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enot [183]
3 years ago
6

Find the solution set for this equation -a2+12a=0

Mathematics
1 answer:
lesantik [10]3 years ago
5 0
Hello,
-a²+12a=0
a²-12a=0
a(a-12)=0
a=0 or a-12=0
a=0 or a=12

The solution are: a=0 or a=12
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(-2,-2) and (3,4)

Step-by-step explanation:

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2 years ago
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M is a degree 3 polynomial with m ( 0 ) = 53.12 and zeros − 4 and 4 i . Find an equation for m with only real coefficients (i.E.
Nitella [24]

Answer:

Therefore the required polynomial is

M(x)=0.83(x³+4x²+16x+64)

Step-by-step explanation:

Given that M is a polynomial of degree 3.

So, it has three zeros.

Let the polynomial be

M(x) =a(x-p)(x-q)(x-r)

The two zeros of the polynomial are -4 and 4i.

Since 4i is a complex number. Then the conjugate of 4i is also a zero of the polynomial i.e -4i.

Then,

M(x)= a{x-(-4)}(x-4i){x-(-4i)}

      =a(x+4)(x-4i)(x+4i)

      =a(x+4){x²-(4i)²}      [ applying the formula (a+b)(a-b)=a²-b²]

      =a(x+4)(x²-16i²)

      =a(x+4)(x²+16)      [∵i² = -1]

      =a(x³+4x²+16x+64)

Again given that M(0)= 53.12 . Putting x=0 in the polynomial

53.12 =a(0+4.0+16.0+64)

\Rightarrow a = \frac{53.12}{64}

      =0.83

Therefore the required polynomial is

M(x)=0.83(x³+4x²+16x+64)

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3 years ago
A student took a total of 4 tests over the course of 8 weeks. How many weeks of school will the student attend to take a total o
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Step-by-step explanation:

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devin began running a month ago to get back in shape.the first day he ran .5 miles. each day after that he ran 10% more than the
SpyIntel [72]
The formula for the sum of the a finite geometric series is this:
Sn = a((1-r^n) / (1-r))

The given values are:
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S30 = 0.5 ((1-1.10^30) / (1-1.10))
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After 30 days, he will have traveled 82.25 miles.
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