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I am Lyosha [343]
2 years ago
10

Find the midpoint of A[5,3) and B(3,-5).

Mathematics
2 answers:
nirvana33 [79]2 years ago
5 0

Answer:

(4,-1)

Step-by-step explanation:

Pani-rosa [81]2 years ago
4 0
(4,-1) is the answer for ur question
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SOMEONE pLZ HELP ME SOLVE THIS IM GETTING so MADDD
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Step-by-step explanation:

I believe this is correct

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Which equation would you use to find how high the bird is flying?
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Answer:

tan45 = x over 1100

Step-by-step explanation:

Because tan = opposite/ adjacent

x is the opposite of the 45 angle in the figure shown

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3 years ago
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Find the Surface Area of the Pyramid. Round to the NEAREST TENTH.
lana66690 [7]

Answer:

Step-by-step explanation:

The side triangles are: 352.35

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2 years ago
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The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am
posledela

Step-by-step walkthrough:

a.

Well a standard half-life equation looks like this.

N = N_0 * (\frac{1}{2})^{t/p

N_0 is the starting amount of parent element.

N is the end amount of parent element

t is the time elapsed

p is a half-life decay period

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Therefore you can create a function based off of the original equation, just sub in the values you already know.

N(t) = 74g * (\frac{1}{2})^{t/2.8days

b.

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2}  =  37g

c.

Now we just gotta do some algebra. Use the original function but this time, replace N(t) with 10g and solve algebraically.

10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}

Take the log of both sides.

log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})

Use the exponent rule for log laws that, log(b^x) = x*log(b)

log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})

\frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = \frac{t}{2.8days}

2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = t

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