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ZanzabumX [31]
2 years ago
13

Factorise or simplify4a² - 8ab²​

Mathematics
1 answer:
Alex73 [517]2 years ago
7 0

Answer: I think it would be 4a(a−2b squared)

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Draw a line through all points with a y-coordinate of -3.
Sedbober [7]

Answer:

See image below

Step-by-step explanation:

We want to draw a line through all points with a y-coordinate of -3; first let's observe some of these points: (-3,-3), (0,-3), (1,-3), (10,-3) are all part of this line.

We can see that what all these points have in common are that they have -3 in the y axis. Therefore, the graph of the line would be y = -3

This line is drawn in the graph below.

3 0
3 years ago
HELPP PLS ILL MARK BRAINLIST
elena-14-01-66 [18.8K]

Answer: g = -2

x = 5

Step-by-step explanation:

3 0
3 years ago
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2.
Fantom [35]

Answer:

the mode remains the same

Step-by-step explanation:

42 is outlier.

mod is the most repetitive number of a number string, and outlier does not change it.

the mode remains the same.

7 0
3 years ago
Show step by stephow to solve this problem 2001 - 1682
expeople1 [14]
So first you subtract 2001-1682=319.
There is just 1 step, and it is hard to show work here.
Hope this helped! :)

                                                                       
3 0
3 years ago
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In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm
tangare [24]

Answer:

AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD

Consider right triangle CAH.By the Pythagorean theorem,

CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8

The length cannot be negative, so CD=8 cm and

CA=\sqrt{3}CD=8\sqrt{3}\ cm

In right triangle ABC, angle B = 90° - 60° = 30°, leg AC is opposite to 30°, and the hypotenuse AB is twice the leg AC. Hence,

AB=2CA=16\sqrt{3}\ cm

By the Pythagorean theorem,

BC^2=AB^2-AC^2\\ \\BC^2=(16\sqrt{3})^2-(8\sqrt{3})^2=256\cdot 3-64\cdot 3=576\\ \\BC=24\ cm

3 0
3 years ago
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