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4 years ago

Find the perimeter of the following shape, rounded to the nearest tenth:

Mathematics

1 answer:

OLga [1]4 years ago

6 0

Answer: the perimeter of the shape is 19.1

Step-by-step explanation:

To determine the length of each side of the quadrilateral, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side

For line AD,

AD² = 2² + 4² = 4 + 16 = 20

AD = √20 = 4.47

For line AB,

AB² = 1² + 5² = 1 + 25 = 26

AB = √26 = 5.1

For line BC,

BC² = 2² + 4² = 4 + 16 = 20

BC = √20 = 4.47

For line CD,

CD² = 1² + 5² = 1 + 25 = 26

CD = √26 = 5.1

The perimeter of a plane figure is the distance around the figure. Therefore

Perimeter = AB + AB + BC + CD

Perimeter = 4.47 + 5.1 + 4.47 + 5.1 =

19.1

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If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ? sin Θ = square root 2 ov

densk [106]

Answer:

$\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}$

Step-by-step explanation:

We have:

$\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}$

For sine use:

$\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x$

Substitute:

$\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}$

θ in IV quadrant, therefore sine is negative.

$\sin\theta=-\dfrac{\sqrt2}{2}$

For tangent use:

$\tan x=\dfrac{\sin x}{\cos x}$

Substitute:

$\tan\theta=\dfrac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=-\dfrac{\sqrt2}{2}\cdot\dfrac{2}{\sqrt2}=-1$

8 0

3 years ago

30,098 in standard form <br>30,098 to four significant figures

fredd [130]

Standard form = 3.0098 × 10 power 4
Significant figures = 30100

4 0

2 years ago

Hat is the value of x in the equation 3 x minus 4 y equals 65, when y equals 4?

kobusy [5.1K]

Answer:

16.3 repeating

Step-by-step explanation:

3x-4(4)=65

3x-16=65

3x=65-16

3x=49

x=49/3

x=16.3 repeating

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3 years ago

Pls help!!! i will give brainliest!!!!! :D

Contact [7]

The answer equals 29.

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3 years ago

Read 2 more answers

Larry and Paul start out running at a rate of 5 mph. Paul speeds up his pace after 5 miles to 10 mph but Larry continues the sam

klasskru [66]

<h2>Hello!</h2>

The answer is:

They will be 10 miles apart after 3 hours.

To calculate how long after they start will they be 10 miles apart, we need to assume that after 1 one hour, they were at the same distance (5 miles), then, calculate the time when they are 10 miles apart, knowing that Paul increased its speed two times, running first at 5mph and then, at 10 mph.

The time that will pass to be 10 miles apart can be calculated using the following equation:

$TotalTime=TimeToReach5miles+TimeToBe10milesApart$

Calculating the time to reach 5 miles for both Larry and Paul, at a speed of 5 mph, we have:

$x=xo+v*t\\\\5miles=0+5mph*t\\\\t=\frac{5miles}{5mph}=1hour$

We have that to reach a distance of 5 miles, they needed 1 hour. We need to remember that at this time, they were at the same distance.

If we want to know how many time will it take for them to be 10 miles apart with Paul increasing its speed to 10mph, we need to assume that after that time, the distance reached by Paul will be the distance reached by Larry plus 10 miles.

So, for the second moment (Paul increasing his speed) we have:

For Larry:

$x_{L}=5miles+5mph*t$

Therefore, the distance of Paul will be equal to the distance of Larry plus 10 miles.

For Paul:

$x{L}+10miles=xo+10mph*t\\\\5miles+5mph*t+10miles=5miles+10mph*t\\\\5miles+10miles-5miles=10mph*t-5mph*t\\\\10miles=5mph*t\\\\t=\frac{10miles}{5mph}=2hours$

Then, there will take 2 hours to Paul to be 10 miles apart from Larry after both were at 5 miles and Paul increased his speed to 10 mph.

Hence, calculating the total time, we have:

$TotalTime=TimeToReach5miles+TimeToBe10milesApart$

$TotalTime=1hour+2hours=3hours$

Have a nice day!

3 0

4 years ago