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Ann [662]
3 years ago
5

What are the solutions to log Subscript 6 Baseline (x squared + 8) = 1 + log Subscript 6 Baseline (x)

Mathematics
1 answer:
lozanna [386]3 years ago
3 0

Answer:

x=2 and x=4

Step-by-step explanation:

i just did it

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Tara is using matrix multiplication to find image points of a translation. She is translating the image 2 units left and 4 units
Ahat [919]
Assuming that this is just on a 2-D coordinate plane, we must convert the expressions on to a 3-D plane since translation cannot be done on a 2-D plane. This is done by adding a dummy coordinate that does not change. Let us use "1" for this case.

Matrix:
| 0 0 -2 |(x) = (x - 2)
<span>| 0 0 4  |(y) = (y + 4)
</span><span>| 0 0 1  |(1) = 1</span>
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3 years ago
The equation of a straight line is y+10=5x
MariettaO [177]

Answer:

y=5x-10

Step-by-step explanation:

y+10=5x

y=5x-10

4 0
3 years ago
How to graph y=2x-4 x+3y=9
NikAS [45]
  • Answer:

Step-by-step explanation:

  • The point form: (3,2)
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  • You solve for y
  • On the Y-axis where the 2 is you go 3 to the right and put your point right there above the 3.
  • There you have your answer on the graph.
4 0
3 years ago
Jan conducts an experiment and tosses a coin 40 times. She gets heads 30 times instead of the expected 20. What could be a reaso
Nady [450]

the answer is B. hope i helped

8 0
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

4 0
2 years ago
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