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If 12p + 4q = 55, then p = (55 - 4q)/12
Wiki User
∙ 7y ago
Wiki User
∙ 7y agoIt is (55 - 4q)/12.
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12 p plus 7 139?
12p + 7*139 = 12p + 973
12p-15 plus 8-4p equals 89?
12p-15+8-4p = 89 12p-4p = 89+15-8 8p = 96 p = 12
What is 17p plus 17q plus p-7q-6p simplified?
17p+17q+p-7q-6p = 12p+17q
How do you solve r equals 4q-5p over 9 solve for p?
Do you mean: r = (4q-5p)/9? If so then: p = (9r-4q)/-5
9 Solve the inequality 12p plus 7 139?
The answer to 12 P plus 7 times 139 equals 39. This is taught in high school math.
What is the answer to p plus 14 plus p equals 12p?
The value of p is 1.4
12 p plus 7 139?
12p + 7*139 = 12p + 973
12p-15 plus 8-4p equals 89?
12p-15+8-4p = 89 12p-4p = 89+15-8 8p = 96 p = 12
What is 17p plus 17q plus p-7q-6p simplified?
17p+17q+p-7q-6p = 12p+17q
What is 12p plus 19q plus p - 5q - 3p?
10p and 14q or 2(5p + 7q)
What is 12p 96 equal?
12p < 96 12p / 12 < 96 / 12 p< 8
What is 3p plus 4q if p is 5 and q is -2?
(3*5)+(4*-2) = 7
How do you solve r equals 4q-5p over 9 solve for p?
Do you mean: r = (4q-5p)/9? If so then: p = (9r-4q)/-5
If the ratio of the roots of the equation x2 plus px plus q0 is equal to the ratio of the roots of x62 plus lx plus m0 prove that mp2ql2.?
Thanks, Ankur. Let x^2+px+q=0 be Eqn 1 (I have shortened q0 to q) Let x^2+lx+m=0 be Eqn 2 I use the ^ to mean raised to the power of i.e. p^2 means p squared I use * to mean multiply The two roots of Eqn 1 (using the quadratic formula) are x =[-p±sqrt(p^2-4q)]/2 or separately [-p+sqrt(p^2-4q)]/2 and [-p-sqrt(p^2-4q)]/2 The two roots of Eqn 2 are x =[-l±sqrt(l^2-4m)]/2 or separately [-l+sqrt(l^2-4m)]/2 and [-l-sqrt(l^2-4m)]/2 Now we are told that the ratio of the roots of the two equations are equal. So {[-p+sqrt(p^2-4q)]/2} / {[-p-sqrt(p^2-4q)]/2} = {[-l+sqrt(l^2-4m)]/2} / {[-l-sqrt(l^2-4m)]/2} Cancel out the denominator 2 and cross multiply and ==>[-p+sqrt(p^2-4q)] * [ -l-sqrt(l^2-4m)] = [-p-sqrt(p^2-4q)] * [-l+sqrt(l^2-4m)] Simplify by multiplying each side ==>lp + p*sqrt(l^2-4m) - l*sqrt(p^2-4q) + [sqrt(p^2-4q)]*(l^2-4m)] = lp - p*sqrt(l^2-4m) + l*sqrt(p^2-4q) + [sqrt(p^2-4q)]*(l^2-4m)] Now cancel out lp as it's on both side and cancel out [sqrt(p^2-4q)]*(l^2-4m) because that's on both sides. Simplify by bring like terms together and ==>2p*sqrt(l^2-4m) = 2l*sqrt(p^2-4q) Cancel out the 2 on each side and square each side. ==>p^2(l^2-4m) = l^2(p^2-4q) Simply by multiplying out ==>p^2*l^2 - 4m*p^2 = p^2l^2 - l^2*4q Cancel out p^2*l^2 ==>-4m*p^2 = -4q l^2 Cancel out the -4 and voila ++> mp^2 = ql^2 If there is a shorter solution, I'd love to see it.
Solve the inequality 12p 7 139 A. p 18 B. p 12 C. p 154 D. p 11?
First of all--- is it 12p -/+ 7=139, but I'll do both anyway. 1. 12p-7=139 this means you + 7 to both sides first, to get the variable(p) by itself. so the equation would be 12p=146. Next divide 12 from both sides and it leaves you with p=12.1666 -repeating of course. 2. 12p+7=139 means you subtract 7 from both sides to get 12p by itself again. the equation would then be 12p=132. divide 12 from both sides and you get- p=11(which is one of your answers) :) so the answer is d.11.
9 Solve the inequality 12p plus 7 139?
The answer to 12 P plus 7 times 139 equals 39. This is taught in high school math.
What is 12 decreased by p?
12p = 12 - p
