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Thepotemich [5.8K]
3 years ago
5

Help me please no links :)

Mathematics
1 answer:
AlekseyPX3 years ago
4 0

Answer:

Domain (Basically means the sets of values of x-axis).Hence, the set of values of domain is given by -3<domain<1. Meaning domain lies between values more than -3 and less than 1.

Range (Basically means the sets of values of y-axis). Hence, the set of values of range is given by -1<range<3. Meaning that range lies between values more than -1 and less than 3.

Function basically means the equation of the given circle. We know that equation of circle is given by;

(x-h)^2+ (y-k)^2 = r^2 [where (h, k) is the center of circle and r is the radius of circle]

From the figure,

(h, k)= (-1 ,1) and r= 2.

So, we can conclude that function is given by equation:

(x-(-1))^2 + (y-1)^2 =  2^2

(x+1)^2  +  (y-1)^2 = 4.

You might be interested in
Is 4/67 grater that 5/777
Anni [7]
Yes because if you divide 4 by 67 you get about 0.0597 and if you divide 5 by 777 then you get about 0.0064 and 0.0597 is greater than 0.0064

6 0
3 years ago
R(-1,-3), S(4,4), T(8,-1)<br><br><br> PLZ HELP ME
Bingel [31]

9514 1404 393

Answer:

  see attachments

Step-by-step explanation:

The side lengths are found using the distance formula.

  d = √((x2-x1)^2 +(y2-y1)^2)

For example, the length of RS in the first triangle is ...

  d = √((4-(-1))^2 +(4-(-3))^2) = √(25 +49) = √74 ≈ 8.602

This same computation is used for all of the sides of the triangles. When the same tedious computation is repeated over and over, I like to use a spreadsheet to do it. In the attached spreadsheet the two points used are the one on the same line as the distance, and the one on the line below. Side lengths are shown in the attachment to 3 decimal places.

__

In order to classify the triangle as acute, right, or obtuse, we can compare the side lengths to those of a right triangle. For the purposes of computation in a spreadsheet, it is convenient to compute the middle length side of a right triangle that has the same longest and shortest sides as the triangle we have.

If the computed side is longer than the actual middle side, then the triangle we have is obtuse. If it is shorter, the triangle we have is acute. If it is exactly the same, then the triangle we have is a right triangle.

In every case here, the computed middle side is shorter than the actual middle-length side, so all of the triangles are acute triangles. (The one of problem 39 is isosceles, since two sides are the same length. The others are scalene.)

__

The triangles are graphed in the second attachment.

  problem 38: RST

  problem 39: ABC

  problem 40: DEF

5 0
3 years ago
Write and solve the equation to find the value of x and the missing angle measures.​
MrRissso [65]

Answer:

x = <u><em>8</em></u>

The missing measures are <em><u>85</u></em> degrees

Step-by-step explanation:

Since 9x + 13 and 11x - 3 are on opposite sides, they are equal, so

9x + 13 = 11x - 3

Subtract 13 from both sides:

9x = 11x - 16

Subtract 11x from both sides:

-2x = -16

Divide both sides by -2:

x = 8

The angle measures are

9(8) + 13 = 72 + 13 = 85

11(8) - 3 = 88 - 3 = 85

6 0
3 years ago
Read 2 more answers
If 8.045 kilometers is about 5 miles, about how many kilometers is 13 miles? Round your answer to the nearest hundredth if neces
Angelina_Jolie [31]

Answer:

20.92 km

Step-by-step explanation:

8.045 km/5 miles = 1.609 km per mils

1.609 x 13 = 20.92 km

3 0
3 years ago
Brown’s time for running a mile in gym class is 9.9 minutes, Ray’s time is 9.47 minutes and Malcolm’s time is 9.76 minutes. Who
GalinKa [24]

Given:

Brown’s time for running a mile in gym class = 9.9 minutes

Ray’s time for running a mile in gym class = 9.47 minutes

Malcolm’s time for running a mile in gym class = 9.76 minutes

To find:

Who ran the mile in less time?

Solution:

It is given that the time taken by Brown, Ray and Malcolm to cover a mile are 9.9 minutes, 9.47 minutes and 9.76 minutes respectively.

9.47 < 9.76 < 9.9

Since, 9.47 minutes is the lowest time, therefore Ray can run a mile in less time.

3 0
3 years ago
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