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lisabon 2012 [21]
3 years ago
11

What is the area of a circle whose radius is x + 2?

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
7 0
The radius is 1 2/2 = 1 
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Is 12 rational or irrational ​
wlad13 [49]

Answer: Rational

Step-by-step explanation: 12 is a rational number because it can be expressed as the quotient of two integers: 12÷ 1

5 0
3 years ago
On a 15 scale drawing of a car, one part is 88 inches long. How long will the actual car part be?
IrinaVladis [17]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

The lenght that <span>will the actual car part be is 40 inches. Below is the solution:
</span>
8" = 8 x 5 
= 40 inches
6 0
3 years ago
in the circl below, AD is diameter and AB is tangent at A. Suppose mADC=228*, find the measures of mCAB and mCAD. Type your nume
Andrew [12]

Given:

AD is diameter of the circle, AB is the tangent, and measure of arc ADC is 228 degrees.

To find:

The m\angle CAB and m\angle CAD.

Solution:

AD is diameter of the circle. So, the measure of arc AD is 180 degrees.

m(arcADC)=m(arcAD)+m(arcDC)

228^\circ=180^\circ+m(arcDC)

228^\circ-180^\circ+=m(arcDC)

48^\circ+=m(arcDC)

The measure inscribed angle is half of the corresponding subtended arc.

m\angle CAD=\dfrac{1}{2}\times m(arcDC)

m\angle CAD=\dfrac{1}{2}\times 48^\circ

m\angle CAD=24^\circ

AB is the tangent. So, m\angle BAD=90^\circ because radius is perpendicular on the tangent and the point of tangency.

m\angle BAD=m\angle CAB+m\angle CAD

90^\circ=m\angle CAB+24^\circ

90^\circ -24^\circ=m\angle CAB

66^\circ=m\angle CAB

Therefore, m\angle CAB=66^\circ and m\angle CAD=24^\circ.

4 0
3 years ago
Graph a triangle (STU) and reflect it over the y-axis to create triangle ST'U'.
lukranit [14]

The x-coordinates of \triangle S'T'U' will be the negation of the x-coordinates of \triangle STU

The line segment from S to the y-axis equals the line segment from S' to the y-axis. Similarly, the line segment from T to the y-axis equals the line segment from T' to the y-axis

See attachment for \triangle STU and \triangle S'T'U'

In order to solve this question, I will make the following assumptions.

Assume that the coordinates of \triangle STU are

S = (4,5)      

T = (5,9)

U=(3,8)

Refer to attachment for illustrations

<u>(1) Reflect </u>\triangle STU<u> over y-axis and describe the transformation</u>

To reflect \triangle STU across the y-axis, the following rule must be followed

(x,y) \to (-x,y)

This means that:

S = (4,5) \to S' = (-4,5)

T = (5,9) \to T' = (-5,9)

U=(3,8) \to U'=(-3,8)

<u>The description of the </u><u>transformation </u><u>is as follows:</u>

Notice that the signs of the x-coordinates \triangle STU and \triangle S'T'U' of both triangles are different.

In other words, if the x-coordinate of one is positive, then the other will have a negative x-coordinate; and vice versa.

<u>(2) Compare the segments and the line of reflection</u>

To reflect across the y-axis means that the reflecting line is the y-axis, itself.

The distance between a point to the y-axis is the absolute value of the x-coordinate.

So, the distance between S and the y-axis is:

S = |4| = 4

The distance between S' and the y-axis is:

S' = |-4| = 4

We can conclude that the two line segments are equal.

This is the same for other point T and T' because of the formula used above.

<u>From T and T' to the y-axis is:</u>

T =|5| =5

T' =|-5| =5

Read more at:

brainly.com/question/938117

8 0
3 years ago
(1,-7); slope=-1 whats the answer please
Anettt [7]

Slope = -1

and it goes through (1, -7)

7 0
3 years ago
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