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stellarik [79]
3 years ago
7

PLEASE ANSWER THIS QUESTION I REALLY NEED IT TODAY AND WHOEVER DOES THIS, SHE/HE WILL GET LUCK! PLS ANSWER

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
8 0

The answer is C I think.

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Which of the following geometric objects could have it length measured
anastassius [24]

Answer:

You need to attach the image

Step-by-step explanation:

3 0
3 years ago
Please explain how to get the answer.
Maru [420]
2r=R

Because 2 of "r" make 1 R
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Compare the weather on two different days using an inequality. Then see if their comparisons change by finding the absolute valu
dusya [7]

Inequalities are used to compare unequal expressions.

The actual comparison is:\mathbf{-3.5 < 5} or \mathbf{5 > -3.5}

The temperatures are given as:

\mathbf{Temperature =  -3.5F,\ \  5F,\ \  1.5F,\ \  -0.5F,\ \ -2F,\  \ 2.5F,\ \  -4F}

From the above list, we have:

\mathbf{Day \ 1=  -3.5F}

\mathbf{Day \ 2=  5F}

By comparison,

\mathbf{-3.5 < 5} or \mathbf{5 > -3.5}

Using absolute values, the inequalities are:

\mathbf{|-3.5| < |5|} or \mathbf{|5| > |-3.5|}

Hence, the actual comparison is:

\mathbf{-3.5 < 5} or \mathbf{5 > -3.5}

Read more about inequalities and absolute values at:

brainly.com/question/24797699

7 0
3 years ago
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Find an equation for the nth term of the arithmetic sequence. a19 = -58, a21 = -164
Elodia [21]
So the formula would be -1856 +98[n-1]
8 0
3 years ago
Identify the center and the radius of a circle that has a diameter with endpoints at (−5, 9) and (3, 5)
marusya05 [52]

Check the picture below, so the circle looks more or less like that one.

well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}

8 0
3 years ago
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