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kow [346]
3 years ago
14

Which postulate or theorem proves that these two triangles are congruent?

Mathematics
1 answer:
avanturin [10]3 years ago
6 0

It is AAS because:

1. They have one shared side WZ which is reflex property   - S

2. Angle XWZ is congruent to angle YWZ because it is given  -A

3. Angle WXZ is congruent to angle WYZ because it is given  -A

SAA=AAS

Hope this helps :)


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What is the solution?. 1/2x2 – x + 5 = 0
AfilCa [17]
1/2 x^2 - x + 5 = 0
Multiply through by 2 to get:
x^2 - 2x + 10 = 0
x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}; where a = 1, b = -2 and c = 10
x=\frac{-(-2)\pm\sqrt{(-2)^2-(4\times1\times10)}}{2\times1}\\=\frac{2\pm\sqrt{4-40}}{2}\\=\frac{2\pm\sqrt{-36}}{2}\\=\frac{2\pm6i}{2}\\=1\pm3i
Therefore, x = 1 + 3i or x = 1 - 3i
4 0
4 years ago
1/3y+1/4=5/12 i need 20 character
lys-0071 [83]
Y/3 + 1/4 = 5/12
4y/12 + 3/12 = 5/12
           4y/12 = 2/12
                4y = 2
                /4     /4
                  y = 1/2

Therefore y = 1/2

3 0
3 years ago
On Sunday, Dan collected 6.8 pounds of aluminum for recycling. On Monday he collected another 8.75 pounds. What is the total num
Tresset [83]
6.8 + 8.75 = 11.55

Dan collected a total of 15.55 pounds of aluminum.
6 0
3 years ago
Read 2 more answers
Georgia knows that in the relationship between two
marysya [2.9K]

Answer:

Maybe try doing it earlier, don’t try doing things last minute it will put too much stress. Thank you :)

Step-by-step explanation:

5 0
3 years ago
PLEASE HELP!!!!!!!!!!!!!!
Lelechka [254]

Answer:

Step-by-step explanation:

(8x²-18x+10)/(x²+5)(x-3)

express the expression as a partial fraction:

(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +bx+c/x²+5

both denominator are equal , so require only work with the nominator

(8x²-18x+10)=(x²+5)A+(x-3)(bx+c)

8x²-18x+10= x²A+5A+bx²+cx-3bx-3c

combine like terms:

x²(A+b)+x(-3b+c)+5A-3c

(8x²-18x+10)

looking at the equation

A+b=8

-3b+c=-18

5A-3c=10

solve for A,b and c (system of equation)

A=2 , B=6, and C=0

substitute in the value of A, b and c

(8x²-18x+10)/[(x^2+5)(x-3)] =A/x-3 +(bx+c)/x²+5

(8x²-18x+10)/[(x^2+5)(x-3)] = 2/x-3 + (6x+0)/(x²+5)

(8x²-18x+10)/[(x^2+5)(x-3)] =

<h2>2/(x-3)+6x/x²+5</h2>

(4x+2)/[(x²+4)(x-2)]

(4x+2)/[(x²+4)(x-2)]= A/(x-2) + bx+c/(x²-2)

(4x+2)=a(x²-2)+(bx+c)(x-2)

follow the same step in the previous answer:

the answer is :

<h2>(4x+2)/[(x²+4)(x-2)]= 5/4/(x-2) + (3/2 -5x/4)/(x²+4)</h2>

8 0
3 years ago
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