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Irina-Kira [14]
2 years ago
15

Needs help with this!!!

Mathematics
1 answer:
barxatty [35]2 years ago
6 0

Answer:

y^1=-1(x-1)+(1+9)

Step-by-step explanation:

y=-1x+1\ \ == > \ \ y^1=-1(x-1)+(1+9)

-1 horizontally = 1 units left (the x-axis)

9 vertically = 9 units up (y-axis)

*Black line: is the preimage

*Red line is the translation

Hope this helps!

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A conjecture and the flowchart proof used to prove the conjecture are shown.
viva [34]

The required proof of parallelogram is given below,

What is parallelogram?

A parallelogram is a straightforward quadrilateral with two sets of parallel sides in Euclidean geometry. The opposing or confronting sides and the opposing angles in a parallelogram are of equal length.

For a parallelogram, opposite sides are always equal and parallel.
So for the given parallelogram, we get AD = BC and AD and BC are parallel.

AD and BC is parallel and DC is common side, therefore, \angle 1 = \angle 3 ......(1)

Given \angle 3 = \angle 2  .......(2)

Therefore, from condition (1) and (2), we get

\angle 1 = \angle 2

If \angle 1 = \angle 2 then DC bisects the angle \angle ADE. [Proved]

To learn more about parallelogram from the given link

brainly.com/question/970600

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4 0
1 year ago
An architect is drawing up plans for a house. The rectangular living room is 4 feet longer than it is wide. The
I am Lyosha [343]

Answer b

Step-by-step explanation:

5 0
3 years ago
30 students in class 1/5 absent how many students were in class
slava [35]
6 students were in class because the 5 represents on how many groups you need to have
3 0
3 years ago
Rationalize the denominator and simplify.
cestrela7 [59]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3151018

——————————
 
     \mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}


Multiply and divide by the conjugate of the denominator, which is  \mathsf{4\sqrt{6}+3\sqrt{2}:}

     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}


Expand those products and eliminate the brackets:
 
     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}

     \mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}} 

     \mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}} 

     \mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

4 0
2 years ago
4-2x divided by 5 plus 3 =9
choli [55]
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5 0
3 years ago
Read 2 more answers
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