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gtnhenbr [62]
2 years ago
14

Find the missing angle in a triangle with the given sides.

Mathematics
2 answers:
melamori03 [73]2 years ago
6 0
The angle would be 1°. can i please get brainliest?
d1i1m1o1n [39]2 years ago
5 0

Answer:

1

Step-by-step explanation:

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starting at home, Omar traveled uphill to the gift store for 30 minutes at just 10mph. He then travelled back home along the sam
SashulF [63]

Answer:

average speed for the entire trip is 15 mph

Step-by-step explanation:

Let x be the uphill speed and y be the downhill speed.

We have been given that

Speed of Omar in uphill to reach gift store = x =10 mph

Speed of Omar in downhill to reach his home = y = 30 mph

We know the formula for average speed for the entire trip

\text{Average speed }=\frac{2xy}{x+y}\\\\=\frac{2\times 10\times 30}{10+30}\\\\=\frac{600}{40}\\\\=15

Therefore, average speed is 15 mph


6 0
3 years ago
Read 2 more answers
Use the quadratic formula to solve the equation.
allochka39001 [22]
=1.3+0.1sqrt29, 1.3-0.1sqrt29
7 0
2 years ago
I need help? Does anyone know how to do this?! Will receive 40 POINTS!
FinnZ [79.3K]
I am not for sure that what im about to type is the right answer but if im not wrong the amswer is 1over4
4 0
3 years ago
A $3000 loan has an annual interest rate of 6.6% on the amount borrowed. How much time has elapsed if the interest is now $1386?
san4es73 [151]

Answer:

7

Step-by-step explanation:

time=<u>interest</u>

           P*R

time=<u>1386*</u>100

   6.6 * 3000

<u>time=7</u>

7 0
2 years ago
Solve the equation 2cosA = 3tanA
zavuch27 [327]
\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)&#10;\\\\\\&#10;tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\&#10;-----------------------------\\\\&#10;&#10;2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)}&#10;\\\\\\&#10;2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A)&#10;\\\\\\&#10;2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2

\bf \\\\\\&#10;0=[2sin(A)-1][sin(A)+2]\implies &#10;\begin{cases}&#10;0=2sin(A)-1\\&#10;1=2sin(A)\\&#10;\frac{1}{2}=sin(A)\\\\&#10;sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\&#10;\frac{\pi }{6},\frac{5\pi }{6}\\&#10;----------\\&#10;0=sin(A)+2\\&#10;-2=sin(A)&#10;\end{cases}

now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1

now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine

so, only the first case are the valid angles for A
8 0
2 years ago
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