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3 years ago

15

Who can follow me on RBX MY USER IS....XXXLAMBOKID \ PLS IM TRYING TO GET LOT OF FOLLOWS

2 answers:

LuckyWell [14K]3 years ago

8 0

Answer:

mkay

Step-by-step explanation:

Kaylis [27]3 years ago

5 0

Answer:

I will!!!!!!

Step-by-step explanation:

ok

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1. Write the nth term of the following sequence in terms of the first term of the sequence. 10, 20, 40, . . . 2.Write the nth te

ozzi

Hello,

1)
$a_{0}=5$
$a_{1}=2*5$
$a_{2}=2^2*5$
$a_{3}=2^3*5$
...
$a_{n}=2^{n-}*5$

2)

$u_{1}=a$
$u_{2}=a+4$
$u_{3}=a+2*4$
$u_{4}=a+3*4$
...
$u_{n}=a+(n-1)*4$

8 0

4 years ago

Is the relation a function if the relation is not a function state why not

aivan3 [116]

Answer:

can you please provide more information

5 0

3 years ago

Find the angle measures for this isosceles triangle.

nasty-shy [4]

Answer:

Step-by-step explanation:

3p + 6p + 6p = 180 deg

15p = 180

p = 12

thus,

36 deg, 72 deg, 72 deg (ANS)

6 0

4 years ago

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

<u><em>Applying the law of sines in the right triangle BDC</em></u>

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

<u><em>Applying the law of sines in the right triangle ABC</em></u>

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

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

<u><em>Find the value of BC</em></u>

$BC=(sin(70\°))\frac{6.6}{sin(20\°)}$

$BC=18.12\ m$

therefore

The height of the tree is 18.12 m

5 0

4 years ago

Plz help real quick

Maru [420]

Answer:

what are I am help you are not explaining so explain me

8 0

2 years ago