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tresset_1 [31]
3 years ago
11

Can someone help me?

Mathematics
1 answer:
nirvana33 [79]3 years ago
6 0
So you can do number one is d
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When the sum of 5 and three times a positive number is subtracted from the square of the number, 0 results. Find the number.
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x-the\ number\\\\x^2-(5+3x)=0\\\\x^2-5-3x=0\\\\x^2-3x-5=0

Use\ the\ quadratic\ formula:\\a^2x+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\if\ \Delta=0\ then\ x_0=\dfrac{-b}{2a}\\\\if\ \Delta < 0\ then\ no\ solution

x^2-3x-5=0\\\\a=1;\ b=-3;\ c=-5\\\\\Delta=(-3)^2-4\cdot1\cdot(-5)+3+20=29 \ \textgreater \  0\\\\\sqrt\Delta=\sqrt{29}\\\\x_1=\dfrac{3-\sqrt{29}}{2\cdot1}=\dfrac{3-\sqrt{29}}{2} \ \textless \  0\\\\x_2=\dfrac{3+\sqrt{29}}{2\cdot1}=\dfrac{3+\sqrt{29}}{2} \ \textgreater \  0

Answer:\ \boxed{x=\dfrac{3+\sqrt{29}}{2}}
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3 years ago
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Step-by-step explanation:

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Because it's decreasing, you use the formula (1-x/100)y with x the percentage decreased which is here 40, and y the number decreased which is here 90
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