HELP ME THIS IS THE LAST ONE

Mathematics

1 answer:

anzhelika [568]2 years ago

4 0

Answer:

100,000m

Step-by-step explanation:

find the square root of 100 squared +1,000squared

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11 hours 18 min 36 sec to hours

zlopas [31]

I'm not really sure about this math but I tried to answer it for you

1)Convert 11 hours to minutes = 660 minutes. 11 hours x 60 mins=660
2)Add the 18 minutes to 660 =678 minutes
3) 678 divided by 60 mins = 11.3 hours, but we still have the 36 seconds remaining so I didn't exactly divide 678 minutes by 60, but I divided 678.36 by 60 which gave me 11.306 hours.

That's it I truly apologise if this is wrong. But if you figure out the right method please show me and others who may have done it this way or another way, so please share. GOOD LUCK.

7 0

3 years ago

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,

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