4m-3n=8, for m wats yeah

ONE MORE question that i got stuck on

Nadya [2.5K]

The slope equals 1/2 (rise over run)

7 0

3 years ago

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Vector V is shown in the graph below.

Lady_Fox [76]

Answer:

v = -8i - 6j

Step-by-step explanation:

The starting point is (9, 9)

The direction of the vector is the direction of the arrowhead.

The i is the horizontal component.

The j is the vertical component.

Therefore, to go from the start to the end of the line, we need to travel negative 8 units (horizontally) and negative 6 units (vertically).

Therefore, v = -8i - 6j

8 0

2 years ago

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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 :-

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