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DochEvi [55]
2 years ago
5

Based upon the information presented in the video, what is the BEST definition of a refugee?

Mathematics
2 answers:
erik [133]2 years ago
8 0

Answer:

Someone who is forced to leave their home country due to war, natural disaster, or persecution.

Step-by-step explanation:

AleksandrR [38]2 years ago
6 0

girl don't be lazy and pay attention to the video

Step-by-step explanation:

1 open video

2 pay attention

3 answer

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A cone is placed inside a cylinder. The cone has half the radius of the cylinder, but the height of each figure is the same. The
Yuliya22 [10]
Here is the answer: 3/4 (pi r^2 h)
5 0
3 years ago
Read 2 more answers
7)
yuradex [85]

Answer:

9 blue marbles

Step-by-step explanation:

12 divided by 4 = 3

3x3=9

7 0
3 years ago
Which statement explains how the lines 2x + y = 4 and y = one halfx + 4 are related?
KatRina [158]

Answer:

They are perpendicular

Step-by-step explanation:

To solve this problem .

we will convert the equations in slope intercept form.

Slope intercept  form of equation is y = mx+c

where m is slope of line and c is y intercept.

________________________________

equation 1 is

2x+y = 4

=> y =4 - 2x or y = -2x + 4

comparing it with y = mx + c

m = -2  , c = 4

_________________________________________

equation 2 is y = one halfx + 4 ( one half is same as 1/2)

so equation is

y = x/2 +4

comparing it with y = mx + c

m = 1/2  , c = 4

_________________________________________

Now lets evaluate options

They are parallel.  wrong option

For lines to be parallel slope should be same.

But here slope are different -2 and 1/2 .

Thus lines are not parallel.

__________________________________________

They are perpendicular.  correct option

For lines to be perpendicular, product of slope should be equal to -1.

-2*1/2 = -1

we can see that product of slope should be equal to -1 .

Thus lines are  perpendicular

______________________________________

They are the same line.  wrong option

For lines to be same both slope and y intercept should  be same.

Y intercept is same but the slopes are different -2 and 1/2  .

Thus lines are not  the same line.

__________________________________________

They are not related.       wrong option

As we have found that the lines are perpendicular .

So this option is intuitively wrong

4 0
3 years ago
Without using trignometery tables find value of , cos70/sin20+cos57.cosec33 - 2cos60
Helen [10]
Here  is some thoughts on this:

cos 57  = sin (90-57) = sin 33 

so cos 57 . cosec 33 = cos 57 / sin 33 = sin 33 / sin33 = 1

2 cos 60 = 2 * 1/2  = 1

so the last 2 parts work out to 0 

now we have to find cos 70 / sin 20

sin 20 = cos 70  so this comes to 1


so finally the answer is 1
4 0
3 years ago
How do you find a vector that is orthogonal to 5i + 12j ?
Rashid [163]
\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies  -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\

\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies  -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}&#10;\\\\\\&#10;\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}&#10;\\\\\\&#10;\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}
4 0
3 years ago
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