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βˆ™ 2009-05-03 18:45:07
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: Q plus r equals s so t plus u equals?
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Evaluate q-r if q equals -13 and r equals -2?


How do you do 12q plus r equals 448?

How many twelves can you put into 448 without going over? 37(37=q). You will get 444. subtract 444 from 448 and you get 4(4=r) So q=37 and r=4.

What is p plus r plus q plus 113 plus 80 plus 54?

p + r + q + 175

What is b plus g plus h plus h plus r plus v plus x plus z plus q plus a plus p plus l plus f plus g plus d equals?


R plus 16 equals 5r-10?

so 26 = 4r so r = 6½

'p q' means p is wife of q and p plus q means p is father of q and p q means p is sister of q than g h plus r d how is g relatied th d?

g h plus r d g is the sister of (h plus (r is a sister of d)) g is the sister of h is the father of r is the sister of d so g is d's aunt.

Pq equals 3p plus 4r minus 2p?

Without prior knowledge of the value of q or r, it is impossible to calculate the answer to this equation.

P 2x-9y q 5y plus 6-4x r 3x plus 3y-5 then p plus q plus r?

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P varies jointly as q r and s One set of values is p equals 70 q equals 7 r equals 5 and s equals 4 Find p when q equals 2 r equals 15 and s equals 7?

If P varies jointly as q, r and s - assume this is in direct proportion, then P ∝ qrs so P = kqrs where k is a constant.70 = k x 7 x 5 x 4 = 140k : k = 140/70 = 0.5When q = 2, r = 15 and s = 7 then,P = 0.5 x 2 x 15 x 7 = 105

Evaluate pq - r when p equals 3 q equals 4 and r equals -6?

pq-r, if p is 3, q is 4, and r is -6 is equal to 3 x 4 - (-6), which is equal to 12 + 6, which is equal to 18.

Pqr when p equals 2 q equals 4 andd r equals 5?

PQR P=2 Q=4 R=5 2 x 4 x 5 = 40

If a equals 1 b equals 2 z equals 26 what is p plus q plus r?

Starting with the information that is given: Two options, due to the lack of punctuation: a=1b=2z=26 -or- a=1 b=2 z=36 p=(not defined) q=(not defined) r=(not defined) Therefore, all we can assert from the given information is that p+q+r = p+q+r (the variables can be rearranged without altering the (unknown) value, because of the commutative property of addition. a, b and z are not used in the equation, so their values are irrelevant. p, q and r are not defined, so we cannot assume any value for any of the variables in the equation. Well, we could assume, but it would be just that - an assumption, and an unfounded assumption at that. There was not enough information provided to draw any more concrete conclusion.

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