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2 years ago

Quadrilateral ABCD is dilated by a scale factor of 2 centered around (2,2).

Mathematics

1 answer:

ANEK [815]2 years ago

6 0

Answer:

B’D’ will run through (2,2) and will be longer than BD

Step-by-step explanation:

B'D' will be twice as long as BD and run through (2, 2).

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Find the equation of the vertical line through (7,-3)

wlad13 [49]

X = 7
Is a vertical line that passes through (7,-3).

4 0

3 years ago

Can anyone solve this? will give brainliest if correct.

miv72 [106K]

Answer:

See below

Step-by-step explanation:

The missing reasons are:

1. k. Given
2. j. Definition of parallelogram
3. d. Definition of linear pair
4. b. Linear pair postulate
5. e/m. Definition of supplementary
6. g. Same side interior angles theorem
7. e/m. Definition of supplementary
8. a/c. Substitution property of congruence
9. i. Subtraction property of congruence
10. f. Alternate interior angles theorem
11. l. Alternate exterior angles theorem
12. h. Angle congruence postulate

8 0

2 years ago

Read 2 more answers

Susan's elevation at the top of a mountain is 1,530 meters. After 4 hours of climbing down her elevation is 680 meters. What is

Nitella [24]

Answer:

C) -212.5

Step-by-step explanation:

I've done this question on a quiz before.

Hope I helped!! :D

4 0

3 years ago

In exercises 21 through 24 is it possible to construct a triangle if the given side lengths? If not explain why.

sattari [20]

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7 0

3 years ago

A farmer has a field in the shape of a triangle. The farmer has asked the manufacturing class at your school to build a metal fe

Finger [1]

Answer:

Fencing required = 1586 m

Step-by-step explanation:

The given statements can be thought of a triangle $\triangle ABC$ as shown in the diagram attached.

A be the 1st vertex, B be the 2nd vertex and C be the 3rd vertex.

Distance between 1st and 2nd vertex, AB = 435 m

Distance between 2nd and 3rd vertex, AC = 656 m

$\angle A =49^\circ$

To find:

Fencing required for the triangular field.

Solution:

Here, we know two sides of a triangle and the angle between them.

To find the fencing or perimeter of the triangle, we need the third side.

Let us use Cosine Rule to find the third side.

Formula for cosine rule:

$cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}$

Where

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

$\Rightarrow cos 49^\circ = \dfrac{656^{2}+435^{2}-BC^{2}}{2\times 435\times 656}\\\Rightarrow BC^2 = 430336+189225-2 (435)(656)cos49^\circ\\\Rightarrow BC^2 = 430336+189225-570720\times cos49^\circ\\\Rightarrow BC^2 =619561-570720\times cos49^\circ\\\Rightarrow BC \approx 495\ m$

Perimeter of the triangle = Sum of three sides = AB + BC + AC

Perimeter of the triangle = 435 + 495 + 656 = 1586 m

Fencing required = 1586 m

4 0

3 years ago