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oksian1 [2.3K]
2 years ago
13

Three equals some number plus five.

Mathematics
1 answer:
Lady bird [3.3K]2 years ago
4 0

Answer:

3 = x + 5

x = - 2

This the answer ^_^

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Hello guys I really need help is anyone available please and ty
Diano4ka-milaya [45]

Answer:

6x5x5

Step-by-step explanation:

the square is a 5x5

and there are 6 of those squares

so it is 6x5x5

4 0
3 years ago
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8% of what number is 16
aliina [53]
200, because 8% is 8 parts of 100%. Then 1 part of that 100% is 16/8=2. Then 100% is the number i.e; 200
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Can anyone help with these maths questions about transformations please?
salantis [7]
A. Reflection over y = 2

B. Reflection over y axis, reflection over y = 1

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Did I do this Math Right
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6 0
3 years ago
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
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