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3 years ago

Draw and label the circle given by the equation (x - 5 + y + 10 = 6

Mathematics

1 answer:

Alik [6]3 years ago

3 0

Answer:

See attachment for drawing

Step-by-step explanation:

Given

$(x -5)^2 + (y + 10)^2 = 6^2$

Required

Draw and label the circle

The equation of a circle is:

$(x -a)^2 + (y - b)^2 = r^2$

Where:

$Center = (a,b)$

$Radius = r$

So, we have:

$Center = (5,-10)$

$Radius = 6$

See attachment for drawing

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Find the distance between the points given. (-3, 2) and (9, -3) √(61) √(37)

cricket20 [7]

Answer:

d=13.

Step-by-step explanation:

d= $\sqrt{(x_{2}-x_{1} ) ^{2} +(y_{2}-y_{1} ) ^{2}}$

From question statement, $x_{1} =-3$ , $x_{2} =9$

$y_{1} =2$ , $y_{2} =-3$

put above values in the distance formula,we obtain

d= $\sqrt{(9-(-3))^{2}+(-3-2)^{2}}$

d= $\sqrt{(12)^{2}+(-5)^{2}}$

d= $\sqrt{144+25}$

d= $\sqrt{169}$

d=13

8 0

3 years ago

Consider the angle. Which TWO statements are true?

svetoff [14.1K]

Answer:

C, E

Step-by-step explanation:

A. INCORRECT

A is wrong because a reflection across the x-axis DOES move the position of the figure (as it is flipped, so the position changes), but it DOES NOT change the angle (since a shift in position doesn't equal to a change in angle measure)

B. INCORRECT

Although a reflection across the x-axis does change the position of the angle, it DOES NOT change the measure of the angle.

C. CORRECT

A reflection across the x-axis does in fact move the position of the figure and does not change the angle measure. Reflections only deal with flipping a figure, not changing it's shape/distorting it so that the angle will change.

D. INCORRECT

A translation right will change the position of the figure but will not change the measure of the angle.

E. CORRECT

Yes, a translation right WILL change the position of the figure but will NOT change the measure of the angle. This is because a translation is simply moving a figure up and down; it has nothing to do with changing the shape of the figure/distorting it so that the angle is different.

8 0

4 years ago

The length of s is 2/3 the length of t. If s has an area of 368 cm squared find the perimeter of the figure

ser-zykov [4K]

$s = \frac{2}{3}t
\\\ \frac{2}{3}t*t = \frac{2}{3}t^2
\\\ \frac{2}{3}t^2 = 368
\\\ t^2 = \frac {3}{2}368
\\\ t^2 = 552
\\\ t \approx 23.5
\\\ s \approx 15.6
\\\ P \approx 78.2$