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Sedbober [7]
2 years ago
12

What is the length of segment AB to the nearest tenth?

Mathematics
1 answer:
ankoles [38]2 years ago
7 0

Answer:

4.9 units

Step-by-step explanation:

Distance between 2 points= sqrt [(x-x)^2 + (y-y)^2

A= -4,2) and B= (1,4)

d = sqrt[(-4-1)^2 + (2-4)^2]

= sqrt (25+4)

= sqrt29

= 4.9 units

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3 0
3 years ago
Read 2 more answers
Enter the correct answer in the box. solve the equation x2 − 16x 54 = 0 by completing the square. fill in the values of a and b
poizon [28]

The roots of the given polynomials exist  $x=8+\sqrt{10}$, and $x=8-\sqrt{10}$.

<h3>What is the formula of the quadratic equation?</h3>

For a quadratic equation of the form $a x^{2}+b x+c=0$ the solutions are

$x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Therefore by using the formula we have

$x^{2}-16 x+54=0$$

Let, a = 1, b = -16 and c = 54

Substitute the values in the above equation, and we get

$x_{1,2}=\frac{-(-16) \pm \sqrt{(-16)^{2}-4 \cdot 1 \cdot 54}}{2 \cdot 1}$$

simplifying the equation, we get

$&x_{1,2}=\frac{-(-16) \pm 2 \sqrt{10}}{2 \cdot 1} \\

$&x_{1}=\frac{-(-16)+2 \sqrt{10}}{2 \cdot 1}, x_{2}=\frac{-(-16)-2 \sqrt{10}}{2 \cdot 1} \\

$&x=8+\sqrt{10}, x=8-\sqrt{10}

Therefore, the roots of the given polynomials are $x=8+\sqrt{10}$, and

$x=8-\sqrt{10}$.

To learn more about quadratic equations refer to:

brainly.com/question/1214333

#SPJ4

3 0
1 year ago
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katen-ka-za [31]

Answer: 21/40

Step-by-step explanation:

I did the test

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3 years ago
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Artyom0805 [142]
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Answer:

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