MATHEMATICS HIGH SCHOOL

(04.01 LC) The vertices of a quadrilateral ABCD are A(1, -3), B(4, -3), C(4, -5), and D(-1, -5). The vertices of another quadrilateral EFCD are E(1, -7), F(4, -7), C(4, -5), and D(-1, -5). Which conclusion is true about the quadrilaterals? (5 points) The shape of the quadrilaterals is same but their areas are different. The ratio of their corresponding sides is not equal. The measures of the corresponding angles are different. The angles and sides overlap when one quadrilateral is placed on the other.

Answers

Answer 1

Answer:

The answer is D: The angles and sides overlap when one quadrilateral is placed on the other

Answer 2

Answer:

Answer:

Taking the test rn and its D!!

Step-by-step explanation:

I will lyk if its right when i submit!!!

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Answers

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Find b, given that a = 20, angle A = 30°, and angle B = 45° in triangle ABC

Answers

Answer:

b = 20√2

Step-by-step explanation:

Given: ΔABC

a = 20 , m∠A = 30° and m∠B = 45°

To find: value of b.

We use Sine result, which state that

$(a)/(sin\,A)=(b)/(sin\,B)$

Substituting given values we, get

$(20)/(sin\,30^(\circ))=(b)/(sin\,45^(\circ))$

we know that $sin\,30^(\circ)=(1)/(2)\:and\:sin\,45^(\circ)=(1)/(√(2))$ , we get

$(20)/((1)/(2))=(b)/((1)/(√(2)))$

$20{*2}=b*√(2)$

$b*√(2)=40$

$b=(40)/(√(2))$

$b=20√(2)$

Therefore, b = 20√2

The side b is opposite to the angle B, applying the law of the sines, we have:

$(a)/(sinA) = (b)/(sinB)$
$(20)/(sin30^0) = (b)/(sin45^0)$
$(20)/( (1)/(2) ) = (b)/( ( √(2) )/(2) )$
$20* ( √(2) )/(2) = b* (1)/(2)$
$(20 √(2) )/(2) = (b)/(2)$
$2*b =2*20 √(2)$
$2b = 40 √(2)$
$b = (40 √(2) )/(2)$
$\boxed{b = 20 √(2) }$

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Answers

Answer:

Can't tell

Step-by-step explanation:

Hear me out. There is no graph hear, so I can't answer. Just letting you know

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Answers

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Answers

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