Step-by-step explanation:
It's an irrational number.
![\sqrt[3]{275:7}=\sqrt[3]{\dfrac{275}{7}}=\dfrac{\sqrt[3]{275}}{\sqrt[3]{7}}=\dfrac{\sqrt[3]{275}\cdot\sqrt[3]{7^2}}{\sqrt[3]{7}\cdot\sqrt[3]{7^2}}=\dfrac{\sqrt[3]{275\cdot49}}{\sqrt[3]{7\cdot7^2}}\\\\=\dfrac{\sqrt[3]{13475}}{\sqrt[3]{7^3}}=\dfrac{\sqrt[3]{13475}}{7}=\dfrac{1}{7}\sqrt[3]{13475}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B275%3A7%7D%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B275%7D%7B7%7D%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B275%7D%7D%7B%5Csqrt%5B3%5D%7B7%7D%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B275%7D%5Ccdot%5Csqrt%5B3%5D%7B7%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B7%7D%5Ccdot%5Csqrt%5B3%5D%7B7%5E2%7D%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B275%5Ccdot49%7D%7D%7B%5Csqrt%5B3%5D%7B7%5Ccdot7%5E2%7D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B13475%7D%7D%7B%5Csqrt%5B3%5D%7B7%5E3%7D%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B13475%7D%7D%7B7%7D%3D%5Cdfrac%7B1%7D%7B7%7D%5Csqrt%5B3%5D%7B13475%7D)
The answer is 60 degrees as well
Answer:
<em>2</em><em>y</em>
Step-by-step explanation:
this is common factor of both the terms
because 2y divides both term
hope it helps
Answer:
x² + (y+3)² = 4²
Step-by-step explanation:
A close look at the diagram reveals that the center of this circle is (0, -3) and that the radius is 4 units.
Starting from the general equation of a circle, (x - h)² + (y - k)² = r²
and substituting the givens (h = 0, k = -3, r = 4), we get:
x² + (y+3)² = 4². This is Answer B.
Answer:
∠1 - 40°
Step-by-step explanation:
∠1 - 40°
b/c it's a right triangle and we have two angles given, 50° and 90°. Add them and subtract by 180° and get 40°.
∠2 - 140°
b/c an exterior (outside) angle is equal to the two most isolated / farthest angles added. The two most is angles are 105° and 35°, add them and get 140°.
∠3 - 40°
b/c ∠'s 1 and 3 are vertical angles meaning they're equal so since ∠1 is 40°, so is ∠3.
∠4 -
b/c ∠' s 2 and 4 are vertical angles meaning they're equal so since ∠2 is 140°, so is ∠4.
∠5 - 35°
b/c we have two angles, 105° and 40°. Add them and subtract by 180° and get 35°.
hope that helps lol >:)
