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Ne4ueva [31]
3 years ago
8

Which expression is equivalent to √ -80?

Mathematics
1 answer:
oksian1 [2.3K]3 years ago
3 0
<h3>Answer: Choice C</h3><h3>which is the expression 4*sqrt(5)*i,  where i = sqrt(-1)</h3>

==================================================

Explanation:

Factor -80 in such a way where we pull out -1 and we also factor out the largest perfect square possible. In this case, it would be 16 because 80 = 16*5. So this means -80 = -1*16*5 which leads to...

sqrt(-80) = sqrt(-1*16*5)

sqrt(-80) = sqrt(-1)*sqrt(16)*sqrt(5)

sqrt(-80) = i*4*sqrt(5)

sqrt(-80) = 4i*sqrt(5)

sqrt(-80) = 4*sqrt(5)*i

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Sara, Kim, and John own a painting company. To paint a particular home, Sara estimates it would take her 4 days. Kim estimates 5
Vitek1552 [10]

Answer:

1.67 days

Step-by-step explanation:

Sara can finish 1/4 work a day

Kim can finish 1/5.5 work a day

John can finish 1/6 work a day

if they work together one day can do: 1/4 + 1/5.5 + 1/6 = 79/132

79/132 * x days = 1

x = 132/79 = 1.67

7 0
4 years ago
Write the equation that describes the line in slope-intercept form.
Thepotemich [5.8K]

Answer:

y = 2x+4

Step-by-step explanation:

slope (m) = 2

point (-1,2)

so,

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y = mx+b

or, y = 2x+4

5 0
3 years ago
Simplify the radical expression <br> <img src="https://tex.z-dn.net/?f=%5Csqrt5%2B6%5Csqrt5" id="TexFormula1" title="\sqrt5+6\sq
dolphi86 [110]

Answer:

Step-by-step explanation:

These are "like terms," as both terms contain √5.  We add the coefficients together but leave the radical term √5 as is:

  √5

+ 6√5

-----------

 7√5

6 0
3 years ago
Read 2 more answers
Uestion
Stella [2.4K]

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JKLM=\sqrt{(~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2} \\\\\\ JKLM=\sqrt{(2 +1)^2 + (-2 - 2)^2} \implies JKLM=\sqrt{( 3 )^2 + ( -4 )^2} \\\\\\ JKLM=\sqrt{ 9 + 16 } \implies JKLM=\sqrt{ 25 }\implies \boxed{JKLM=5}

now, let's check the other path, JM and KL

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JMKL=\sqrt{(~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2} \\\\\\ JMKL=\sqrt{(3 +2)^2 + (1 +1)^2} \implies JMKL=\sqrt{( 5 )^2 + ( 2 )^2} \\\\\\ JMKL=\sqrt{ 25 + 4 } \implies \boxed{JMKL=\sqrt{ 29 }}

so the red path will be  5~~ + ~~\sqrt{29} ~~ \approx ~~ \blacksquare~~ 10 ~~\blacksquare

3 0
2 years ago
If 8x=16y then<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7By%7D%20" id="TexFormula1" title=" \frac{x}{y} " alt=" \f
Rama09 [41]

<em>Answer is 2</em>

<em>If 8x = 16y </em>

<em>on rearranging question,</em>

<em> x/y = 16/8 = 2</em>

8 0
3 years ago
Read 2 more answers
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