Answer:
4. Divide both sides by l.
Step-by-step explanation:
You are trying to find out the answer for "w" so you need to get it alone on one side. Because "l" and "w" are multiplied together, you will have to divide "l" from both sides in order to get "w" by itself.
Answer:
The answer is
Step-by-step explanation:
First you have to find the radius of sphere by dividing diameter by 2 :
Let diameter = 14,
Next you have to use the volume of sphere formula :
Let r = 7,
if we have a number like say hmm 4, and we say hmmm √4 is ±2, it simply means, that if we multiply that number twice by itself, we get what's inside the root, we get the 4, so (+2)(+2) = 4, and (-2)(-2) = 4, recall that <u>minus times minus = plus</u>.
so, any when we're referring to even roots like ![\bf \sqrt[2]{~~},\sqrt[4]{~~},\sqrt[6]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B2%5D%7B~~%7D%2C%5Csqrt%5B4%5D%7B~~%7D%2C%5Csqrt%5B6%5D%7B~~%7D....)
, the positive number, that can multiply itself an even amount of times, will produce a valid value, BUT the negative number that multiply itself an even amount of times, will also produce a valid value.
now, that's is not true for odd roots like ![\bf \sqrt[3]{~~},\sqrt[5]{~~},\sqrt[7]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B~~%7D%2C%5Csqrt%5B5%5D%7B~~%7D%2C%5Csqrt%5B7%5D%7B~~%7D....)
, because the multiplication of the negative number will not produce a valid value, let's put two examples on that.
![\bf \sqrt[3]{27}\implies \sqrt[3]{3^3}\implies 3\qquad because\qquad (3)(3)(3)=27 \\\\\\ however\qquad (-3)(-3)(-3)\ne 27~\hspace{8em}(-3)(-3)(-3)=-27 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt[3]{-125}\implies \sqrt[3]{-5^3}\implies -5\qquad because\qquad (-5)(-5)(-5)=-125 \\\\\\ however\qquad (5)(5)(5)\ne -125~\hspace{10em}(5)(5)(5)=125](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B27%7D%5Cimplies%20%5Csqrt%5B3%5D%7B3%5E3%7D%5Cimplies%203%5Cqquad%20because%5Cqquad%20%283%29%283%29%283%29%3D27%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%28-3%29%28-3%29%28-3%29%5Cne%2027~%5Chspace%7B8em%7D%28-3%29%28-3%29%28-3%29%3D-27%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B-125%7D%5Cimplies%20%5Csqrt%5B3%5D%7B-5%5E3%7D%5Cimplies%20-5%5Cqquad%20because%5Cqquad%20%28-5%29%28-5%29%28-5%29%3D-125%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%285%29%285%29%285%29%5Cne%20-125~%5Chspace%7B10em%7D%285%29%285%29%285%29%3D125)
so, when the root is an odd root, you will always get only one number that will produce the radicand.
These involve the rules for the power of a point. For questions 6-8, we use the theorem that the square of the length of the tangent is equal to the length of the secant multiplied by the length of its external segment.
6. 6^2 = (3)(x + 3)12 = x + 3x = 9 units
7. 4^2 = (2)(x + 2)8 = x + 2x = 6 units
8. 24^2 = (12)(x + 12)48 = x + 12x = 36 units
9. The included angle is half the difference of the larger and smaller arcs. Since the larger arc is 85 degrees, the smaller is 25 degrees, and the difference is 60. Half this difference is x = 30 degrees.
10. The angles at the intersection are vertical angles, so both equal to 65 degrees. Then the sum of the intercepted arcs must be equal to double of the vertical angle: 2 x 65 = 130 degrees. Since one is 95 degrees, the other is x = 130 - 95 = 35 degrees.
