The answer would be 6(2x^4y^3-6x^3y^2+x^2y)
I must assume that you meant the following:
10(Q-3R)
P = ----------------
R
Multiplying both sides by R, we get PR = 10R(Q-3R), or PR = 10RQ - 30R^2.
Rearranging these terms so that powers of R are in descending order:
30R^2 - 10RQ + PR = 0
Factoring out R, we get
R(30R - 10Q + P) = 0. This has two solutions:
R = 0, and 10Q - P
30R = 10Q - P, so that R = --------------
30
Answer:
36pi
Step-by-step explanation:
volume = (4/3) · π · r3
in this case r = 3
4/3 x pi x r^3
3 cubed = 27
27 x pi x 4/3
27 x 4/3 = 36
36
is the answer
<u>General method</u><u>:</u>
Given numbers are 1.3 bar and 1.4 bar
- 1.3 bar = 1.333...
- 1.4 bar = 1.444...
Two rational numbers between them
= 1.3222... and 1.43333...
<u>Mean Method</u><u>:</u>
Let X = 1.333... → → →eqn(i)
since the periodicity is 1 then multiply eqn(i) with 10
⇛10×X = 1.333...×10
⇛10X = 13.333... → → →eqn(ii)
Subtract eqn(ii)-eqn(i)
10X = 13.333...
X = 1.333...
(-)
____________
9X = 12.000...
____________
⇛9X = 12
⇛X = 12/9
⇛X = 4/3
and
Let X = 1.444... → → →eqn(i)
since the periodicity is 1 then multiply eqn(i) with 10
⇛10×X = 1.444...×10
⇛10X = 14.444...→ → →eqn(ii)
Subtract eqn(ii)-eqn(i)
10X = 14.444...
X = 1.444...
(-)
____________
9X = 13.000...
____________
⇛9X = 13
⇛X = 13/9
Now we have 12/9 and 13/9
The rational number between them by mean method (a+b)/2
⇛{(12/9)+(13/9)}/2
⇛(25/9)/2
⇛25/18
and Second rational number
⇛{(12/9)+(25/18)}/2
⇛{(24+25)/18}/2
⇛(49/18)/2
⇛49/36
<u>Answer</u><u>:</u> The two rational numbers between them are 25/18 and 49/36.
<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u> INSERT TWO RATIONAL NUMBERS BETWEEN 2 AND 3. How to find them?
brainly.com/question/85169?referrer
