well, you already know an absolute value expression has a ± siblings, so let's proceed without much fuss.
![\bf |2x-5|=4\implies \begin{cases} +(2x-5)=4\implies 2x=9\implies x=\cfrac{9}{2}\\[-0.5em] \hrulefill\\ -(2x-5)=4\implies 2x-5=-4\\[1em] 2x=1\implies x=\cfrac{1}{2} \end{cases}](https://tex.z-dn.net/?f=%20%5Cbf%20%7C2x-5%7C%3D4%5Cimplies%20%20%5Cbegin%7Bcases%7D%20%2B%282x-5%29%3D4%5Cimplies%202x%3D9%5Cimplies%20x%3D%5Ccfrac%7B9%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20-%282x-5%29%3D4%5Cimplies%202x-5%3D-4%5C%5C%5B1em%5D%202x%3D1%5Cimplies%20x%3D%5Ccfrac%7B1%7D%7B2%7D%20%5Cend%7Bcases%7D%20)
Answer:
50=(3x)x
Step-by-step explanation:
50=(3x)x. The question is only asking for the inequality of the problem. Since the area of a rectangle is L x W, you would write L as 3x and W as x, and since the area of the rectangle is given, you can set the area as 50.
Answer:
is there a passage or sum?
Step-by-step explanation:
Answer:
See explanation
Step-by-step explanation:
Solution:-
- We will use the basic formulas for calculating the volumes of two solid bodies.
- The volume of a cylinder ( V_l ) is represented by:

- Similarly, the volume of cone ( V_c ) is represented by:

Where,
r : The radius of cylinder / radius of circular base of the cone
h : The height of the cylinder / cone
- We will investigate the correlation between the volume of each of the two bodies wit the radius ( r ). We will assume that the height of cylinder/cone as a constant.
- We will represent a proportionality of Volume ( V ) with respect to ( r ):

Where,
C: The constant of proportionality
- Hence the proportional relation is expressed as:
V∝ r^2
- The volume ( V ) is proportional to the square of the radius. Now we will see the effect of multiplying the radius ( r ) with a positive number ( a ) on the volume of either of the two bodies:

- Hence, we see a general rule frm above relation that multiplying the result by square of the multiple ( a^2 ) will give us the equivalent result as multiplying a multiple ( a ) with radius ( r ).
- Hence, the relations for each of the two bodies becomes:

&
