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Ipatiy [6.2K]
3 years ago
12

Y is directly proportional to the square root of x if y = 60 when x = 36 find x when y = 80

Mathematics
2 answers:
bearhunter [10]3 years ago
7 0

Answer:

x = 64

Step-by-step explanation:

Given y is directly proportional to \sqrt{x} then the equation relating them is

y = k\sqrt{x} ← k is the constant of proportion

To find k use the condition y = 60 when x = 36, then

60 = k \sqrt{36} = 6k ( divide both sides by 6 )

10 = k

y = 10\sqrt{x} ← equation of proportion

When y = 80 , then

80 = 10\sqrt{x} ( divide both sides by 10 )

8 = \sqrt{x} ( square both sides )

64 = x

butalik [34]3 years ago
6 0
Y = x^2
80 = x^2
x = √80
x = 8.94
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Suppose a and b are both non zero real numbers. Find real numbers c and d such that 1/a+ib= c+id
Thepotemich [5.8K]

\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}

Explanation

\frac{1}{a+bi}=c+di

Step 1

multiplicate by the conjugate

\begin{gathered} \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(bi)^2} \\ \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(-b^2)}=\frac{a-bi}{a^2+b^2} \end{gathered}

notice that

\begin{gathered} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i \\ \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di \\ so \\  \end{gathered}\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}

I hope this helsp you

6 0
1 year ago
6/8 is equal to 15/ what? NEED HELP ASAP PLZ will choose brainliest
irga5000 [103]

Answer:

No

Step-by-step explanation:

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3 0
3 years ago
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Keith_Richards [23]
3x - 18 = 42
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3 years ago
The temperature on the planet Mercury alternates depending
Sindrei [870]

Answer:

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Step-by-step explanation:

950+346=1296

Hope this helps! : )

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Ierofanga [76]

Answer:

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4x+16=2x‐18

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x= – 17

I hope I helped you^_^

3 0
3 years ago
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