Write the two consecutive integers between which the given rational lies

Let's solve the equation for x! 3x-3=180 *

Trava [24]

I think this is the answer

3 0

3 years ago

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What is the length of a diagonal of a cube with a side length of 10 cm? 200 cm 210cm 300 cm 320cm

Scilla [17]

Answer:

The length of the diagonal of the cube = √(3 × 10²) = √300 cm

Step-by-step explanation:

* Lets revise the properties of the cube

- It has six equal faces all of them are squares

- It has 12 vertices

- The diagonal of the cube is the line joining two vertices in opposite

faces (look to the attached figure)

- To find the length of the diagonal do that:

# Find the diagonal of the base using Pythagoras theorem

∵ The length of the side of the cube is L

∵ The base is a square

∴ The length of the diagonal d = √(L² + L²) = √(2L²)

- Now use the diagonal of the base and a side of a side face to find the

diagonal of the cube by Pythagoras theorem

∵ d = √(2L²)

∵ The length of the side of the square = L

∴ The length of the diagonal of the cube = √[d² + L²]

∵ d² = [√(2L²)]² = 2L² ⇒ power 2 canceled the square root

∴ The length of the diagonal of the cube = √[2L² + L²] = √(3L²)

* Now lets solve the problem

∵ The length of the side of the square = 10 cm

∴ The length of the diagonal of the cube = √(3 × 10²) = √300 cm

- Note: you can find the length of the diagonal of any cube using

this rule Diagonal = √(3L²)

7 0

3 years ago

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Rewrite the expression in terms of the given function 1/1-sinx - sinx/1+sinx

Feliz [49]

Your question seems a bit incomplete, but for starters you can write

$\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}$

Expanding where necessary, recalling that $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ , you have

$\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}$

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

$\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x$

Now, since $1+\tan^2x=\sec^2x$ , the final form could also take

$\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1$

or

$\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x$

7 0

3 years ago

15 POINTSSS<br> Need Anwsered Asap!

german

The equation we will use here is A^2+B^2=C^2, which is also know as the Pythagorean Theorem.
The given values are 6 and 9, where they can represent any value, there true values in the equation would be 36(6), and 81(9), so you must select a value that makes the equation true, given the constraints.

with that being said 3, doesnt work because
·36(6)+9(3)≠81(9)
·9(3)+81(9)≠36(6)
·36(6)+81(9)≠9(3)

10 doesnt work either because
·36(6)+81(9)≠100(10)
·81(9)+100(10)≠36(6)
·100(10)+36(6)≠81(9)

12 doesnt work either because
·144(12)+36(6)≠81(9)
·36(6)+81(6)≠144(12)
·81(9)+144(12)≠36(6)

If you see where this is going you would know that there is no valid solution here, however rounding is always a possibility, when you actually do the math 81(9)+36(6)=117, and when squared you get your answer of 10.8, and the closest answer is 10, there fore your answer would be 10

-I hope this is the answer you are looking for, feel free to post your questions on brainly at any time.

5 0

3 years ago

Need help! pls and thank youu...

tatuchka [14]

C dilation and rotation

5 0

3 years ago

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