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3 years ago

Find the midpoint of the line segment whose endpoints are (2.6, 5.1) and (3, 4.7).

2 answers:

5 0

$(2.6, 5.1) \Rightarrow x_1=2.6,y_1=5.1
\\(3,4.7)\Rightarrow x_2=3,y_2=4.7
\\
\\x= \frac{x_1+x_2}{2}= \frac{2.6+3}{2}=2.8
\\
\\y= \frac{y_1+y_2}{2}= \frac{5.1+4.7}{2}=4.9
\\
\\M(2.8,4.9)
$

Nimfa-mama [501]3 years ago

3 0

Answer:

(2.8, 4.9)

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

X^2 - 10x - 20 = 0, which number would have to be added to "complete the square"?

Airida [17]

Answer:

Step-by-step explanation:

When you're completing the square you're adding (b/2)^2 which in this case is (-5)^2 or 25. This is because when you rewrite it as a square binomial you write it in the form (x+(b/2))^2 which is going to result in x^2+b+(b/2)^2. Since you need to reach 25 you need to add 45 since the last value is at -20.

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1 year ago

Read 2 more answers

I need help with this question

Novay_Z [31]

Answer:

$\frac{\sqrt{3} - 1}{2\sqrt{2}}$

$\frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

$- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Step-by-step explanation:

Given $\frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6}$

(A) $sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4} + \frac{\pi}{6})$

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2}$

$$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

(B) Cos(A + B) = CosAcosB - SinASinB

$cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}})$

$\implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2}$

$\implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6})$

$$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

From the above obtained values this can be calculated and the value is $- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ .

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3 years ago

The distance between (2, 0) and (5, -1) is:<br> 9.<br> 0.<br> 10.<br> <img src="https://tex.z-dn.net/?f=%20%5Csqrt10" id="TexFor

Sidana [21]

Distance = square root ((x2 - x1)^2 + (y2 - y1)^2)
(2,0)...x1 = 2 and y1 = 0
(5,-1)..x2 = 5 and y2 = -1
sub and solve
d = sqrt ((5 - 2)^2 + (-1 - 0)^2)
d = sqrt ((3^2) + (-1^2))
d = sqrt (9 + 1)
d = sqrt 10 <==

4 0

2 years ago