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Alex17521 [72]
3 years ago
8

In AFGH, h = 9.6 inches, f = 2.5 inches and 2G=78°. Find the length of g, to the

Mathematics
1 answer:
erastovalidia [21]3 years ago
8 0

Answer:

20.9 ft

This is a right triangle trigonometry question because N is 90 degrees. MN is adjacent to M and LM is the hypotenuse. Adjacent any hypotenuse use the cosine function. 

plug in known values

switch cos(20) and x using the products property

plug into calculator to get 20.9 ft

Step-by-step explanation:

I tried my hardest sorry if this is wrong

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Range: all y values between the ends of a line on the y-axis

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Two equation Ax+By=1 and (3,-1)(-2,-2). Find A and B​
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1)

(-2+\sqrt{-5})^2\implies (-2+\sqrt{-1\cdot 5})^2\implies (-2+\sqrt{-1}\sqrt{5})^2\implies (-2+i\sqrt{5})^2 \\\\\\ (-2+i\sqrt{5})(-2+i\sqrt{5})\implies +4-2i\sqrt{5}-2i\sqrt{5}+(i\sqrt{5})^2 \\\\\\ 4-4i\sqrt{5}+[i^2(\sqrt{5})^2]\implies 4-4i\sqrt{5}+[-1\cdot 5] \\\\\\ 4-4i\sqrt{5}-5\implies -1-4i\sqrt{5}

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let's recall that the conjugate of any pair a + b is simply the same pair with a different sign, namely a - b and the reverse is also true, let's also recall that i² = -1.

\cfrac{6-7i}{1-2i}\implies \stackrel{\textit{multiplying both sides by the denominator's conjugate}}{\cfrac{6-7i}{1-2i}\cdot \cfrac{1+2i}{1+2i}\implies \cfrac{(6-7i)(1+2i)}{\underset{\textit{difference of squares}}{(1-2i)(1+2i)}}} \\\\\\ \cfrac{(6-7i)(1+2i)}{1^2-(2i)^2}\implies \cfrac{6-12i-7i-14i^2}{1-(2^2i^2)}\implies \cfrac{6-19i-14(-1)}{1-[4(-1)]} \\\\\\ \cfrac{6-19i+14}{1-(-4)}\implies \cfrac{20-19i}{1+4}\implies \cfrac{20-19i}{5}\implies \cfrac{20}{5}-\cfrac{19i}{5}\implies 4-\cfrac{19i}{5}

7 0
2 years ago
Read 2 more answers
A 46 gram sample of a substance that is used to sterilize surgical instruments has a k-value of 0.1374. Find the substance's hal
mars1129 [50]

Answer:

  t=5.0days

Step-by-step explanation:

Using the formula for the exponential decay that is N=N_{0}e^{-kt}, we have N=\frac{1}{2}{\times}46=23, N_{0}=46 and k=0.1374.

Thus, N=N_{0}e^{-kt} becomes

23=46{\times}e^{-0.1374t}

\frac{23}{46}=e^{-0.1374t}

\frac{1}{2}=e^{-0.1374t}

Taking log on both sides, we get

ln(\frac{1}{2})={-0.1374t}

t=\frac{ln\frac{1}{2}}{-0.1}

t=\frac{-0.6931}{-0.1374}

t=5.0days

4 0
2 years ago
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