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2 years ago

-9×+1=-x+17 solve equation

Mathematics

1 answer:

gavmur [86]2 years ago

3 0

-9×+1=-×+17
-9×=-×+16 (subtract 1 from both sides)
-8×=16 ( add x to both sides)
×=-2 (divide -8 from both sides)

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On a coordinate plane, triangles R S T and L M N are shown. Triangle R S T has points (5, 5), (2, 1), and (1, 3). Triangle L M N

ivolga24 [154]

Answer:

LMN has a larger area

Step-by-step explanation:

Given

RST

$R =(5,5)$

$S= (2,1)$

$T = (1,3)$

LMN

$L = (0,-1)$

$M= (2,-4)$

$N= (-2,-3)$

Required

Compare both areas

The area of triangle is:

$Area = \frac{1}{2}|x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2 + x_3y_1 - x_1y_3|$

For RST, we have:

$Area = \frac{1}{2}|5*1 - 2*5 + 2*3 - 1*1 + 1*5 - 5*3|$

$Area = \frac{1}{2}|-10|$

Remove absolute signs

$Area = \frac{1}{2}*10$

$Area = 10$

For LMN, we have:

$Area = \frac{1}{2}|x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2 + x_3y_1 - x_1y_3|$

$Area = \frac{1}{2}|0*-4-2*-1+2*-3--2*-4--2*-1-0*-3|$

$Area = \frac{1}{2}|-14|$

Remove absolute signs

$Area = \frac{1}{2}*14$

$Area = 7$

<em>By comparing both areas, we can conclude that LMN has a larger area</em>

3 0

2 years ago

What is the answer to 7 inches more than normal

dsp73

Answer: 12 ...............

7 0

3 years ago

Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?

Ede4ka [16]

Answer:

The solution is $x=2,\ y=0,\ z=-1.$

Step-by-step explanation:

You are given the system of three equations:

$\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.$

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

$\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.$

So,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.$

Multiply the third equation by 23 and subtract it from the second equation:

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.$