In a 30-60-90 triangle, the hypotenuse is the shorter leg times the square root of two.

1 answer:

7 0

False!

In a 30-60-90 triangle, the hypotenuse (c) is twice the length of the shorter leg (a): c = 2a
In a 30-60-90 triangle, the longer leg (b) is equal to the shorter leg (a) multiplied by √3: b = a√3

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Which represents a quadratic function y=2(x+2)(x-1) in standard form

ale4655 [162]

answer: y= 2x^2+2x-4
steps:
y= (2x+4)(x-1)
*FOIL method* y= (2x^2-2x+4x-4)
*simplify* y= 2x^2+2x-4

4 0

3 years ago

I need help w/ this!! thank you

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 :-

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