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Vsevolod [243]
3 years ago
10

The value of -9(square root) is not-3 because

Mathematics
2 answers:
love history [14]3 years ago
7 0

Answer:

A)  (-3)^{2} \neq -9

Step-by-step explanation:

Vlad [161]3 years ago
6 0

Answer:

A) is correct

Step-by-step explanation:

(-3)**2 = 9,  not equal -9

The issue is negative numbers can only be imaginary numbers. The sq root of -9 would be 3i, where i is defined as the sq root of -1.

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3 0
3 years ago
Question 3 of 10
mestny [16]

Answer:

㋡

Check Answer

♣ Qᴜᴇꜱᴛɪᴏɴ :

If tan θ = \sf{\dfrac{1}{\sqrt{7}}}

7

1

, Show that \sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

We know :

\large\boxed{\sf{tan\theta=\dfrac{Height}{Base}}}

tanθ=

Base

Height

So comparing this formula and value of tan θ from question, we get :

Height = 1

Base = √7

Now we need to Prove the value of : \sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

Also :

\large\boxed{\sf{cosec\theta=\dfrac{Hypotenuse}{Height}}}

cosecθ=

Height

Hypotenuse

\large\boxed{\sf{sec\theta=\dfrac{Hypotenuse}{Base}}}

secθ=

Base

Hypotenuse

From this we get :

\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}}

cosec

2

θ=(

Height

Hypotenuse

)

2

\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}}

sec

2

θ=(

Base

Hypotenuse

)

2

But we have Height and Base, we dont have Hypotenuse.

Hypotenuse can be found by using Pythagoras Theorem

Pythagoras Theorem states that :

Hypotenuse² = Side² + Side²

For our question :

Hypotenuse² = Height² + Base²

Hypotenuse² = 1² + √7²

Hypotenuse² = 1 + 7

Hypotenuse² = 8

√Hypotenuse² = √8

Hypotenuse = √8

➢ Let's find value's of cosec²θ and sec²θ

________________________________________

First cosec²θ :

\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}}

cosec

2

θ=(

Height

Hypotenuse

)

2

\sf{cosec^2\theta=\left(\dfrac{\sqrt{8}}{1}\right)^2}cosec

2

θ=(

1

8

)

2

\sf{cosec^2\theta=\dfrac{8}{1}}cosec

2

θ=

1

8

cosec²θ = 8

________________________________________

Now sec²θ :

\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}}

sec

2

θ=(

Base

Hypotenuse

)

2

\sf{sec^2\theta=\left(\dfrac{\sqrt{8}}{\sqrt{7}}\right)^2}sec

2

θ=(

7

8

)

2

\sf{sec^2\theta=\dfrac{8}{7}}sec

2

θ=

7

8

sec²θ = 8/7

________________________________________

Now Proving :

\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

Taking L.H.S :

\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=\sf{\dfrac{8 - sec ^2\theta}{8 + sec^2\theta }}=

8+sec

2

θ

8−sec

2

θ

=\sf{\dfrac{8 - \dfrac{8}{7}}{8 + \dfrac{8}{7} }}=

8+

7

8

8−

7

8

=\sf{\dfrac{\dfrac{48}{7}}{\dfrac{64}{7} }}=

7

64

7

48

\sf{=\dfrac{48\times \:7}{7\times \:64}}=

7×64

48×7

\sf{=\dfrac{48}{64}}=

64

48

\bf{=\dfrac{3}{4}}=

4

3

= R.H.S

Hence Proved !!!

7 0
3 years ago
The two rectangles below are similar. What is the measurement of the missing side of the rectangle?
r-ruslan [8.4K]
B

You would divide 45 mm/15mm to get 3. Multiply 3 by 27 to get 81 mm.
5 0
3 years ago
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