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mezya [45]
4 years ago
14

Remove all perfect squares from inside the square root. \sqrt{15y^3}

Mathematics
1 answer:
USPshnik [31]4 years ago
6 0

y\sqrt{15y} is the term after removing perfect square from  \sqrt{15y^3}  .

<u>Step-by-step explanation:</u>

Here we have , an expression as \sqrt{15y^3} or \sqrt{15y^3} . We need to remove all perfect squares inside the square root . Let's find out:

We know that a perfect square is the term whose degree is two , or can be written in form a^2 , Where a is any term or integer or value or variable . Now ,

⇒  \sqrt{15y^3}

⇒  \sqrt{15y(y)(y)}

⇒  \sqrt{15y(y^2)}

⇒  \sqrt{15y}(\sqrt{(y^2)})

⇒  \sqrt{15y}((y^2)^{\frac{1}{2}})

⇒  \sqrt{15y}(y)

⇒  y\sqrt{15y}

Therefore , y\sqrt{15y} is the term after removing perfect square from  \sqrt{15y^3}  .

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4 years ago
an object is launched at 25 m/s from a height of 80 m. the equation for the height (h) in terms of time (t) is given by h(t)=4.9
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3 0
3 years ago
Answer soon please! Question and answers are in the photograph attached.
-BARSIC- [3]

Answer:

Δ MLP  ~  Δ HKG

By S-A-S similarity Postulate.

Step-by-step explanation:

Given:

In Δ MLP

MP = 20 ,

LP = 15

∠ P = 42°

In Δ HKG

HG = 16 ,

KG = 12

∠G = 42°

To Prove:

Δ MLP ~Δ HKG

Proof:

In Δ MLP and Δ HKG

\dfrac{MP}{HG}= \dfrac{20}{16}=\dfrac{5}{4} ..........( 1 )

\dfrac{LP}{KG}= \dfrac{15}{12}=\dfrac{5}{4}  ..............( 2 )

∴ \dfrac{MP}{HG}=\dfrac{LP}{KG} ...From 1 and 2

       ∠P ≅ ∠ G = 42°          ...............Given

∴ Δ MLP ~Δ HKG ......{ By S-A-S similarity test}............Proved

4 0
3 years ago
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