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ipn [44]
3 years ago
13

Determine the value of 0

Mathematics
1 answer:
MrMuchimi3 years ago
7 0

Answer:

θ = 76.3°

Step-by-step explanation:

using the <u>SOHCAHTOA </u>method,

Here hypotenuse is 38, adjacent is 9

Using cosine formula:

cos(x ) = \frac{adjacent}{hypotenuse}

cos(x) = \frac{9}{38}

x = cos^{-1} (\frac{9}{38} )

x = 76.3°

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mixas84 [53]

Answer:V

Step-by-step explanation:

im going to guess c = degrees and its always the angle with a higher degree

6 0
3 years ago
Help please thank you
Naddika [18.5K]

Positive that its C.7

5 0
4 years ago
Read 2 more answers
PLEASE HELP                                                                                                                    
sweet [91]
To model and solve our situation we are going to use the equation: s= \frac{d}{t}
where
s is speed
d is distance 
t is time 

1. We know that the distance between the cities is 2400 miles, so d=2400. We also know that the speed of the plane is 450 mi/h. Since we don't know the speed of the air, S_{a}=?. We don't know how much the westward trip takes, so t_{w}=?, and we also don't know how much the eastward trip takes, so t_{e}=?.

Going westward. Here the plane is flying against the air, so we need to subtract the speed of the air from the speed of the plane:
450-S_{a}= \frac{2400}{t_{w} }
Going eastward. Here the plane is flying with the the air, so we need to add the speed of the air to the speed of the plane:
450+S_{a}= \frac{2400}{t_{e} }

2. We know for our problem that the round trip takes 11 hours; so the total time of the trip is 11, t_{t}=11. Notice that we also know that the total time of the trip equals time of the tip going westward plus time of the trip going eastward, so t_{t}=t_{w}+t_{e}. Since we know that the total trip takes 11 hours, we can replace that value in our total time equation and solve for t_{w}:
11=t_{w}+t_{e}
t_{w}=11-t_{e}

Now we can replace t_{w} in our going westward equation to model our round trip with a system of equations:
450-S_{a}= \frac{2400}{t_{w}}
450-S_{a}= \frac{2400}{11-t_{e} } equation (1)
450+S_{a}= \frac{2400}{t_{e}} equation (2)

3. To solve our system of equations, we are going to solve for t_{e} in equations (1) (2):

From equation (1)
450-S_{a}= \frac{2400}{11-t_{e} }
11-t_{e}= \frac{2400}{450-S_{a} }
-t_{e}= \frac{2400}{450-S_{a} } -11
t_{e}=11- \frac{2400}{450-S_{a} }
t_{e}= \frac{4950-11S_{a} -2400}{450-S_{a} }
t_{e}= \frac{2550-11S_{a} }{450-S_{a} } equation (3)

From equation (2):
450+S_{a}= \frac{2400}{t_{e} }
t_{e}= \frac{2400}{450+S_{a} } equation (4)

Replacing (4) in (3)
\frac{2400}{450+S_{a}} = \frac{2550-11S_{a}}{450-S_{a} }
Now, we can solve for S_{a} to find the speed of the wind:
2400(450-S_{a})=(450+S_{a})(2550-11S_{a})
1080000-2400S_{a}=1147500-4950S_{a}+2550S_{a}-11(S_{a})^{2}
11(S_{a})^{2}-67500=0
11(S_{a})^{2}=67500
(S_{a})^{2}= \frac{67500}{11}
S_{a}=+/-  \sqrt{ \frac{67500}{11} }
Since speed cannot be negative, the solution of our equation is:
S_{a}= \sqrt{ \frac{67500}{11} }
S_{a}=78.33

We can conclude that the speed of the wind is 78 mph.

3 0
4 years ago
What is equivalent to -1/3t=4/3​
stepan [7]

<u>Answer:</u>

The equivalent of -1/3t = 4/3​ is t = -4

<u>Solution:</u>

Need to determine such value of t so that when it is multiplied by -1/3, we should get 4/3

That is we need to solve following expression for value of t.

-\frac{1}{3} t=\frac{4}{3}

On multiplying both the sides by 3 we get

\begin{array}{l}{-\frac{1}{3} t \times 3=\frac{4}{3} \times 3} \\\\ {=>-t=4}\end{array}

Now multiplying both the sides by -1 we get

\begin{array}{l}{-t \times-1=4 \times-1} \\ {=>t=-4}\end{array}

Let’s recheck the value of t

-\frac{1}{3} t=-\frac{1}{3} \times-4=\frac{4}{3}

Hence for t = -4, expression -1/3 t is equal to 4/3.

6 0
3 years ago
TRUE FALSE The supplement of 789 is 12​
Burka [1]
The answer is False :)
7 0
3 years ago
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