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anonymous

  • one year ago

CAN SOMEONE PLEASE HELP!! Just need to finish just 1 and I'm done!!

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  1. anonymous
    • one year ago
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    hello

  2. anonymous
    • one year ago
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    ummmmmm

  3. anonymous
    • one year ago
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    Calculate the area of triangle WXY with altitude YZ, given W (2,-1), X (6,3), Y (7,0) and Z (5,2) 8 square units 9.4 square units 7.7, square units 12 square units

  4. anonymous
    • one year ago
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    @mumustar78612

  5. campbell_st
    • one year ago
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    well I'd say you need the distance formula to find the base length and altitude the base would be WX, so find the distance from (2, -1) tp (6,3) and the altitude YZ is the distance from (7,0) to (5,2) then use the normal area of a triangle A =1/2bh

  6. anonymous
    • one year ago
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    I got 7.7 idk if that's right

  7. campbell_st
    • one year ago
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    I think you need to check your answer

  8. anonymous
    • one year ago
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    Is it 9.4?

  9. campbell_st
    • one year ago
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    \[WX = \sqrt{(6 -2)^2 + (3 - (-1))^2}\] \[YZ = \sqrt{(7 - 5)^2 + (0 - 2)^2}\] find those 2 values..

  10. anonymous
    • one year ago
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    I got 4

  11. anonymous
    • one year ago
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    Well 32/8

  12. anonymous
    • one year ago
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    I calculated everything

  13. campbell_st
    • one year ago
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    \[WX = \sqrt{32} = 4\sqrt{2}~~~and~~~YZ = \sqrt{8} = 2\sqrt{2}\] then the area is \[A = \frac{1}{2} \times WX \times YZ\] which is \[A = \frac{1}{2} \times 4 \sqrt{2} \times 2 \sqrt{2}\]

  14. anonymous
    • one year ago
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    12??

  15. campbell_st
    • one year ago
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    so you have \[A = \frac{1}{2} \times 4 \sqrt{2} \times 2\sqrt{2} = 2 \times 2 \times \sqrt{2} \times \sqrt{2}\] and you need to know \[\sqrt{a} \times \sqrt{a} = a\] hope that helps

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