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Evgen [1.6K]
4 years ago
6

Evaluate 3 over 2y-3 + 5 over 3z when y=6 and z=3.

Mathematics
1 answer:
anastassius [24]4 years ago
3 0

Answer:

\frac{8}{9}Step-by-step explanation:[tex]\frac{3}{2y-3}  + \frac{5}{3z}    y = 6, z = 3

Substitute values into the expression:

\frac{3}{2(6)-3}  + \frac{5}{3(3)}

Simplify both fractions:

\frac{3}{9}  + \frac{5}{9}

Add the fractions:

[tex]\frac{8}{9}

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YA is the angle bisector of ZXYZ. If mZXYZ = 52°, what is mZZYA?
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Given:

YA is the angle bisector of \angle XYZ.

m\angle XYZ=52^\circ

To find:

The measure of m\angle ZYA.

Solution:

It is given that YA is the angle bisector of \angle XYZ. It means

m\angle XYA=m\angle ZYA            ...(i)

Now,

m\angle XYA+m\angle ZYA=m\angle XYZ

m\angle ZYA+m\angle ZYA=52^\circ            [Using (i)]

2m\angle ZYA=52^\circ

Divide both sides by 2.

m\angle ZYA=\dfrac{52^\circ}{2}

m\angle ZYA=26^\circ

Therefore, the required value is m\angle ZYA=26^\circ.

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3 years ago
a trapezoid has one base of 10 inches,an altitude of 6 inches and an area of 51 square inches, find the length of the other base
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\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\ ---------\\ a=10\\ h=6\\ A=51 \end{cases}\implies 51=\cfrac{6(10+b)}{2} \\\\\\ 51=3(10+b)\implies \cfrac{51}{3}=10+b\implies 17=10+b\implies 7=b

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