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Anestetic [448]
3 years ago
10

How do you simplify x4·(y2)3·y−2 and (3x)2·x2 Look at picture

Mathematics
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

9x^{4} for (3x)^2· x^{2} and (x^{4})·(y^{4}) for x^4·(y^2)^3·y^{-2}

Step-by-step explanation:

(3x)^{2} (x^{2} ) Multiply in

(9x^{2})(x^{2}) Multiply together

9x^{4}

(x^{4})(y^{2})^{3} (y^{-2})

(x^{4})(y^{6})(\frac{1}{y^{2}})  Multiply in

(x^{4})(\frac{y^{6}}{y^{2}}) Multiply together

(x^{4})(y^{4})

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Find the value of C that makes the trinomial a perfect square<br><br> HELP PLS SHOW WORK 30 POINTS
lesya692 [45]

Answer:

\frac{9}{4}

Step-by-step explanation:

{(x +   \frac{3}{2}  )}^{2}  =   {x}^{2}  + 3x +  \frac{9}{4}

4 0
2 years ago
Hii please help i’ll give brainliest!!
castortr0y [4]
I am going to guess C but I’m not entirely sure
3 0
3 years ago
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Write the slope x+y=4 I have no idea how to do this
shutvik [7]
That’s the answer
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4 0
3 years ago
(19-4) times (14+4)+4 to the power of 2
Volgvan

Answer:

75,076

Step-by-step explanation:

19-4 = 15

14+4 = 18

15 x 18 = 270, then add 4 which equals 274.

274 x 274 = 75,076

8 0
4 years ago
Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess
Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

4 0
3 years ago
Read 2 more answers
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