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Tpy6a [65]
3 years ago
12

What’s is the solution to the equation x to the power of 2 =16/25

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer:

x=\dfrac{4}{5}

Step-by-step explanation:

The given equation is :

x^2=\dfrac{16}{25}

We need to find the value of x.

When the exponent 2 removes from LHS, it will shift to RHS as square root as follows :

x=\sqrt{\dfrac{16}{25}}

We know that, 4×4= 16 and 5×5 = 25

So,

x=\sqrt{\dfrac{4\times 4}{5\times 5}}\\\\x=\dfrac{4}{5}

So, the solution of the given equation is 4/5.

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Step-by-step explanation:

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1: a 24 ounce box of cornflakes cost $4.59?
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The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC
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Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x²  ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

∴  10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴  10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

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PA = 2sqrt5 in  and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

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