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My name is Ann [436]
2 years ago
9

Solve for x in the equation p(m+qx)=rx+n

Mathematics
1 answer:
katrin [286]2 years ago
6 0

p(m+qx)=rx+n\\\\\implies pm+pqx = rx+n\\\\\implies pqx-rx = n-pm\\\\\implies x(pq -r) = n-pm\\\\\implies x = \dfrac{n-pm}{pq-r}

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what is the answer for 1/3x+y=4 and if you use big ideas math program and you got the right answer lemme know
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Answer:

x=-3y+12

Step-by-step explanation:

Yes.

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the price of one share of ABC company decreased a total of $45.75 in 5 days what was the acerage change of the price of one shar
yulyashka [42]

Just divide 40 divided by 5 and you should get $9.15
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3 years ago
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A right triangle has a leg of 11 cm and a hypotenuse of 17 cm.
Vedmedyk [2.9K]
The answer would be the second option, 13.0 cm.

To find the missing length of the leg, you would use the pythagorean theorem and substitute your inputs in.

a^2+b^2=c^2
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Step-by-step explanation:

7 0
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Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\
motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in \sqrt{7}, 1 - \sqrt{6} and -3. The last root gives us the factor (x+3). Hence, our polynomial is

P(x) =(x+3)q(x)

where q is a polynomial with rational coefficients and roots \sqrt{7} and 1 - \sqrt{6}. The root \sqrt{7} gives us a factor x-\sqrt{7}, but in order to obtain rational coefficients we must consider the factor x^2-7.

An analogue idea works with 1 - \sqrt{6}. For convenience write  x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}. This gives the factor (x-1)^2-6. Hence,

P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105

Notice that P(-1)=24. So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type x-\sqrt7 will introduce in the expression, we need to multiply by its conjugate x+\sqrt7. Hence, we will obtain x^2-7 that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.

4 0
3 years ago
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