Please help. here is a screenshot of the question.

Mathematics

1 answer:

Feliz [49]3 years ago

6 0

Hello,

1) the line parallele to ox passing by (1,4) has for equation y=4

f(x)=4 if -1<=x<=1

2) the line pasing through (3,2) and (5,4) has for equation:

y-2=(4-2)/(5-3)(x-3)
==>y=x-3+2
==>y=x-1

f(x)=x-1 if 3<=x<=5.

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What is the area of rhombus ABCD? Enter your answer in the box. Do not round at any steps. put the answer in units²

AlladinOne [14]

Answer:

Therefore the Area of Rhombus ABCD is 36 unit².

Step-by-step explanation:

Given:

ABCD is a Rhombus

A = (-1,0)

B = (5,-3)

C = (-1,-6)

D = (-7 ,-3)

To Find:

Area of Rhombus ABCD = ?

Solution:

We know that Area of Rhombus is given as

$\textrm{Area of Rhombus}=\dfrac{1}{2}\times d_{1}\times d_{2}$

Where ,

d₁ and d₂ are the Diagonals.

We have,

Diagonals as AC and BD,

Using Distance Formula we get

$l(AC) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}$

Substituting coordinates A and C we get

$l(AC) = \sqrt{((-1-(-1))^{2}+(-6-0)^{2} )}=\sqrt{36}\\l(AC)=6\ unit$

Similarly for BD we have,

$l(BD) = \sqrt{((5-(-7))^{2}+(-3-(-3))^{2} )}=\sqrt{144}\\l(BD)=12\ unit$

Now Substituting AC and BD in Formula we get

$\textrm{Area of Rhombus}=\dfrac{1}{2}\times AC\times BD$

$\textrm{Area of Rhombus}=\dfrac{1}{2}\times 6\times 12=36\ unit^{2}$

Therefore the Area of Rhombus ABCD is 36 unit².

6 0

3 years ago

What is the value of -2xy if x = -1 and y = 6? 8 12 26 -12

Elena L [17]

12 would be your answer
-2*-1= 2
2*6=12
so 12

3 0

3 years ago

PLEASE can someone explain why The exact value of sec ( π/6) is 2 /√ 3 ??

Umnica [9.8K]

$\bf sec\left( \frac{\pi }{6} \right)\qquad recall\to sec(\theta)=\cfrac{1}{cos(\theta)}\qquad thus
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sec\left( \frac{\pi }{6}\right)\iff \cfrac{1}{cos\left( \frac{\pi }{6}\right)}
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\textit{check your Unit Circle for what the }cos\left( \frac{\pi }{6}\right)\ is
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