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Mathematics

1 answer:

Darya [45]3 years ago

8 0

Answer:

First & Third options

Step-by-step explanation:

$\bigg( {5}^{ \frac{3}{8}} \: . {5}^{ \frac{1}{8}} \bigg)^{3} \\ \\ = \bigg( {5}^{ \frac{3}{8} + \frac{1}{8}} \bigg)^{3} \\ \\ = \bigg( {5}^{ \frac{4}{8}} \bigg)^{3} \\ \\ = \bigg( {5}^{ \frac{1}{2}} \bigg)^{3} \\ \\ = {5}^{ \frac{3}{2}} \\ \\ = \sqrt{ {5}^{3} } \\ \\ \huge\red{ \boxed{\therefore \bigg( {5}^{ \frac{3}{8}} \: . {5}^{ \frac{1}{8}} \bigg)^{3}= \sqrt{ {5}^{3} } }}$

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Graph the circle (x+6)^2 + (y+5)^2 =9

Andrei [34K]

Answer:

This a circle centered at the point $(-6,-5)$ , and of radius "3" as it is shown in the attached image.

Step-by-step explanation:

Recall that the standard formula for a circle of radius "R", and centered at the point $(x_0,y_0)$ is given by:

$(x-x_0)^2+(y-y_0)^2=R^2$

Therefore, in our case, by looking at the standard equation they give us, we extract the following info:

1) $R^2 = 9\,\,\,then\,\,\,R=3$ since the radius must be a positive number and ( $R=-3$ ) is not a viable answer.

2) $x_0=-6$ for ( $x-x_0$ ) to equal $(x+6)$

3) $y_0=-5$ for ( $y-y_0$ ) to equal $(y+5)$

Therefore, we are in the presence of a circle centered at the point $(-6,-5)$ , and of radius "3". That is what we draw as seen in the attached image.