If the negative square root is found to be one of your solutions, then that is indicative of a pair of imaginary roots (the imaginary i). According to the conjugate rule, if you have one solution that is imaginary, you will have another but with the opposite sign. For example, if a solution to a quadratic is found to be 2 - i, then its conjugate, 2 + i is also a solution. They will ALWAYS go in pairs. Same thing with radical solutions. If one solution is found to be 
then 
will also be a solution.
Answer:
D.
Step-by-step explanation:
It is simply asking for the number that makes the equation untrue. So, all you need to do is plug in the various numbers below for w, and find out the incorrect number. Example with option D:
w-10≤16
27-10≤16
17≤16.
This makes the equation untrue, as 17 is not less than or equal to 16.
Hope this helps!
Answer:
12
Step-by-step explanation:
From the above question, we ate told that:
Becky needs to put 4 cups of flour into her bread maker to make one loaf of bread.
The size of her measuring cup = 1/3 cup
The number of 1/3 cups of flour she will need to make a loaf of bread is calculated as:
4 cups of flour ÷ 1/3 cup of flour
= 4 cups of flour × 3/1 cup of flour
= 12
Answer:
x = 3.87
Step-by-step explanation:
Using the right angle altitude theorem, we know that all three triangles are congruent, so the lengths of corresponding sides of the triangles are in proportion..
AD/DB = DB/DC
15/x = x/1
x² = 15
x = √15
x = 3.87
Answer:
a) ![v = \frac{[L]}{[T]} = LT^{-1}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B%5BL%5D%7D%7B%5BT%5D%7D%20%3D%20LT%5E%7B-1%7D)
b) ![a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5BL%7D%7BT%7D%5E%7B-1%7D%5D%7D%7B%7BT%7D%7D%3D%20L%20T%5E%7B-1%7D%20T%5E%7B-1%7D%3D%20L%20T%5E%7B-2%7D)
c) ![\int v dt = s(t) = [L]=L](https://tex.z-dn.net/?f=%20%5Cint%20v%20dt%20%3D%20s%28t%29%20%3D%20%5BL%5D%3DL)
d) ![\int a dt = v(t) = [L][T]^{-1}=LT^{-1}](https://tex.z-dn.net/?f=%20%5Cint%20a%20dt%20%3D%20v%28t%29%20%3D%20%5BL%5D%5BT%5D%5E%7B-1%7D%3DLT%5E%7B-1%7D)
e) ![\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bda%7D%7Bdt%7D%3D%20%5Cfrac%7B%5BL%5D%5BT%5D%5E%7B-2%7D%7D%7BT%7D%20%3D%20%5BL%5D%5BT%5D%5E%7B-2%7D%20%5BT%5D%5E%7B-1%7D%20%3D%20LT%5E%7B-3%7D)
Step-by-step explanation:
Let define some notation:
[L]= represent longitude , [T] =represent time
And we have defined:
s(t) a position function
Part a
If we do the dimensional analysis for v we got:
Part b
For the acceleration we can use the result obtained from part a and we got:
Part c
From definition if we do the integral of the velocity respect to t we got the position:
And the dimensional analysis for the position is:
Part d
The integral for the acceleration respect to the time is the velocity:
And the dimensional analysis for the position is:
Part e
If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:
