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3 years ago

Help me with this question please

Mathematics

1 answer:

jasenka [17]3 years ago

7 0

Answer:

its blurd soryy

i cant see it

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Help someone to help me :( in number 2 and 3

sveticcg [70]

So the things that you need to know are in the picture I have to see if I can help you with the other one hope this helps.

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2 years ago

What is the perimeter of the quadrilateral that is the length of 7 height of 8

marta [7]

Answer:

Step-by-step explanation:

What you need to do is add all the sides when you want to find perimeter so for this one you would want to do 8+8+7+7 and that should give you 30. I hope I could help!!

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2 years ago

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Joel is mailing a large envelope to his cousin. The envelope has pictures inside, so he doesn't want to bend it.

Nuetrik [128]

Answer: Put a cardboard piece inside the envelope then mail it

Step-by-step explanation: If you make the cardboard piece the exact size as the pictures then put it behind all of the pictures it would make it hard for the pictures to bend if it has support.

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3 years ago

Write an equation parallel to the line determined by the points (15, -6) and (-3, 13), through: (4, 2)

gulaghasi [49]

Answer:

The answer is

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

To find the equation of the parallel line we must first find the slope of the original line

That's

Slope of the through points

(15, -6) and (-3, 13) is

Since the lines are parallel their slope are also the same

So slope of parallel line = - 19/18

Equation of the line using point (4,2) and slope -19/18 is

We have the final answer as

Hope this helps you

7 0

2 years ago

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Find cotθ, cosθ, and secθ, where θ is the angle shown in the figure.

Darina [25.2K]

Answer:

$\cos( \theta) = \frac{5}{8} \\ {8}^{2} = {5}^{2} + {opp}^{2} \\ {opp}^{2} = 64 - 25 = 39 \\ opp = \sqrt{39} \\ \sin(\theta) = \frac{\sqrt{39}}{8} \\ \tan(\theta) = \frac{ \sqrt{39} }{5} \\ \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{\frac{ \sqrt{39} }{5}} \\ \cot(\theta) = \frac{5}{ \sqrt{39} } \\ \csc( \theta) = \frac{1}{\sin(\theta)} = \frac{1}{\frac{\sqrt{39}}{8}} \\ \csc( \theta) = \frac{8}{ \sqrt{39} } \\ \sec( \theta) = \frac{1}{\cos( \theta) } = \frac{1}{ \frac{5}{8} } \\ \sec( \theta) = \frac{8}{5}$

5 0

2 years ago