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sladkih [1.3K]
3 years ago
15

For a dilation with a scale factor less than 1, how do the angles and side lengths of the preimage relate to the angles and side

lengths of the image?
A. Angle are proportional;side lengths are larger

B. Angles are proportional;side lengths are smaller

C. Angles are congruent;side lengths are larger

D. Angles are congruent; side lengths are smaller
Mathematics
2 answers:
stiks02 [169]3 years ago
4 0

Answer:

A

Step-by-step explanation:

kakasveta [241]3 years ago
3 0

Answer: option A

Step-by-step explanation:

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How do you find the area of the shaded sector of the circle? The radius is 10in and the sector is 45 degrees
Step2247 [10]

Given:

Radius of the circle = 10 in

Central angle of the sector = 45 degrees

To find:

The area of the sector.

Solution:

Area of a sector is

A=\dfrac{\theta}{360^\circ}\pi r^2

Where, \theta is the central angle in degrees.

Putting r=10 and \theta=45^\circ, we get

A=\dfrac{45^\circ}{360^\circ}\pi (10)^2

A=\dfrac{1}{8}\pi (100)

A=12.5\pi

Therefore, the area of the sector is 12.5π sq. inches.

7 0
2 years ago
Use the function rule to complete the table.<br> -10x+y=4<br><br> X: -2 -1 0 1 2 <br> Y:
harina [27]
Y = 10x + 4
 x                      y
-2  10(-2)+4 = -16
-1  10(-1)+4 = -6
0    10(0)+4 = 4
1    10(1)+4  = 14
2    10(2)+4 = 24
5 0
3 years ago
Anyone tell me the answer to b please
insens350 [35]

Answer:

total amount of students in the class = 24

the amount of boy students in the class = 8

the amount of girl students in the class = ?

a) we know how many boys are in so B=8/24

b) know you calculate the amount of girl students in the class

totally they are 24 students are found we know the boys then the last are girls to know how many girls are in there

24 - 8 = 16 girls.

G=16/24

8 0
3 years ago
In the circle below, suppose m FEH=272º and m EFG=116º. Find the following.​
adoni [48]

Answer:

m∠FEH = 44°

m∠EHG =  64°

Step-by-step explanation:

1) The given information are;

The angle of arc m∠FEH = 272°, the measured angle of ∠EFG = 116°

Given that m∠FEH = 272°, therefore, arc ∠HGF = 360 - 272 = 88°

Therefore, angle subtended by arc ∠HGF at the center = 88°

The angle subtended by arc ∠HGF at the circumference = m∠FEH

∴ m∠FEH = 88°/2 = 44° (Angle subtended at the center = 2×angle subtended at the circumference)

m∠FEH = 44°

2) Similarly, m∠HGF is subtended by arc m FEH, therefore, m∠HGF = (arc m FEH)/2 = 272°/2 = 136°

The sum of angles in a quadrilateral = 360°

Therefore;

m∠FEH + m∠HGF + m∠EFG + m∠EHG = 360° (The sum of angles in a quadrilateral EFGH)

m∠EHG = 360° - (m∠FEH + m∠HGF + m∠EFG) = 360 - (44 + 136 + 116) = 64°

m∠EHG =  64°.

5 0
3 years ago
Please help! 56 points. Greatly appreciated!
VashaNatasha [74]

Answer:

The first thing you should do for this case is to draw the ordered pairs in the plane and join the points.

 The area you are looking for is the area of a rectangle plus the area of a triangle.

 Thus, the total area will be

 At = (8) * (4.5) + (1/2) * (4) * (2.5) = 41

 answer

 the area of the city is 41 units^2

Step-by-step explanation:

41 square miles If you draw the outline of the city, you'll realize that if you draw a line from point B to point D, that you can subdivide the city into a triangle and a trapezoid. After performing the division, you can then calculate the area of both polygons and the add their areas together. So first, let's deal with the trapezoid ABDE. The area of a trapezoid is the average of the length of the parallel sides multiplied by the height. The parallel sides are AB and DE. So: ((18-10)+(14-10))*(9-4.5)/2 =(9 + 4)*(4.5)/2 = 12*4.5/2 = 27 Now for the area of triangle BCD. The area of a triangle is 0.5*b*h where b is the base and h the height. I'll use BC as the base and the distance from BC to D as the height. So: (9-2)*(18-14)/2 = 7*4/2 = 14 And now to add the areas. 27 + 14 = 41 So the area of the city is 41 square miles. Note: The subdivision used is not the only possible subdivision, just one of the easier ones. I could have divided the city area into 3 triangles ABE, BDE, and BCD and solved it that way instead. It was just a happy coincidence that AB and DE were parallel and as such I was able to use trapezoid ABDE instead of the two triangles ABE and BDE.

6 0
2 years ago
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