Nolan polled the 2 fastest swimmers on the swim team.

Mathematics

1 answer:

torisob [31]2 years ago

4 0

Here's link to the answer:

cutt.us/tWGpn

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X.7=14 plz help me 4th grade work

OlgaM077 [116]

x*7=14

14/7=2

x=2

brainliest please

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3 years ago

0=4t-16t^2 Solve this please

vaieri [72.5K]

Answer:

$\large\boxed{t=0\ \vee\ t=\dfrac{1}{4}}$

Step-by-step explanation:

$4t-16t^2=0\qquad\text{divide both sides by 4}\\\\\dfrac{4t}{4}-\dfrac{16t^2}{4}=\dfrac{0}{4}\\\\t-4t^2=0\\\\t(1-4t)=0\iff t=0\ \vee\ 1-4t=0\\\\1-4t=0\qquad\text{subtract 1 from both sides}\\\\-4t=-1\qquad\text{divide both sides by (-4)}\\\\t=\dfrac{1}{4}$

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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,

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