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sale on watches 10% off marked price + 20% off the discounted price what will a watch marked at $1,200.00cost?

Sagot :

Answer:

[tex]

[tex] \large{\boxed{\sf Hypotenuse = 16.97\ cm }}[/tex]

Hypotenuse=16.97 cm

Step-by-step explanation:

Here it is given that the area of a right isosceles ∆ is 72 cm² . Let us assume that each equal side is x . Therefore the height and the base of the ∆ will be same that is x .

p

[tex] \begin{gathered}\sf\qquad\longrightarrow [/tex]

Area =\dfrac{1}{2}(base)(height)\\\end{gathered}⟶Area=21(base)(height)

[tex] \begin{gathered}\sf\qquad\longrightarrow[/tex]

[tex] 72cm^2=\dfrac{1}{2}(x)(x)\\\end{gathered} [/tex]

⟶72cm2=21(x)(x)

[tex] \begin{gathered}\sf\qquad\longrightarrow x^2=[/tex]

144cm^2\\\end{gathered}⟶x2=144cm2

[tex] \begin{gathered}\sf\qquad\longrightarrow x[/tex]

[tex] =\sqrt{144cm^2}\\\end{gathered}⟶x=144cm2 [/tex]

[tex]\sf\qquad\longrightarrow \pink{x = 12cm }⟶x=12cm [/tex]

Hence we may find hypotenuse using Pythagoras Theorem as ,

[tex] \sf\qquad\longrightarrow h =\sqrt{ p^2+b^2}⟶h=p2+b2 [/tex]

Here p = b = 12cm ,

[tex] \begin{gathered}\sf\qquad\longrightarrow h =\sqrt{ (12cm)^2+(12cm)^2}\\\end{gathered} [/tex]

⟶h=(12cm)2+(12cm)2

[tex] \begin{gathered}\sf\qquad\longrightarrow h [/tex]

[tex] =\sqrt{144cm^2+144cm^2}\\\end{gathered}[/tex]

⟶h=144cm2+144cm2

[tex] \begin{gathered}\sf\qquad\longrightarrow h[/tex]

=\sqrt{288cm^2}\\\end{gathered}⟶h=288cm2

[tex] \sf\qquad\longrightarrow \pink{ hypotenuse=[/tex]

16.97cm }⟶hypotenuse=16.97cm

Hence the hypotenuse is 16.97 cm .

[/tex]

Answer:

$864

Step-by-step explanation:

100-10 = 90/100= 0.9 (multiplier)

1200*0.9= 1080

100-20 = 80/100= 0.8 (2nd multiplier)

1080*0.8= 864

hope this helps

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