Answer:
![3\sqrt[4]{5}](https://tex.z-dn.net/?f=3%5Csqrt%5B4%5D%7B5%7D)
Step-by-step explanation:
Given: ![\frac{\sqrt[4]{810} }{\sqrt[4]{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B4%5D%7B810%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%20%7D)
Top and bottom has fourth root. so we take fourth root in common
![\sqrt[4]{\frac{810}{2} }](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B810%7D%7B2%7D%20%7D)
When we divide 810 by 2. we will get 405
![\sqrt[4]{405}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B405%7D)
Now we simplify the fourth root
405 can be written as 3*3*3*3*5
![\sqrt[4]{3*3*3*3*5}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B3%2A3%2A3%2A3%2A5%7D)
![3\sqrt[4]{5}](https://tex.z-dn.net/?f=3%5Csqrt%5B4%5D%7B5%7D)
Answer:
3/4
Step-by-step explanation:
If he/she ate 1/4 then there would still be 3/4
Answer:
4/8
Step-by-step explanation:
There are an infinite number of fractions equal to 1/2, but the number line in the question has 8 spaces, and shows 1 divided into 8 pieces. Therefore, the answer to this question is likely 4/8.
(may I ask why exactly this is put as a college level problem?)
Answer: B.), D.), and E.)
Step-by-step explanation:
That’s the answer on USAtestprep
<h3>
Answer: 80 degrees</h3>
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Explanation:
I'm assuming that segments AD and CD are tangents to the circle.
We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.
By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.
----------------------------
Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.
Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.
We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.
---------------------------
Here's what we have so far for quadrilateral DAEC
- angle A = 90
- angle E = 100
- angle C = 90
- angle D = unknown
Now we'll use the idea that all four angles of any quadrilateral always add to 360 degrees
A+E+C+D = 360
90+100+90+D = 360
D+280 = 360
D = 360-280
D = 80
Or a shortcut you can take is to realize that angles E and D are supplementary
E+D = 180
100+D = 180
D = 180-100
D = 80
This only works if AD and CD are tangents.
Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.