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frutty [35]
3 years ago
7

Find the LCM of k^2 + 3k + 2 and 2k^2 + 14k + 20.

Mathematics
1 answer:
Airida [17]3 years ago
3 0

Answer:

  C.  2(k +2)(k +5)(k +1)

Step-by-step explanation:

The LCM will be the product of unique factors.

  (k^2 + 3k + 2)=(k+1)(k+2)\\(2k^2 + 14k + 20)=2(k+2)(k+5)

The unique factors are 2, (k+1), (k+2), (k+5), so the LCM is their product:

  2(k+1)(k+2)(k+5) . . . . matches choice C

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Please answer this mathematical problem.
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x = total amount of gumballs

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\stackrel{total}{x}-\stackrel{\textit{to jaysen}}{\cfrac{x}{2}}\implies \stackrel{\textit{what's left}}{\cfrac{x}{2}}~\hfill \stackrel{\textit{half of what's left to Marinda}}{\cfrac{~~ \frac{x}{2}~~}{2}\implies \cfrac{x}{4}} \\\\\\ \stackrel{\textit{what was left minus Marinda's}}{\cfrac{x}{2}-\cfrac{x}{4}\implies \stackrel{\textit{what's left}}{\cfrac{x}{4}}}~\hfill ~\hfill \stackrel{\textit{a third of what's left to Zack}}{\cfrac{~~ \frac{x}{4}~~}{3}\implies \cfrac{x}{12}}

\stackrel{\textit{what was left minus Zack's}}{\cfrac{x}{4}-\cfrac{x}{12}\implies \stackrel{\textit{what's left}}{\cfrac{x}{6}}}~\hfill \stackrel{\textit{her sister gets 5 balls of what's left}}{ \cfrac{x}{6}-5 }

and we also know that after all that has been subtracted, she's only left with 5, so we can say that

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5 0
2 years ago
In ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45. What is the value of the cosine of ∠X to the nearest hundredth?
kotegsom [21]

The value of ∠X = 58.11°, If ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45.

Step-by-step explanation:

The given is,

                   In ΔWXY, ∠Y=90°

                        XW = 53

                         YX = 28

                        WY = 45

Step:1

             Ref the attachment,

             Given triangle XWY is right angled triangle.

             Trigonometric ratio's,

                              Cos ∅  = \frac{Adj}{Hyp}    

             For the given attachment, the trigonometric ratio becomes,

                              Cos ∅  = \frac{XY}{XW}.....................................(1)

             Let, ∠X = ∅

             Where, XY = 28

                         XW =  53

             Equation (1) becomes,

                                 Cos ∅  = \frac{28}{53}

                                 Cos ∅ = 0.5283

                                        ∅ = cos^{-1} (0.5283)

                                        ∅ = 58.109°

Result:

          The value of ∠X = 58.11°, If ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45.

             

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