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3 years ago

2 times 2 plus 10 plus 200 plus 56

2 answers:

5 0

2×2 + 10 + 200 + 56
The order of operations requires the multiplication be done first.
= 4 + 10 + 200 + 56
The addition is done left to right.
= 14 + 200 + 56
= 214 + 56
= 270

Illusion [34]3 years ago

3 0

Answer:

270

Step-by-step explanation:

2*2+10+200+56=270

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Which statement is true about the box plots? Both the ranges and the interquartile ranges for the data sets are the same. Neithe

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Answer:

Neither the ranges nor the interquartile ranges for the data sets are the same.

Step-by-step explanation:

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The graph of an inverse trigonometric function passes through the point (1, pi/2). Which of the following could be the equation

rodikova [14]

Answer: C) $y=sin^-1 x$

Step-by-step explanation:

Since, the graph of an inverse trigonometric function will pass through the point $(1,\frac{\pi}{2})$ ,

If this point satisfies the function,

For the function $y=cos^{-1} x$

If x = 1

$y=cos^{-1}1=0$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=cos^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=cot^{-1}x$

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$y=cot^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cot^{-1}x$ ,

⇒ The graph of $y=cot^{-1}x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=sin^{-1} x$

If x = 1

$y=sin^{-1}1=\frac{\pi}{2}$

Thus, $(1,\frac{\pi}{2})$ is satisfying function $y=sin^{-1} x$ ,

⇒ The graph of $y=sin^{-1} x$ is passing through the point $(1,\frac{\pi}{2})$ .

For the function $y=tan^{-1}x$

If x = 1

$y=tan^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=tan^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$ .

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3 years ago

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AB and EF

Step-by-step explanation:

They are the same length

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mojhsa [17]

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