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1. If r varies directly as s and inversely as the square of u and r=2 when s=18 and u=2, FIND
A. r when u=3 and s=27
B. s when u=2 and r=4
C. u when r=1 and s=36

2. If x varies as the square of y and inversely as z and x=12, when y=3 and z=6, find x when y=9 and z=6

3. w varies directly as x and y and inversely as v² and w=1200, when x=4, y=9 and v=6, find w when x=3, y=12 and v=9

answer please. thankyouuuuuuuuuuuu!!!!♥♡♥ with solution ^____^

Sagot :

CebuMath

Answer:

1)

  • A. [tex]r=\frac{4}{3}[/tex]
  • B. [tex]s=36[/tex]
  • C. [tex]u=4[/tex]

2) [tex]x=108\\[/tex]

3) [tex]w=\frac{1600}{3}[/tex]

Step-by-step explanation:

Variation, in mathematics, is the relationship between two or more variables. One is constant and the other one is changing. The process of changing variables is called variation.

Two Types of Variation

1. Varies Directly

2. Varies Inversely

Varies directly is the relationship between two variables when both of them go in the same direction. For example, one variable is increasing, the other one should also be increasing.

Varies inversely or indirectly is the relationship between two variables when they don't tend in the same direction. For example, one variable goes increasing and the other one goes decreasing.

Definition 1: We say that [tex]y[/tex] varies directly with [tex]x[/tex], then there exist a [tex]k[/tex] such that [tex]y=kx[/tex].

Definition 2: We say that [tex]y[/tex] varies inversely with [tex]x[/tex], then there exists a [tex]k[/tex] such that [tex]y=\frac{k}{x}[/tex].

For number 1, since [tex]r[/tex] varies directly as [tex]s[/tex] and inversely as the square of [tex]u[/tex], then the mathematical representation is [tex]r=\frac{ks}{u^2}[/tex]. Substitute the values of [tex]u=2[/tex], [tex]r=2[/tex], and [tex]s=18[/tex] into the model and solve for [tex]k[/tex].

                                     [tex]\begin{aligned}r&=\frac{ks}{u^2}\\2&=\frac{k(18)}{2^2}\\2&=\frac{18k}{4}\\4\left(2\right)&=4\left(\frac{18k}{4}\right)\\8&=18k\\\frac{8}{18}&=\frac{18k}{18}\\\frac{8}{18}&=k\\\frac{4}{9}&=k\end{aligned}[/tex]

  • A. Find [tex]r[/tex], when [tex]u=3[/tex] and [tex]s=27[/tex].

                    [tex]\begin{aligned}r&=\frac{ks}{u^2}\\r&=\frac{\frac{4}{9}(27)}{3^2}\\r&=\frac{4(3)}{9}\\r&=\frac{12}{9}\\r&=\frac{4}{3}\end{aligned}[/tex]

  • B. Find [tex]s[/tex], when [tex]u=2[/tex] and [tex]r=4[/tex].  

                    [tex]\begin{aligned}r&=\frac{ks}{u^2}\\4&=\frac{\frac{4}{9}s}{2^2}\\4&=\frac{\frac{4}{9}s}{4}\\4(4)&=4\left(\frac{\frac{4}{9}s}{4}\right)\\16&=\frac{4}{9}s\\\frac{9}{4}(16)&=\frac{9}{4}\left(\frac{4}{9}s\right)\\9(4)&=s\\36&=s\end{aligned}[/tex]

  • C. Find [tex]u[/tex], when [tex]r=1[/tex] and [tex]s=36\\[/tex].

                        [tex]\begin{aligned}r&=\frac{ks}{u^2}\\1&=\frac{\frac{4}{9}(36)}{u^2}\\1&=\frac{4(4)}{u^2}\\1&=\frac{16}{u^2}\\u^2(1)&=u^2\left(\frac{16}{u^2}\right)\\u^2&=16\\\sqrt{u^2}&=\sqrt{16}\\u&=4\end{aligned}[/tex]

For number 2, since [tex]x[/tex] varies directly as [tex]y^2[/tex] and inversely as [tex]z[/tex], then the mathematical representation is [tex]x=\frac{ky^2}{z}[/tex]. Substitute the values of [tex]x=12[/tex], [tex]y=3[/tex], and [tex]z=6[/tex] into the model and solve for [tex]k[/tex].

                                  [tex]\begin{aligned}x&=\frac{ky^2}{z}\\12&=\frac{k(3^2)}{6}\\12&=\frac{9k}{6}\\12(6)&=6\left(\frac{9k}{6}\right)\\72&=9k\\\frac{72}{9}&=\frac{9k}{9}\\8&=k\end{aligned}[/tex]

  • Find [tex]x[/tex], when [tex]y=9[/tex] and [tex]z=6[/tex].

                                   [tex]\begin{aligned}x&=\frac{ky^2}{z}\\x&=\frac{8(9^2)}{6}\\x&=\frac{648}{6}\\x&=108\end{aligned}[/tex]

                                     

For number 3, since [tex]w[/tex] varies directly as [tex]x[/tex] and [tex]y[/tex], and varies inversely as [tex]v^2[/tex] then the mathematical representation is [tex]w=\frac{kxy}{v^2}[/tex]. Substitute the values of [tex]x=4[/tex], [tex]w=1200[/tex], [tex]y=9[/tex], and [tex]v=6[/tex] into the model and solve for [tex]k[/tex].

                             [tex]\begin{aligned}w&=\frac{kxy}{v^2}\\1200&=\frac{k(4)(9)}{6^2}\\1200&=\frac{36k}{36}\\1200&=k\end{aligned}[/tex]

  • Find [tex]w[/tex], when [tex]x=3[/tex], [tex]y=12[/tex], and [tex]v=9[/tex].

                               [tex]\begin{aligned}w&=\frac{kxy}{v^2}\\w&=\frac{1200(3)(12)}{9^2}\\w&=\frac{43,200}{81}\\w&=\frac{1600}{3}\end{aligned}[/tex]

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To learn more about variations, please click the following links:

https://brainly.ph/question/2476944

https://brainly.ph/question/845317

https://brainly.ph/question/1922362

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