3 years ago

Please help: Find the class width of this graph

Mathematics

1 answer:

maksim [4K]3 years ago

4 0

Answer:

The biggest one is 32 to 38

Step-by-step explanation:

hope it helps

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Write 1/5 as a percent.

Softa [21]

Answer:

0.2

Step-by-step explanation:

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3 years ago

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Easy question.. giving brainliest<br> x/16 = 10/36<br> x=_

irina [24]

Answer:

x=40/9

Step-by-step explanation:

$\frac{x}{16} = \frac{10}{36} |first \: you \: cross \: multiply| \\ x(36) = 16(10) |open \: bracket | \\ 36x = 160 |divide \: both \: side \:by \: 36 | \\ \frac{36x}{36} = \frac{160}{36} \\ x = \frac{40}{9}$

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3 years ago

Find the vertices and foci of the hyperbola with equation quantity x plus 1 squared divided by 225 minus the quantity of y plus

miss Akunina [59]

Answer:

The vertices are (-16,-5) and (14,-5)

The foci are (-26,-5) and (24,-5)

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3 years ago

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NISA [10]

77 feet I according to my calculations :)

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3 years ago

Someone please let me know the answer

erastova [34]

Answer:

The answer is 30°

Step-by-step explanation:

$when \: \: \sqrt{3} \sin( \alpha ) - \cos( \alpha ) = 0 \\ \sqrt{3} \sin( \alpha ) = \cos( \alpha ) \\ multiply \: by \: \frac{1}{ \sqrt{3} \cos( \alpha ) } \\ then \\ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{1}{ \sqrt{3} } \\ \tan( \alpha ) = \frac{1}{ \sqrt{3} } \\ then \: \alpha = 30$

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3 years ago