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WITCHER [35]
3 years ago
9

Determine the solution set of (x - 3)2 = 49. {6, -8} {4, 10} {10, -4}

Mathematics
1 answer:
never [62]3 years ago
5 0

Answer:

{10, -4}

Step-by-step explanation:

(x-3)^2 = 49

Take the square root of each side

sqrt((x-3)^2) = sqrt(49)

x-3 = ±7

Add 3 to each side

x-3+3 = 3±7

x =  3±7

x = 3+7   and x = 3-7

x = 10        x = -4

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What are the real and complex solutions of the polynomial equation? X4-41x2=-400
stich3 [128]
For the answer to the question above, I'll provide a solution for my answer below.
x^4 - 41x^2 = - 400 
<span>=> x^4 - 41x^2 + 400 = 0 </span>
<span>=> x^2 = (1/2) [41 ±√(1681 - 100)] </span>
<span>=> x^2 = (1/2) (41 ± 9) </span>
<span>=> x^2 = 16 or 25 
</span>
So, therefore, the answers for your problem are
<span>=> x = ± 4 or ± 5.

I hope my answer helped you. </span>
8 0
3 years ago
What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form? Quadratic formula: x = StartFraction neg
Harman [31]

Answer:

x=\frac{-5(+/-)\sqrt{5}} {2}

x = StartFraction negative 5 plus or minus StartRoot 5 EndRoot Over 2 EndFraction

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

f(x)=x^{2} +5x+5  

equate the function to zero

x^{2} +5x+5=0  

so

a=1\\b=5\\c=5

substitute in the formula

x=\frac{-5(+/-)\sqrt{5^{2}-4(1)(5)}} {2(1)}

x=\frac{-5(+/-)\sqrt{5}} {2}

therefore

x = StartFraction negative 5 plus or minus StartRoot 5 EndRoot Over 2 EndFraction

6 0
3 years ago
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hodyreva [135]

Answer:

48 units

Step-by-step explanation:

75%=36

or 3/4=36

then 1/4=12 and 4/4=48

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