Will give brainliest answer

Mathematics

1 answer:

Jet001 [13]3 years ago

7 0

Radius is 6.5 and the diameter is 13

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What is the line of symmetry on this graph?

Kisachek [45]

Answer:

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Step-by-step explanation:

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What's -8 + (-1 1/8)

gogolik [260]

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3 years ago

Here are the endpoints of the segments BC, FG, and JK.<br> B, −67

yulyashka [42]

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{[-4 - (-6)]^2 + [4 - 7]^2}\implies BC=\sqrt{(-4+6)^2+(-3)^2} \\\\\\ BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}$

$F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}$

$JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill$

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2 years ago

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dangina [55]

I think its 15+5X not shur

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3 years ago

Help me please !!!! (:

riadik2000 [5.3K]

3x=x+40
Subtract x from both sides
2x=40
Divide both sides by 2nto get x by itself
x=2
Plug 2 in for x
m∠4 = (3x)
m∠4 = 3(2)
Solve
m∠4 = 6

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3 years ago