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FromTheMoon [43]
3 years ago
7

When A = 300, solve the equation

id="TexFormula1" title=" {x }^{2} - 40x + a = 0" alt=" {x }^{2} - 40x + a = 0" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
lilavasa [31]3 years ago
7 0

Answer:

Solution given:

a=300

and equation is:

x²-40x+a=0

when a=300

x²-40x+300=0

making it a perfect square

x²-2*20x+20²-20²+300=0

(x+20)²-400+300=0

(x+20)²=100

x+20=±10

taking positive

x+20=10

x=10-20

x=-10

taking negative

x+20=-10

x=-10-20

z=-30

when A=300

x=-10 or -30

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Step-by-step explanation:

Many ways to do this in the details.

All start by writing three equations with three unknowns

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subtract equation 2 from equation 1

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subtract equation 3 from equation 1

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substitute 4 into 5

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substitute a and b into any of 1, 2, or 3

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We can use a plotting calculator to confirm our result.

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145 degrees it is thank you have a wonderful day
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The volume of the cube shown below is _____________ inches. Round to the nearest tenth.​
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The area of a square is 300 feet. What is the side length of the square? Round to the nearest tenths place.
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