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horrorfan [7]
3 years ago
14

How to show your work to solve  3r^2-16r-7=5

Mathematics
2 answers:
Irina18 [472]3 years ago
7 0
OK so   3r^2-16r-7=5
Subtract five from both sides
3r^2-16r-7-5=0
Add like terms
3r^2-16r-12=0
Factorise it, this means putting it into brackets
(3r + 2)(r - 6)
Find out what multiplies to make negative twelve but adds to make negative sixteen
However none of the terms do this so we put negatives next to the "r"s
(-3r + 2)(-1r - 6)
The answer is r=6
alukav5142 [94]3 years ago
3 0

Answer:

Step 1: Subtract 5 from both sides.

3r2−16r−7−5=5−5

3r2−16r−12=0

Step 2: Factor left side of equation.

(3r+2)(r−6)=0

Step 3: Set factors equal to 0.

3r+2=0 or r−6=0

r=−2/3

or r=6

Answer:

r=−2/3

or r=6

Step-by-step explanation:

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Consider substituting u=4+xy, so that \mathrm du=x\,\mathrm dy. Then

\displaystyle\int_{x=a}^{x=b}\int_{y=a}^{y=b}\frac x{(4+xy)^2}\,\mathrm dy\,\mathrm dx=\int_{x=a}^{x=b}\int_{u=4+ax}^{u=4+bx}\frac1{u^2}\,\mathrm du\,\mathrm dx
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=\ln\sqrt[a]{\dfrac{4+ab}{4+a^2}}+\ln\sqrt[b]{\dfrac{4+ab}{4+b^2}}
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