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Ierofanga [76]
3 years ago
8

Given that triangle ABC ~ triangle DEF, solve for x and y

Mathematics
1 answer:
Marina CMI [18]3 years ago
4 0
Y would be 8 X would be 10
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M is the midpoint of RS, and
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What is the net decimal equivalent of the series 30/25/15. round to the nearest thousandths?
Brut [27]
<span>We can determine, a 30/25/15 series trade discount would be calculated as follows:

let X = retail price.

X - 0.30 </span>× X = Y    ⇒ first discounted price.
<span>
Y - 0.25 </span>× Y = Z    ⇒ second discounted price.
<span>
Z - 0.15*Z = T        </span>⇒ total discounted price.
<span>
X - 0.30 </span>× X = (1 - 0.30) × X = 0.70 × X = Y
<span>
Y - 0.25 </span>× Y = (1 - 0.25) × Y = 0.75 × Y = Z
<span>
Z - 0.15 </span>× Z  = (1 - 0.15) × Z  = 0.85 × Z = T
<span>
since Z = 0.85 </span>× Y, then 0.85 × Z = 0.85 × 0.75 × Y
<span>
since Y = 0.70 </span>× X, then 0.85 × Z = 0.85 × 0.75 × 0.70 × X
<span>
based on the above, then T = total discount


                                               = 0.85 </span>× 0.75 × 0.70 × X


                                               = 0.44625X<span>

30/25/15 series discount is equivalent to a total discount of 44.625%
</span>
7 0
3 years ago
Find the length of UV. v(2,-1) u(1,-9)
fiasKO [112]

Answer: \sqrt{65}

Step-by-step explanation:

<u>Concept:</u>

Here, we need to know the concept of the distance formula.

The distance formula is the formula, which is used to find the distance between any two points.

If you are still confused, please refer to the attachment below for a clear version of the formula.

<u>Solve:</u>

Find the length of UV, where:

  • U (1, -9)
  • V (2, -1)

Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Distance = \sqrt{(1-2)^2+(-9+1)^2}

Distance = \sqrt{(-1)^2+(-8)^2}

Distance = \sqrt{1+64}

Distance = \sqrt{65}

Hope this helps!! :)

Please let me know if you have any questions

8 0
3 years ago
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