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Luba_88 [7]
4 years ago
15

plzzzz answer right away will mark BRAINLIST AND FIVE STARS PLUS THANKS What is the constant of proportionality in the equation

y = StartFraction x over 9 EndFraction? 0 StartFraction 1 over 9 EndFraction StartFraction 8 over 9 EndFraction 1
Mathematics
2 answers:
guajiro [1.7K]4 years ago
7 0

Answer:

{\sf \frac{1}{9}}

Step-by-step explanation:

y and x are the directly proportional quantities.

\sf y=kx

The constant of proportionality is k.

The equation is:

\sf y=\frac{x}{9}

\sf y=\frac{1}{9} x

The value of k in this equation is:

\sf k= \frac{1}{9}

o-na [289]4 years ago
4 0

Answer:

Proportionality Constant = k = \frac{1}{9}

Step-by-step explanation:

Equation is:

y = \frac{x}{9}

=> y = (\frac{1}{9} ) x

Comparing it with y = kx, where k is the proportionality constant, We get:

Proportionality Constant = k = \frac{1}{9}

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Answer:

y=231x

Step-by-step explanation:

To get y by itself just subtract 2 from both sides so, y=231x

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Can you help me calculate the angles from this kite in the attachment?<br><br><br>AY =, XZ =,
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Answer:

AY=8, XZ=20

Step-by-step explanation:

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J=UC solve for c please​
mart [117]

»»————- ★ ————-««

I believe your answer is:  

c=\frac{j}{u}; u\neq 0

»»————- ★ ————-««  

Here’s why:  

⸻⸻⸻⸻

\boxed{\text{Solving for 'c'...}}}\\\\j = uc  \\------------\\\rightarrow \frac{j=uc}{u}\\\\\rightarrow \frac{j}{u}=c\\\\\rightarrow \boxed{c=\frac{j}{u}}}; u\neq 0

⸻⸻⸻⸻

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.  

8 0
3 years ago
Let R be the triangular region in the first quadrant with vertices at points (0,0), (a,0), and (0,b), where a and b are positive
Salsk061 [2.6K]

Answer:

volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( \frac{-b}{a}x + b )² dx

Step-by-step explanation:

Given the data in the question and as illustrated in the image below;

R is in the region first quadrant with vertices; 0(0,0), A(a,0) and B(0,b)

from the image;

the equation of AB will be;

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divide through by a

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so R is bounded by  y = \frac{-b}{a}x + b and y =0, 0 ≤ x ≤ a

The volume of the solid revolving R about x axis is;

dv = Area × thickness

= π( Radius)² dx

= π ( \frac{-b}{a}x + b )² dx

V = π ₀∫^a ( \frac{-b}{a}x + b )² dx

Therefore,  volume of the solid generated when region R is revolved about the x-axis is π ₀∫^a ( \frac{-b}{a}x + b )² dx

6 0
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