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4 years ago

Find the tangent of

Mathematics

1 answer:

Kamila [148]4 years ago

3 0

A² + b² = c²
RQ² + 60² = 68²
RQ² = 68² - 60²
RQ² = 1024
RQ = √1024
RQ = 32

Tan(R) = opp/adj
Tan(R) = 60/32
Tan(R) = 15/8

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An architect is sketching a blueprint of a patio for a new fence. On the blueprint, C is the midpoint of segment AD. Point B is

Slav-nsk [51]

The length of AD is 32 feet

Step-by-step explanation:

An architect is sketching a blueprint of a patio for a new fence

On the blueprint, C is the midpoint of segment AD
Point B is the midpoint of segment AC
BC = 8 feet

We need to find the length of AD

A mid-point divides a line segments into two equal parts in length

∵ C is the mid-point of AD

- The mid-point C divides the line segments AD into two equal parts

∴ AC = CD

∵ B is the mid point of AC

∴ AB = BC

∵ BC = 8 feet

∴ AB = 8 feet

- AC contains AB and BC

∵ AC = AB + BC

∴ AC = 8 + 8 = 16

∵ AC = CD

∵ AC = 16

∴ CD = 16

- AD contains AC and CD

∵ AD = AC + CD

∴ AD = 16 + 16 = 32

The length of AD is 32 feet

Learn more:

You can learn more about the mid-point in brainly.com/question/3269852

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4 years ago

Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10

Leona [35]

Answer:

( $\frac{2}{3} , \frac{1}{5}$ )

Step-by-step explanation:

The fastest way to do this is to convert both equations into slope-intercept form and graph it to find the solution point. If you wanted to do this algebraically, you might want to start out by getting rid of the fractions and using either substitution or elimination to find x and y.

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3 years ago

PLEASE HELP!!!!!!!<br> (do not answer if you do not know, so the points go to people who do know.)

zheka24 [161]

Answer:

1a = 8 cu. in.

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1c = 15 cu. ft.

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2c = 90 cu. ft.

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Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re

fiasKO [112]

Answer: $R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}$

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts $38^{\circ}$ angle to each other

The resultant of the two forces is given by

$\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}$

Insert the values

$\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb$

Resultant makes an angle of

$\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}$

So, the resultant makes an angle of $22.57^{\circ}$ with 9 lb force

Angle made with 13 lb force is $38^{\circ}-22.57^{\circ}=15.43^{\circ}$

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3 years ago

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Black_prince [1.1K]

Rewritten: y=15x + 50
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y intercept: (0,50)

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3 years ago