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3 years ago

WILL MARK BRAINLIEST Solve for x. 3 1 2 140

Mathematics

1 answer:

Elina [12.6K]3 years ago

5 0

Answer:

x=1

Step-by-step explanation:

141x-1=2(70)

141x=140+1

141x=141

x=141/141

x=1

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Solve using law of sines or law of cosines!

malfutka [58]

Answer:

Part 5) The length of the ski lift is $1.15\ miles$

Part 6) The height of the tree is 18.12 m

Step-by-step explanation:

Part 5)

Let

A -----> Beginning of the ski lift

B -----> Top of the mountain

C -----> Base of mountain

we have

$b=0.75\ miles$

$A=20\°$

$C=180\°-50\°=130\°$ ----> by supplementary angles

Find the measure of angle B

Remember that the sum of the interior angles must be equal to 180 degrees

$B=180\°-A-C$

substitute

$B=180\°-20\°-130\°=30\°$

Applying the law of sines

$\frac{b}{sin(B)}=\frac{c}{sin(C)}$

substitute

$\frac{0.75}{sin(30\°)}=\frac{c}{sin(130\°)}$

$c=\frac{0.75}{sin(30\°)}(sin(130\°))$

$c=1.15\ miles$

Par 6)

see the attached figure with letters to better understand the problem

Applying the law of sines in the right triangle BDC

In the right triangle BDC 20 degrees is the complement of 70 degrees

$\frac{BC}{sin(70\°)}=\frac{x}{sin(20\°)}$

$BC=(sin(70\°))\frac{x}{sin(20\°)}$ -----> equation A

Applying the law of sines in the right triangle ABC

In the right triangle ABC 50 degrees is the complement of 40 degrees

$\frac{BC}{sin(40\°)}=\frac{x+15}{sin(50\°)}$

$BC=(sin(40\°))\frac{x+15}{sin(50\°)}$ -----> equation B

Equate equation A and equation B and solve for x

$(sin(70\°))\frac{x}{sin(20\°)}=(sin(40\°))\frac{x+15}{sin(50\°)}\\\\2.7475x=0.8391(x+15)\\\\2.7475x=0.8391x+12.5865\\\\2.7475x-0.8391x=12.5865\\\\x=6.60\ m$

Find the value of BC

$BC=(sin(70\°))\frac{6.6}{sin(20\°)}$

$BC=18.12\ m$

therefore

The height of the tree is 18.12 m

5 0

4 years ago

A 30° 60° 90° triangle is shown below. Find the length of the side labeled y.

bulgar [2K]

Answer:

Part 1) $y=3\ units$

Part 2) The length of the hypotenuse is 6 units

Step-by-step explanation:

Part 1) we know that

In the right triangle of the figure

$sin(60^o)=\frac{y}{2\sqrt{3}}$ ----> by SOH (opposite side divided by the hypotenuse)

solve for y

$y=sin(60^o)2\sqrt{3}$

Remember that

$sin(60^o)=\frac{\sqrt{3}}{2}$

substitute

$y=(\frac{\sqrt{3}}{2})2\sqrt{3}$

$y=3\ units$

Part 2)

Let

h ----> the length of the hypotenuse

we know that

In the right triangle of the figure

$sin(45^o)=\frac{3\sqrt{2}}{h}$ ----> by SOH (opposite side divided by the hypotenuse)

$h=\frac{3\sqrt{2}}{sin(45^o)}$

Remember that

$sin(45^o)=\frac{\sqrt{2}}{2}$

substitute

$h=3\sqrt{2}:\frac{\sqrt{2}}{2}$

$h=6\ units$

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3 years ago

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Lera25 [3.4K]

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i’m sorry if it’s wrong but i’m pretty sure it should be right.

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Can someone help me with the math on my page i will mark them brainliest pls

Lera25 [3.4K]

Answer:

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Step-by-step explanation:

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