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liraira [26]
3 years ago
11

Suppose that y is inversely proportional to square root of x.

Mathematics
1 answer:
FromTheMoon [43]3 years ago
7 0

Answer:

i) k = 36.8

ii)y = 36.8/square root of x

iii)y = 1.4

Step-by-step explanation:

If y is inversely proportional to square root of x;

then the two variables are related by,

y = k/ square root of x

i)Value of k when...x = 8 and y = 13,

substitute into the equation above..so as to find the value of k

13 = k/square root of 8

;k = 13 × square root of 8

;k = 36.8

ii)Then substitute the value of k with...36.8

y = 36.8/square root of x

iii)If x = 27...then y =;

y = 36.8/27

y = 1.4

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\boxed{ \bold{ \boxed{ \sf{1. \:  \:  \:  \:  \: x = 12}}}}

\boxed{ \bold{ \boxed{ \sf{2. \:  \:  \:  \:  \: x = 3}}}}

Step-by-step explanation:

\sf{1. \:  \:  \:  \:  \frac{x}{3}  + 2 =  \frac{x}{4}  + 3}

Move x / 4 to left hand side and change its sign

Similarly, move 2 to right hand side and change it's sign

\longrightarrow{ \sf{ \frac{x}{3}  -  \frac{x}{4}  = 3 - 2}}

Take the L.C.M of 3 and 4

L.C.M of 3 and 4 = 12

\longrightarrow{ \sf{ \frac{x \times 4 - x \times 3}{12}  = 3 - 2}}

\longrightarrow{ \sf{ \frac{4x - 3x}{12}  = 3 - 2}}

\longrightarrow{ \sf{ \frac{x}{12}  = 3 - 2}}

Subtract 2 from 3

\longrightarrow{ \sf{ \frac{x}{12}  = 1}}

Do cross multiplication

\longrightarrow{ \sf{x  = 1 \times 12}}

\longrightarrow { \sf{x = 12}}

--------------------------------------------------------------------

\sf{2. \:  \:  \:  \: 3(4 + 2x) = 33 - x }

Distribute 3 through the parentheses

\longrightarrow{ \sf{12 + 6x = 33 - x}}

Move x to left hand side and change it's sign

Similarly, move 12 to right hand side and change it's sign

\longrightarrow{ \sf{6x +  x = 33 - 12}}

Collect like terms

\longrightarrow{ \sf{7x = 33 - 12}}

Subtract 12 from 33

\longrightarrow{ \sf{7x = 21}}

Divide both sides by 7

\longrightarrow{ \sf{ \frac{7x}{7}  =  \frac{21}{7} }}

Calculate

\longrightarrow{ \sf{x = 3}}

Hope I helped!

Best regards! :D

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