Answered

find the equation and solve for the constant of the variation k.
1. A varies directly as T and S = 12 when T-8
2. E is inversely proportional to the square of F and E = 4 when F-5
3. varies jointly as Q and R, and P=12 when Q-8 and R-3
4. Z varies directly as X and inversely as Y and Z=10 when X-5 and Y=12

this is for my ate, I am answering her modules but I don't know how to answer these, she is sick and her module is due very soon. please help me, I even watched yt videos to answer this. please explain if you can, I really need it.​

Sagot :

ARMYLGO

Find the equation and solve for the constant of the variation k.

[tex] \\ [/tex]

1. A varies directly as T and A= 12 when T=8

[tex] \\ [/tex]

Equation:

[tex]A = kT \\ [/tex]

-Evaluate the values

[tex]12 = k(8) \\ [/tex]

-Divide both sides to 8, then cancel both 8 from the right side

[tex] \frac{12}{8} = \frac{k(8)}{8} \\ \frac{12}{8} = \frac{k( \cancel8)}{ \cancel8} \\ [/tex]

-Divide 12 by 8

[tex] \green{ \boxed{ \boxed{ \: \: k = \frac{3}{2 \: \: }}}} \\ \\ [/tex]

2. E is inversely proportional to the square of F and E = 4 when F=5

[tex] \\ [/tex]

Equation:

[tex] E = \frac{k}{ {F}^{2} } \\ [/tex]

-Evaluate the given values

[tex]4 = \frac{k}{ {5}^{2} } \\ [/tex]

-Square 5

[tex]4 = \frac{ k}{25} \\ [/tex]

-Multiply 25 to both sides, then cancel both 25 from the right side

[tex](25)(4) = \frac{k}{25} (25) \\ (25)(4) = \frac{k}{( \cancel{25)}} ( \cancel{25}) \\ [/tex]

-Multiply 25 by 4

[tex] \green{ \boxed{ \boxed{ \: \: k = 100 \: \: }}} \\ \\ [/tex]

3. P varies jointly as Q and R, and P=12 when Q=8 and R= 3

Equation:

[tex]P = kQ R \\ [/tex]

-Evaluate the values

[tex]12 = k(8)(3) \\ [/tex]

-Multiply 8 by 3

[tex]12 = k(24) \\ [/tex]

-Divide 24 by both sides, then cancel both 24 from the right side

[tex] \frac{12}{24} = \frac{k(24)}{24} \\ \frac{12}{24} = \frac{k( \cancel{24})}{ \cancel{24}} \\ [/tex]

-Divide 12 by 24

[tex] \green{ \boxed{ \boxed{ \: \: k = \frac{1}{2 } \: \: }}} \: \: \\ \\ [/tex]

4. Z varies directly as X and inversely as Y and Z=10 when X=5 and Y=12

[tex] \\ [/tex]

Equation:

[tex]Z = \frac{kX }{ Y} \\ [/tex]

-Evaluate the values

[tex]10 = \frac{k (5)}{12} \\ [/tex]

-Multiply 12 by both sides, then cancel both 12 from the right side.

[tex](12)(10) = \frac{k(5)}{12} (12) \\ (12)(10) = \frac{k(5)}{ \cancel{12}} ( \cancel{12}) \\ [/tex]

-Multiply 12 by 10

[tex]120 = k(5) \\ [/tex]

-Divide 5 by bith sides, then cancel both 5 from the right side.

[tex] \frac{120}{5} = \frac{k(5)}{5} \\ \frac{120}{5} = \frac{k( \cancel{5})} {\cancel{5}} \\ [/tex]

-Divide 120 by 5

[tex] \green{ \boxed{ \boxed{ \: \: k = 24 \: \: }}} \\ \\ \\ [/tex]

Go Educations: Other Questions