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mamaluj [8]
2 years ago
10

From the top of a lighthouse the angle of depression of a speed

Mathematics
1 answer:
otez555 [7]2 years ago
5 0

Answer:

164m (to 3 s.f.)

Step-by-step explanation:

Please see the attached picture for the full solution.

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Emily took 300 US dollars to the bank to exchange it for Canadian dollars. The exchange rate on that day was 1.25 Canadian dolla
Paha777 [63]

Emily wanted to exchange her 300 US Dollars for Canadian Dollars.

The exchange rate for it was 1.25 Canadian Dollars for 1 US Dollar.

This means that 1 US Dollar = 1.25 Canadian Dollars.

So if we were to follow this pattern,

2 US Dollars = 2 × 1.25 = 2.5 Canadian Dollars

3 US Dollars = 3 × 1.25 = 3.75 Canadian Dollars

4 US Dollars = 4 × 1.25 = 5 Canadian Dollars

So for 300 US Dollars, we'll need to multiply 300 by 1.25.

300 US Dollars = 300 × 1.25 = 375 Canadian Dollars.

Hope it helps. :)

4 0
2 years ago
What is 363 rounded to the nearest ten
klio [65]
I think the answer is 360
6 0
3 years ago
Read 2 more answers
Plzzzzz ill make u brainliest like frfr
Luden [163]

Answer:

3,520 yds in 2 hours

Step-by-step explanation:

5,280:1 ; 1 mile an hour is 5,280 feet in an hour

10,560:2 ; 2 miles in 2 hours is 5,280+5,280 (10,560) feet in 2 hours

3 ft = 1 yd ; every 3 feet is a yard, so divide 10,560 by 3

10,560/3 = 3520 yds ; Jorge can walk 3,520 yds in 2 hours

6 0
3 years ago
An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each oth
dedylja [7]

Answer:

(a) D(t) = 250t -500 miles

(b) Controller has 2 hours, but including time for pilots to divert course or altitude.

Step-by-step explanation:

Given:

two planes at same altitude heading in a collision course.

Plane A at 400 miles from collision point at 200 mph

Plane B at 300 miles from collision point at 150 mph.

Theoretical collision happens in

t = 400/200 = 300/150 = 2 hours

Distance ya of plane A from collision point as a function of time in hours

ya(t) = 400 -200t

Distance yb of plane B from collision point as a function of time in hours

yb(t) = 300-150t

(a) Distance between two planes,

Since the two planes are on courses perpendicular to each other, will need using pythagorean theorem

D(t) = sqrt(ya(t)^2+yb(t)^2)

= sqrt((400-200t)^2+(300-150t)^2)

= 250(t-2)

D(t) = 250t -500 miles

b. time available

Time until D(t) = 0

solve D(t) = 0

D(t) = 0

250(t-2) = 0

t = 2  (two hours)

3 0
3 years ago
What are the solution(s) for the quadratic equation?
MariettaO [177]

Let's see

\\ \rm\rightarrowtail x^2+12x+36=0

\\ \rm\rightarrowtail x^2+6x+6x+36=0

\\ \rm\rightarrowtail x(x+6)+6(x+6)=0

\\ \rm\rightarrowtail (x+6)(x+6)=0

\\ \rm\rightarrowtail x=-6

3 0
1 year ago
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