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tatuchka [14]
3 years ago
7

The value of y varies directly with x, and y = 12 when x = 16.

Mathematics
1 answer:
Wewaii [24]3 years ago
3 0
\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------\\\\
\textit{we also know that }
\begin{cases}
y=12\\
x=16
\end{cases}\implies 12=k16\implies \cfrac{12}{16}=k
\\\\\\
\cfrac{3}{4}=k\qquad therefore\qquad \boxed{y=\cfrac{3}{4}x}
\\\\\\
\textit{when x = 6, what is \underline{y}?}\qquad y=\cfrac{3}{4}(6)
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svp [43]
Let's simplify the equation:
V=πrh^2 - (πh^3)/3
πrh^2=V + (πh^3)/3
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3 years ago
Please help me I need help please​
MakcuM [25]

Answer:

p = 4

Step-by-step explanation:

Given equation:

x^2+(p-3)y^2-4x+6y-16=0

<u>Standard equation of a circle:</u>

(x-a)^2+(y-b)^2=r^2

(where (a,b) is the centre of the circle, and r is the radius)

If you expand this equation, you will see that the coefficient of y^2 is always one.

Therefore, p-3=1

\implies p=1+3=4

<u>Additional information</u>

To rewrite the given equation in the standard form:

\implies x^2+y^2-4x+6y-16=0

\implies x^2-4x+y^2+6y=16

\implies (x-2)^2-4+(y+3)^2-9=16

\implies (x-2)^2+(y+3)^2=16+4+9

\implies (x-2)^2+(y+3)^2=29

So this is a circle with centre (2, -3) and radius √29

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2 years ago
If AB has endpoints A(-15,9) and B(0,3) what is the slope of AB
AysviL [449]

Answer:

-2/5

Step-by-step explanation:

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= -2/5

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Answer:

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Step-by-step explanation:

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