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3 years ago

Which are perfect squares? Check all that apply. 49 17 g 143 225

Mathematics

2 answers:

Vika [28.1K]3 years ago

5 0

<span>49 17 g 143 225
</span><span>
you have g in here
is g is 9?

perfect squares
49 = 7^2
225 = 15^2</span>

Ksenya-84 [330]3 years ago

4 0

49,
9,
and
225
are all perfect squares.
I hope this helps you!

Brainliest if satisfied :D

Would mean a lot

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A jewelry sells gold and platinum rings. Each ring is available in eight styles and is fitted with one of ten gemstone

katrin [286]

Answer: There are 160 different rings.

Step-by-step explanation:

The question is missing, i guess you want to know the number of different rings that the jewelry has:

For the material we have 2 options.

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Now, the total number of possible combinations is equal to the product of the number of options for each selection, this is:

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4 years ago

Find at least three nonzero terms (including a 0 and at least two cosine terms and two sine terms if they are not all zero) of

Ainat [17]

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3 years ago

A boat is being pulled into a dock with a rope attached to the boat at water level. When the boat is 12 feet from the dock, the

Natali5045456 [20]

Given :

A boat is being pulled into a dock with a rope attached to the boat at water level. When the boat is 12 feet from the dock, the length of the rope form the boat to the dock is 3 feet longer than twice the height of the dock above the water.

To Find :

The height of the dock.

Solution :

This will make a right angle triangle as given in link below .

Now , applying Pythagoras theorem :

$(2h+3)^2=h^2+12^2\\\\4h^2+9h+9=h^2+144\\\\h^2+4h-45=0\\\\(h-5)(h+9)=0$

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Hence , this is the required solution .

8 0

4 years ago

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just olya [345]

The "9" in the second term is a coefficient.

5 0

3 years ago

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Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$

=> The clause is correct

20. Do the same (18)

21. Left = $\frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA} } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}$

Right = $cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA} \\$

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

5 0

3 years ago