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Elis [28]
3 years ago
11

Solve using elimination or substitution method

Mathematics
1 answer:
Jlenok [28]3 years ago
6 0

\left\{\begin{array}{ccc}y=\dfrac{2}{3}x+3\\2x-3y=-9\end{array}\right\\\\\text{substitute}\ y=\dfrac{2}{3}x+3\ \text{to the second equation}\\\\2x-3\left(\dfrac{2}{3}x+3\right)=-9\qquad|\text{use distributive property}\\\\2x+(-3)\left(\dfrac{2}{3}x\right)+(-3)(3)=-9\\\\2x-2x-9=-9\qquad|\text{add 9 to both sides}\\\\2x-2x=0\\\\0=0\qquad TRUE\\\\Answer:\ \text{Infinitely many solutions}\\\\\left\{\begin{array}{ccc}x\in\mathbb{R}\\y=\dfrac{3}{2}x+3\end{array}\Rightarrow\left(x,\ \dfrac{2}{3}x+3\right)\right

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4/6 = 2/3

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If y varies inversely with x and y= 17<br> when z = 11<br> when y = 11<br> find y when x=19<br> .
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Where’s your z? incomplete question
8 0
3 years ago
15 POINTSSS<br> Need Anwsered Asap!
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The equation we will use here is A^2+B^2=C^2, which is also know as the Pythagorean Theorem.
The given values are 6 and 9, where they can represent any value, there true values in the equation would be 36(6), and 81(9), so you must select a value that makes the equation true, given the constraints.

with that being said 3, doesnt work because
·36(6)+9(3)≠81(9)
·9(3)+81(9)≠36(6)
·36(6)+81(9)≠9(3)

10 doesnt work either because
·36(6)+81(9)≠100(10)
·81(9)+100(10)≠36(6)
·100(10)+36(6)≠81(9)

12 doesnt work either because 
·144(12)+36(6)≠81(9)
·36(6)+81(6)≠144(12)
·81(9)+144(12)≠36(6)

If you see where this is going you would know that there is no valid solution here, however rounding is always a possibility, when you actually do the math 81(9)+36(6)=117, and when squared you get your answer of 10.8, and the closest answer is 10, there fore your answer would be 10

-I hope this is the answer you are looking for, feel free to post your questions on brainly at any time.



5 0
3 years ago
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