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Yeah that makes more sense...your logic is all good!

Ohh sorryy !

I meant that angle DCE + angle DEC = 76

I get the same answer as you but your first step confuses me...Angle DCE is not 76 degrees, it's 38 degrees... could you explain that a bit more?

Thanks!

" ...Sikorsky UH-60A helicopter ... "

Thank youuu!!!

I get it know :)

Yes that is great.    

x=45

So angle DAB=90  and you know that angle ABC=90

So pow can you use that to determine if  AD and BC are parallel?

This is what I got 

360-90 = 270˚
270˚= 3x + 2x + x
270˚/3= 90
90/2=45
x=45˚

Please show me your answer, with working , for part a

WE are not here to do all your homework

Thanks, but apparantley that was only worth 1/3 marks for that question, any idea why that could be?

Thanks, but apparantley that was only worth 1/3 marks for that question do you know why that could be?

15/12*200 = 250

k = 5/4

Здравствуйте, товарищ,

что УДИВИТЕЛЬНО?

  !

You would never be able to create a sensible graph with this data by plotting N/m^2 with 1 N/m^2 grid lengths on the x-axis!  You could do it if you scaled the values by 10^7 or 10^8 perhaps.

Thanks for the response! I meant like having squares of 1 by 1 instead of 1*10⁸ by 1*10^-3 . That's what caused so much confusion for me. 

Let one angle be x.  The other is 17x.

So:   x + 17x = 90

You can take it from here,

"Noa drove from the Dead Sea up to Jerusalem, and her altitude increased at a constant rate of 740 meters per hour. When she arrived in Jerusalem after 1.5 hours of driving, her altitude was 710 meters above sea level. Let A represent Noa's altitude (in meters) relative to sea level after t hours. Complete the equation for the relationship between the altitude and number of hours."

Let t = time in hours:

A = 740*t + k

710 = 740*1.5 + k

710 = 1110 + k

k = 710 - 1110 = -400

A = 740*t - 400

There are 10 students in a class: 6 boys and 4 girls. If the teacher picks a group of 3 at random,

what is the probability that everyone in the group is a girl?

Hypergeometric distribution:
\(\text{$N$ is the number of $\mathbf{students} = 10$}, \\ \text{$b$ is the number of $\mathbf{boys} = 6$}, \\ \text{$g$ is the number of $\mathbf{girls}= 4$}, \\ \text{$k$ the teacher picks a $\mathbf{boy} = 0 $}, \\ \text{$n$ the teacher picks a $\mathbf{girl} = 3 $}.\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{ \dbinom{b}{k} \dbinom{ g }{n} }{ \dbinom{N}{n+k} }} \\\\ &=& \dfrac{ \dbinom{6}{0}_{\text{boys}} \dbinom{4}{3}_{\text{girls}} } { \dbinom{10}{3}_{\text{all}} } \\\\ && \boxed{\dbinom{6}{0} = 1,\\ \dbinom{4}{3}=\binom{4}{4-3}=\binom{4}{1}=4,\\ \dbinom{10}{3}=\dfrac{10}{3}\cdot \dfrac{9}{2}\cdot \dfrac{8}{1} = 5\cdot 3 \cdot 8 } \\\\ &=& \dfrac{ 1\cdot 4 } { 5\cdot 3 \cdot 8 } \\\\ &=& \dfrac{ 1 } { 5\cdot 3 \cdot 2 } \\\\ &=&\mathbf{ \dfrac{ 1 } { 30 }} \\\\ &=& 0.03333333333\quad (=3.3\ \%) \\ \hline \end{array}\)

Compute the sum \(\dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \cdots\)

Hi guest!

The answer is C. 

There are 3 black squares and 8 squares in total. So the ratio is 3/8. 

The only choice that doesn't work is C, because 3*7=21, but 8*7 does not = 21.

Post ONE question.

Think about the answer.

Think about your next question.

If necessary post your next question

I am locking this question.

Go away, we are not here to do your assignment for you!

Since you are too lazy to present your question properly I am quite sure you are too lazy to learn from any answer.

Because you asked 3 questions.

Answerers are here to help. They are not here to do all your homework for you.

Plus asking more than one question on a post causes confusion.

Finding the vertex of       y  =  x² - 2x + 3                           y  =  x² + 4x + 10

                                   y - 3  =  x² - 2x                          y - 10  =  x² + 4x

                             y - 3 + 1  =  x² - 2x + 1              y - 10 + 4  =  x² + 4x + 4

                                    y - 2  =  (x - 2)²                           y - 6  =  (x + 2)²  

                                    Vertex:  (2, 2)                              Vertex:  (-2, 6)

To find the distaces between the vertices:  distance  =  sqrt( (x₂ - x₁)² + (y₂ - y₁)² )                   

                                                                                    =  sqrt( (-2 - 2)² + (6 - 2)² )

                                                                                    =  sqrt( (-4)² + (4)² )

                                                                                     =  sqrt( 16 + 16 )

                                                                                     =  sqrt( 32 )