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iVinArrow [24]
3 years ago
7

I am very confused - please help! thx

Mathematics
2 answers:
ivanzaharov [21]3 years ago
8 0
A trapezoid has four sides, and two of them are parallel. You should have no trouble finding the one shape in the picture that fits that description.
Ira Lisetskai [31]3 years ago
7 0
The answer here should be C

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A line passes through (−1, 7) and (2, 10).
Murljashka [212]
The formula is y=mx+b
To get the slope or m, use this formula
(the second y minus the first y)/(the second x minus the first x)
Now set it up.
(10-7)/(2--1)
3/3=slope is 1.
y=1x+b

Insert one of the points for x and y.
i will do (-1,7)
7=1(-1)+b
7=-1+b
8=b
Insert this into the final equation:
y=1x+8

Try it out. If you're not sure, try both points. If it works, then you set it up correctly.
6 0
3 years ago
Read 2 more answers
Find the equation of the axis of symmetry of the following parabola algebraically.<br> y=2x^2+16.
Inessa05 [86]

Answer:

x = 0

axis of symmbery = -b/2a or x vertex on quadratic equation

0/4=0

x=0

Step-by-step explanation:

3 0
3 years ago
Is 32.5 greater or less than 30.8<br><br>PLEASE HELP QUICKLY AS POSSIBLE THANK YOU :)​
ivann1987 [24]

Answer:

32.5 is greater than 30.8

6 0
3 years ago
Read 2 more answers
Hey guys what is this type of equation/ question called 1 &lt; x &lt; 30
dimulka [17.4K]

Answer:

2,3,4,5,6,7,8,9,10,...29

Step-by-step explanation:

6 0
3 years ago
How to prove this???
swat32
\cos^3 2A + 3 \cos 2A \\&#10;\Rightarrow \cos 2A (\cos^2 2A + 3) \\&#10;\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3)  \\&#10;\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\&#10;\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\&#10;\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\&#10;\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ &#10;\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) &#10;

go to right side now

4( \cos^6 A - \sin^6 A)\\&#10;\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)

use x^3 - y^3 = (x-y)(x^2 + xy + y^2) and x^3 + y^3 = x^2 - xy + y^2

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\&#10;\Rightarrow  4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\&#10;~\quad  \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)

so \sin^2 A + \cos^2 A = 1

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\&#10; \Rightarrow Left hand side
4 0
3 years ago
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