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3 years ago

PLEASE help me with this! (50+ Points)

Mathematics

1 answer:

____ [38]3 years ago

8 0

Answer:

Step-by-step explanation:

slope=∆y/∆x

thus any values of x and y in the table can be used.

slope=(11-9)÷(4-3)

slope= 2.

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HELP ASAP, LINKS AND ABSURD ANSWERS WILL BE REPORTED! I WILL MARK BRAINLIEST!!

STALIN [3.7K]

Answer:

6/5

Step-by-step explanation:

brainly plz

but if one whole is 5 then one has 5 and the other has 1 and 5+1=6 and one whole=5 thus is 6/5

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3 years ago

What is the equation of the line that is parallel to the line 5x + 2y = 12 and passes through the point (-3, 2)

qwelly [4]

2y = 12 - 5x
y = 6 - 5/2x

2 = -5/2(-3) + b
2 = 15/2 + b
2 - 15/2 = b
b = -11/2

y = 15/2x - 11/2

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3 years ago

Please help I need an answer please

erma4kov [3.2K]

15 girls 12 boys 5x3=15 and 4x3=12 so 15 girls 12 boys

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3 years ago

A copier was purchased for $12,000. Its value each year is 75% of what it was the preceding

bagirrra123 [75]

Answer:

Step-by-step explanation:

it will take 1.5 years, because a and b are too little. D is too much a value. C is the right answer

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3 years ago

Solve the system of equations.

emmainna [20.7K]

$x = \frac{ 419 }{ 113 } ~,~y = -\frac{ 208 }{ 113 } ~,~z = \frac{ 21 }{ 113 }$

<h2>Explanation:</h2>

We have the following system of three linear equations:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\5~ x&+~~8~ y&+~~~~ z&~=~4\\4~ x&+~~5~ y&+~~2~ z&~=~6\end{array}$

Let's use elimination method in order to get the solution of this system of equation, so let's solve this step by step.

Step 1: Multiply first equation by $-5/2$ and add the result to the second equation. So we get:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\4~ x&+~~5~ y&+~~2~ z&~=~6\end{array}$

Step 2: Multiply first equation by −2 and add the result to the third equation. So we get:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\&-~~~3~ y&-~~~62~ z&~=~-6\end{array}$

Step 3: Multiply second equation by −32 and add the result to the third equation. So we get:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\&&+~~\frac{ 113 }{ 2 }~ z&~=~\frac{ 21 }{ 2 }\end{array}$

Step 4: solve for z.

$\begin{aligned} \frac{ 113 }{ 2 } ~ z & = \frac{ 21 }{ 2 } \\ z & = \frac{ 21 }{ 113 } \end{aligned}$

Step 5: solve for y.

$\begin{aligned}-2y-79z &= -11\\-2y-79\cdot \frac{ 21 }{ 113 } &= -11\\y &= -\frac{ 208 }{ 113 } \end{aligned}$

Step 6: solve for x by substituting $y=-\frac{208}{113}$ and $z = \frac{21}{113}$ into the first equation:

$2x+4(-\frac{208}{113})+32(\frac{21}{113})=6 \\ \\ 2x-\frac{832}{113}+\frac{672}{113}=6 \\ \\ 2x=6+\frac{832}{113}-\frac{672}{113} \\ \\ 2x=\frac{838}{113} \\ \\ x=\frac{319}{113}$

Finally:

$x = \frac{ 419 }{ 113 } ~,~y = -\frac{ 208 }{ 113 } ~,~z = \frac{ 21 }{ 113 }$

<h2>Learn more:</h2>

Solving System of Equations: brainly.com/question/13121177

#LearnWithBrainly

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4 years ago