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3 years ago

The vertices of a quadrilateral in the coordinate plane are known. How can the perimeter of the figure be found?

Mathematics

2 answers:

bonufazy [111]3 years ago

7 0

Use the distance formula to find the length of each side, and then add the lengths.

iren2701 [21]3 years ago

6 0

Answer:

Step-by-step explanation:

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Use the parabola tool to graph the quadratic function. f(x)=2x2+12x+16

yKpoI14uk [10]

We have that
<span>f(x)=2x</span>²<span>+12x+16

using a graph tool
see the attached figure</span>

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3 years ago

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“solve each triangle to the nearest tenth of a unit”

Hitman42 [59]

Answer:

Step-by-step explanation:Solve each triangle. Round side lengths to the

nearest tenth and angle measures to the nearest

degree.

SOLUTION:

Use the Law of Cosines to find the missing side

length.

Use the Law of Sines to find a missing angle

measure.

Find the measure of .

ANSWER:

A ≈ 36°, C ≈ 52°, b ≈ 5.1

SOLUTION:

Use the Law of Cosines to find the measure of the

largest angle, .

Use the Law of Sines to find3. a = 5, b = 8, c = 12

SOLUTION:

Use the Law of Cosines to find the measure of the

largest angle, .

Use the Law of Sines to find the measure of angle,

Find the measure of .

ANSWER:

A ≈ 18°, B ≈ 29°, C ≈ 133°

4. B = 110°, a = 6, c = 3

SOLUTION:

Use the Law of Cosines to find the missing side

length.

Use the Law of Sines to find a missing angle

measure.

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ANSWER:

A ≈ 48°, C ≈ 22°, b ≈ 7.6

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2 years ago

What is decimal of 1/10 of .08

Zinaida [17]

.008 is the answer you are looking for :)

6 0

3 years ago

Find the length of the missing leg of a right triangle whose hypotenuse measures 15 cm and whose other leg measures 13 cm. geome

Jet001 [13]

Answer:

you can always use sin or cos or tan in these situations

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3 years ago

The equation tan(x- pi/3) is equal to _____.

kherson [118]

Answer:

Step-by-step explanation:

Using the expansion

tan(x - y) = $\frac{tanx-tany}{1+tanxtany}$ , then

tan(x - $\frac{\pi }{3}$ )

= $\frac{tanx-tan(-\frac{\pi }{3}) }{1+tanxtan(-\frac{\pi }{3}) }$

note that tan( - $\frac{\pi }{3}$ ) = - tan( $\frac{\pi }{3}$ ) = - $\sqrt{3}$

= $\frac{tanx-(-\sqrt{3}) }{1+tanx(-\sqrt{3}) }$

= $\frac{tanx+\sqrt{3} }{1-\sqrt{3}tanx }$ → B

6 0

3 years ago

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