Rational Equations
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TIME REMA
22:4
What is the solution to the equation h-5
216
h+5 1225
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option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

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step-by-step explanation:

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629