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Maurinko [17]
3 years ago
9

Name a diameter for the circle. A.) DH B.) DG C.) EF D.) FH

Mathematics
1 answer:
Phantasy [73]3 years ago
4 0
D. FH

Diameter-
a straight line passing from side to side through the center of a body or figure, especially a circle or sphere.
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The two polygons above are similar. The scale factor of the smaller polygon to the larger polygon is 3. The area of the smaller
Softa [21]
You would think it was 3 * 12 = 36. Not so. All polygons have to be broken down into some figure that will give 2 dimensions that are at right angles to each other. That would mean that
d1 * d2 = Area for the small polygon
3d1 * 3d2 = area of the larger polygon

What that means is that the area of the larger one is 9 times the smaller one.
Area large = 12 * 9 = 108 square units. <<<<< answer.

If you find this hard to be leave try it with a square.
Suppose you have a square (the small one) that is 3 cm by 3 cm
The small one has an area of 3*3 cm^2 = 9 cm

Now you have another square that is 3 times larger. That means that each side is 3*3 = 9
So s = 9
Area = s^2
Area = 9^2 = 81 cm^2
81 is 9 times larger than 9 just as you would think. 


4 0
3 years ago
The pool has a diameter of 5 feet and a height of 3 feet
BigorU [14]

Answer:

what are you asking

Step-by-step explanation:

cool?

6 0
3 years ago
Read 2 more answers
What is the probability of rolling two dice, and having a total of 8 dots showing? Express your answer as a percent
Solnce55 [7]

Answer:

Step-by-step explanation:

8 0
3 years ago
Let Upper A equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 4 2nd Row 1st Column 1 2nd Column 3
Anastaziya [24]

Answer:

See explanation

Step-by-step explanation:

Given:

A=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right]

B=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right]

A. Find AB:

AB=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right]\cdot \left[\begin{array}{cc}-2&1\\3&7\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+4\cdot 3&-2\cdot 1+4\cdot 7\\1\cdot (-2)+3\cdot 3&1\cdot 1+3\cdot 7\end{array}\right]=\left[\begin{array}{cc}16&26\\7&22\end{array}\right]

B. Find BA:

BA=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right]\cdot \left[\begin{array}{cc}-2&4\\1&3\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+1\cdot 1&-2\cdot 4+1\cdot 3\\3\cdot (-2)+7\cdot 1&3\cdot 4+7\cdot 3\end{array}\right]=\left[\begin{array}{cc}5&-5\\1&33\end{array}\right]

C. Answers are not the same

D. Matrices multiplication is not commutastive in general, so

AB\neq BA

7 0
3 years ago
Consider the set of points defined by z = 5 cos(theta) + 3i sin(theta).
vitfil [10]

Answer:

This is an ellipse.

Step-by-step explanation:

z=5cos(\theta)+3isin(\theta)

Applying the angle values in intervals of right angles (90°), the resulting set of values for z are:

\theta = 0,90,180,270,360

z=5,3i,-5,-3i,5 (again)

If you look at the plot, you will observe how these values represent points in the xy plain. All of them belong to either the x or the y axis.

The resulting figure for the whole set of values of \theta is in fact an ellipse.

4 0
2 years ago
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