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Fantom [35]
4 years ago
11

(i need help ;-;)

Mathematics
1 answer:
navik [9.2K]4 years ago
6 0
Really bro haha why cant you just look it up on google ??? haha the answer is 4:5 im sure 
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4n - 2n = 4<br> Can someone help me solve this?
Semmy [17]

Answer:

n= 2

Step-by-step explanation:

i know this is right

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Given the lengths of the sides which of the following would NOT be a right triangle?
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Which equation is equivalent to the equation x^2+y^2= 25?
satela [25.4K]

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Answer:

  D.  y = ±√(25 -x²)

Step-by-step explanation:

Subtract x² and take the square root to solve for y.

  x² +y² = 25 . . . . . . . given

  y² = 25 -x² . . . . . . . . subtract x²

  y = ±√(25 -x²) . . . . . take the square root

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3 years ago
A stem-and-leaf plot for the number of homeruns hit by 30 high school teams is shown. find the probability that a team chosen at
pantera1 [17]
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There are 10 teams at 28 or less, so 20 teams are above 28.

The number to the left of the | tells us the tens place of the set of number that come after it.

For example 2|4 7 8 stands for 24, 27, and 28.

If 20 out of 30 were above 28 homeruns, then your probability is 20/30 or 2/3 or 0.667.
4 0
3 years ago
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

If x = 1

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}).

Hence, Option C is correct.

3 0
3 years ago
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