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frozen [14]
3 years ago
8

Please Help ASAP (show work if you want its not needed i just need the answer)

Mathematics
1 answer:
cricket20 [7]3 years ago
4 0

\cos(60)  =  \frac{12}{x}  \\

\frac{1}{2}  =  \frac{12}{x}  \\

x = 24

_____________________________________________

\sin(60)  =  \frac{y}{x}  \\

\frac{ \sqrt{3} }{2}  =  \frac{y}{24}  \\

y = 12 \sqrt{3}

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I really need help with this right now :(
erma4kov [3.2K]
Lol sry nobody came to help you earlier...
honestly u shoould just put all the values in a calculator.
The second one, fourth one, and last one
4 0
3 years ago
Whats the area of the lawn in this shape
Licemer1 [7]

Answer:

208 ft

Step-by-step explanation:

in order to do this we need to find the area of the retangle:

(l x w) 14x20= 280ft

then the area of the rectangle:

a^2 + b^2 = c^2 ( hypotenus lenght )

c= 8.5

now we can plug in to find the area of the triangle, which is 12 (11.99)

then the area of the trapezium= 60

finally we can find the answer

280 - (60+12)= 208 ft :)

7 0
3 years ago
Which statement is true ?
8090 [49]

Answer:

f(x)=4x^2

f(2)=4(2)^2

f(2)=4(4)

f(2)=16

2. g(x)=4*2^x

g(2)=4*2^2

g(2)=4*4

g(2)=16

f(2)=g(2)

Both is equal to 16

I hope this help you :)

Step-by-step explanation:

8 0
3 years ago
Which number has the same value as 7 X 1/1000
Yuki888 [10]
Thw answer would be 0.007
5 0
3 years ago
Tutorial Exercise A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due
ANTONII [103]

Answer:

Both the boats will closet together at 2:21:36 pm.

Step-by-step explanation:

Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).

Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),

        Formula :   d=v*t

at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)

the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)

      Formula : D=\sqrt{(x2-x1)^2+(y2-y1)^2}

⇒                     D = \sqrt{20^2t^2+15(t-1)^2}

Now let           F(t) = D^2(t)

                ∵    F'(t) = 800t + 450(t-1) =  1250t -450\\F'(t) =0

⇒                     t= 450/1250

⇒                     t= .36 hours

⇒                       = 21 min 36 sec

Since F"(t)=0,

∴ This time gives us a minimum.

Thus, The two boats will closet together at 2:21:36 pm.

3 0
3 years ago
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