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SashulF [63]
3 years ago
13

For kite JKLM, find m∠JML.

Mathematics
1 answer:
Simora [160]3 years ago
5 0

Answer:

∠JML = 76°

Step-by-step explanation:

since it is a kite, ∠1 = 38°

∠JML = 2(38°) = 76°

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Factor: 4b^3 - 12b^2 - 9b + 27
Mashcka [7]

Answer:

(2b - 3)(2b + 3)(b - 3)

Step-by-step explanation:

4b³ - 9b - 12b² + 27

b(4b² - 9) - 3(4b² - 9)

(4b² - 9)(b - 3)

(2b - 3)(2b + 3)(b - 3)

8 0
3 years ago
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Find an ordered pair (x, y) that is a solution to the equation. -x+4y= 5​
xenn [34]
The pair in the equation is (0,5)
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3 years ago
12. In ∆PQR, if m∠P is 14 less than five times x, m∠Q is five less than x, and m∠R is nine less than twice x, find x and the mea
pychu [463]

Answer:

12)x=26.125°

∠P  = 115.625°

∠Q =21.125°

∠R=43.25°

13)x=52°

∠L=52°

∠J =45°

∠K =69°

14)x=11°

∠C = 36°

∠D =36°

∠B = 108°

15)x=7

WX = 52

WY=52

XY =29

Step-by-step explanation:

12)In ∆PQR

∠P is 14 less than five times x = 5x-15

∠Q is five less than x=x-5

∠R is nine less than twice x=2x-9

Angle sum property of triangle : The sum of all angles of triangle is 180°

So,5x-15+x-5+2x-9=180

8x-29=180

8x=180+29

x=\frac{180+29}{8}

x=26.125

So,∠P  = 5x-15 =5(26.125)-15=115.625°

∠Q =x-5=26.125-5=21.125°

∠R=2x-9=2(26.125)-9=43.25°

13)  In ∆JKL

Let the angle L be x

∠L=x

∠J is seven less than ∠L=x-7

∠K is 21 less than twice ∠L=2x-21

Angle sum property of triangle : The sum of all angles of triangle is 180°

So, x+x-7+2x-21=180°

4x-28=180°

4x=180+28

x=\frac{180+28}{4}

x=52

∠L=x=52°

∠J =x-7=52-7=45°

∠K =2x-21=2(45)-21=69°

14)In ∆BCD

BC ≅ BD(Given)

Opposite angles of equal sides are equal

So, ∠C =∠D

∠C = 5x – 19

∠D = 2x + 14

So, 5x-19=2x+14

3x=33

x=11

∠C = 5x – 19=5(11)-19=36°

∠D = 2x + 14=2(11)+14=36°

∠B = 13x – 35=13(11)-35=108°

15) In ∆WXY

∠X ≅ ∠Y(Given)

Opposite sides of equal angles are equal

WY=WX

WX = 9x – 11

WY = 7x – 3

So, 9x-11=7x-3

2x=14

x=7

So, WX = 9x – 11=9(7)-11=52

WY=52

XY = 4x + 1=4(7)+1=29

3 0
3 years ago
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Well if the alloy makes the 1000 grams, but you are dealing with copper that isn't even 100% copper, than you either need 55% more copper, which isn't given, which means you really need 550 grams of alloy copper.
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Can someone answer these two questions please
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