What is the answer I’ve been stuck on this for way too long

Mathematics

2 answers:

sukhopar [10]2 years ago

8 0

Please check attached image for the numbers.

Hope this helps...

velikii [3]2 years ago

4 0

$\text{Let's write it like}\\x+y=8\\+\:\:\:+\\z-x=6\\\downarrow\:\:\:\:\:\:\downarrow\\13\:\:\:\:8\\=>\\\displaystyle\\\left \{ {{x+z=13} \atop {z-x=6}} \right. \\\left \{ {{x+z=13} \atop {z=6+x}} \right. \\\downarrow\\x+x+6=16\\2x=7\\x=3,5=>z=9,5\\\text{Now we need to find y}\\3,5+y=8\\y=4,5\\\\\text{so we have}\\3,5+4,5=8\\+\:\:\:\:\:\:\:\:\:+\\9,5-3,5=6\\\downarrow\:\:\:\:\:\:\:\:\:\:\:\:\downarrow\\13\:\:\:\:\:\:\:\:\:\:8\\$

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Find the quadratic function that fits the following data. which one function fits.

Oksana_A [137]

Answer:

Step-by-step explanation:

The general rule for the quadratic function is

$y=ax^2+bx+c$

Use the data from the table:

$y(50)=130\Rightarrow 130=a\cdot 50^2+b\cdot 50+c\\ \\y(70)=130\Rightarrow 130=a\cdot 70^2+b\cdot 70+c\\ \\y(90)=200\Rightarrow 200=a\cdot 90^2+b\cdot 90+c$

We get the system of three equations:

$\left\{\begin{array}{l}2500a+50b+c=130\\ \\4900a+70b+c=130\\ \\8100a+90b+c=200\end{array}\right.$

Subtract these equations:

$\left\{\begin{array}{l}4900a+70b+c-2500a-50b-c=130-130\\ \\8100a+90b+c-2500a-50b-c=200-130\end{array}\right.\Rightarrow \left\{\begin{array}{l}2400a+20b=0\\ \\5600a+40b=70\end{array}\right.$