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yKpoI14uk [10]
3 years ago
8

Find all rational zeros of the polynomial

Mathematics
1 answer:
rjkz [21]3 years ago
4 0

x^3+x^2-5x+3=x^3+3x^2-2x^2-6x+x+3\\\\=(x^3-2x^2+x)+(3x^2-6x+3)=x(x^2-2x+1)+3(x^2-2x+1)\\\\=(x^2-2x+1)(x+3)=\underbrace{(x^2-2(x)(1)+1^2)}_{(a-b)^2=a^2-2ab+b^2}(x+3)=(x-1)^2(x+3)\\\\P(x)=0\iff(x-1)^2(x+3)=0\iff(x-1)^2=0\ \vee\ (x+3)=0\\\\x-1=0\ \vee\ x-1=0 \vee\ x+3=0\\\\x=1\ \vee\ x=1\ \vee\ x=-3\\\\Answer:\ \boxed{x_1=1,\ x_2=1,\ x_3=3}

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'2aa4b' is a 5 digit number and when its divided by '36' it gives '11' as remainder. What is the sum of the values 'a' can have.
OverLord2011 [107]

The only 5-digit numbers that do the job are  21,143  and  28,847 .

So, either 
a = 1  and
b = 3

or
a = 8  and
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The sum of the values that 'a' can have is (1 + 7) = <em>8</em> .

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7 0
3 years ago
Right triangles to find the exact length of:<br> T<br> 30°<br> 14 in<br> a) TI = in<br> 60°<br> R
ira [324]
<h3>Answer:</h3>

a) TI = 7√3 in

b) IR = 7 in

<h3>Explanation:</h3>

<u>Using cosine rule</u>:

\sf cos(x) = \dfrac{adjacent}{hypotenuse}

\hookrightarrow \sf cos(30) = \dfrac{TI}{14}

\hookrightarrow \sf TI =  14cos(30)

\hookrightarrow \sf TI =  7\sqrt{3}

<u>Using sine rule</u>:

\sf sin(x) = \dfrac{opposite}{hypotenuse}

\hookrightarrow \sf sin(30) = \dfrac{IR}{14}

\hookrightarrow \sf  IR = 14sin(30)

\hookrightarrow \sf  IR = 7

5 0
2 years ago
If F(x)= integral from 0 to x of square root of (t^3 +1), then F'(2)=
stealth61 [152]
By the fundamental theorem of calculus,

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which means

F'(2)=\sqrt{2^3+1}=\sqrt{8+1}=\sqrt9=3
7 0
4 years ago
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8 0
3 years ago
What is the agree of the circle?
andre [41]

Answer:

11.6

Step-by-step explanation:

6 0
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