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Alexandra [31]
3 years ago
14

Please help!!!!Evaluate the function rule for the given value. y=6x4^x for x = –3

Mathematics
1 answer:
Shkiper50 [21]3 years ago
7 0
Y=6*4^x = 6*4^-3=6*0.015625=0.09375
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HELP HELP HELP IM BEGGIN YU
marishachu [46]

Answer:

Option C

Step-by-step explanation:

If two triangles are congruent then their corresponding sides and angles are congruent.

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<u><em>hope it helps you..</em></u>

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5 0
3 years ago
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If A(3,4) and B(13,11), what are the coordinates of the point 3/5from A to B ?
Goshia [24]

Answer:

The correct answer is B: (9,5.75



4 0
3 years ago
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Which method do you prefer to use to divide polynomials? Long division or Synthetic division? Explain why.
elena-s [515]

Answer: I prefer long division because that is the method I understand

Step-by-step explanation:

I prefer long division because that is the method I understand .for example

(x^2+3x+2) ➗ (x+2)

x+1

|````````````````````````

x+2 x^2+3x+2

-(x^2+2x)

````````````````````````````

x+2

- (x+2)

````````````````````````

0

`````````````````````````

therefore (x^2+3x+2) ➗ (x+2)=x+1

7 0
3 years ago
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PLEASE HELP
kirza4 [7]

Answer:

17.50

Step-by-step explanation:

7*3 is 21 and 7/2 is 3.7 so its 5*3 (15) plus 2.50, so 17.50

Hope this helped!

5 0
3 years ago
Can anyone figure this out?
Verizon [17]

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill


\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}


now that we know how long each one is, let's plug those in Heron's Area formula.


\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18

5 0
4 years ago
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