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2 years ago

If your bike 8 miles south and then 6 miles east, what is the distance you are from your starting point?

Mathematics

1 answer:

Lady bird [3.3K]2 years ago

8 0

Answer:

14 miles

Step-by-step explanation:

It's still 14 miles no matter which way you go.

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A 51-foot wire running from the top of a tent pole to the ground makes an angle of 58° with the ground. If the length of the ten

AfilCa [17]

Answer:

35.11 ft

Step-by-step explanation:

This given situation can be thought of as triangle $\triangle PQR$ where PQ is the length of pole.

PR is the length of rope.

and QR is the distance of bottom of pole to the point of fastening of rope to the ground.

And $\angle Q \neq 90^\circ$

Given that:

PQ = 44 ft

PR = 51 ft

$\angle R = 58^\circ$

To find:

Side QR = ?

Solution:

We can apply Sine Rule here to find the unknown side.

Sine Rule:

$\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC}$

Where

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

$\dfrac{PR}{sinQ}=\dfrac{PQ}{sinR}\\\Rightarrow sin Q =\dfrac{PR}{PQ}\times sinR\\\Rightarrow sin Q =\dfrac{51}{44}\times sin58^\circ\\\Rightarrow \angle Q =79.41^\circ$

Now,

$\angle P +\angle Q +\angle R =180^\circ\\\Rightarrow \angle P +58^\circ+79.41^\circ=180^\circ\\\Rightarrow \angle P = 42.59^\circ$

Let us use the Sine rule again:

$\dfrac{QR}{sinP}=\dfrac{PQ}{sinR}\\\Rightarrow QR =\dfrac{sinP}{sinR}\times PQ\\\Rightarrow QR =\dfrac{sin42.59}{sin58}\times 44\\\Rightarrow QR = 35.11\ ft$

So, the answer is 35.11 ft.

3 0

3 years ago

Bee + ant= 10, ant - bee = 8, ant × bee= ?

Alona [7]

The Correct Answer Is...

9.

Any Questions? Comment Below!

-AnonymousGiantsFan

7 0

2 years ago

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$