine the distance between p(2,8 and q(5,3)

defon

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
$d = \sqrt{(5 - 2)^{2} + (3 - 8)^{2}}\$
$d = \sqrt{(3)^{2} + (-5)^{2}}$
$d = \sqrt{9 + 25}$
$d = \sqrt{34}$

7 0

3 years ago