Answer:
x = 3 (see below)
Step-by-step explanation:
To solve for x, you need to isolate the variable to one side of the equation. To do that, you need to use reverse operations. For example, the reverse operation of subtraction is addition.
In this case, we have:
15 = 5x
5x is the same thing as 5 (x) or 5 times x. This means the reverse operation is division. So, we need to divide both sides of the equation by 5:
15 = 5x
----- ----
5 5
--------------
3 = x
x = 3
Answer: No
Step-by-step explanation:
A variable is always denoted with a symbol (commonly x), and a variable means that it can change based on what you plug into the symbol
Constants must always stay the same, so variables can't be constants and constants can't be variables
Cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.
<u>Solution:
</u>
Need to calculate
and then multiply the result by ![\sqrt[3]{14903}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D)
Let us first evaluate ![\sqrt[3]{1728}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1728%7D)
![\Rightarrow \sqrt[3]{1728}=\sqrt[3]{12 \times 12 \times 12}=12](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csqrt%5B3%5D%7B1728%7D%3D%5Csqrt%5B3%5D%7B12%20%5Ctimes%2012%20%5Ctimes%2012%7D%3D12)
As need to multiply 12 by ![\sqrt[3]{14903}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D)
![\Rightarrow 12 \times \sqrt[3]{14903}](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D)
On solving
, we get
![\sqrt[3]{14903}=24.608](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D%3D24.608)
![\Rightarrow 12 \times \sqrt[3]{14903}=12 \times 24.608=295.306](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D%3D12%20%5Ctimes%2024.608%3D295.306)
Hence cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.
Answer:
for the first one I got y=-2x-2 the second part i got y=-x-1 or -1x-y-1=0 Correct me if I am wrong
Step-by-step explanation:
Use the slope formula
-4-4 =-8
/ =-2
1 - -3=4
2. The second part use this formula y-y1=m(x-x1)
y--4=-1(x-3)
y+4=-x+3
-4 -4
y+0=-x-1
y=-x-1 or -1x-y-1=0 I am not sure