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kykrilka [37]
3 years ago
9

As cars drive up a ramp at a multi-storey car park they go up 2 meters. the length of the ramp is 10 meters. find the angle betw

een the ramp and the horizontal

Mathematics
1 answer:
daser333 [38]3 years ago
7 0
This is a right angle triangle problem
drawing a vertical line at from the point where the ramp touches the car park leaves a right angle triangle with the
opposite being 2m
hypothenus being 10m
adjacent unknown
we could use sine
SineO equal to opposite over hypothenus
SineO equal to 2/10
SineO equal to 0.2
O equal to Sine^1(0.2)
O equal to 11 .5
The angle between the ramp and the horizontal is 11.5 degrees
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Researchers are monitoring two different radioactive substances. They have 300 grams of substance A which decays at a rate of 0.
Korolek [52]

Answer:

231.59 years

Step-by-step explanation:

To model this situation we are going to use the exponential decay function:

f(t)= a (1-b)^t

where f(t) is the final amount remaining after t years of decay

a is the final amount

b is the decay rate in decimal form

t is the time in years

For Substance A:

Since  we have 300 grams of the substance, a=300. To convert the decay rate to decimal form, we are going to divide the rate by 100%:

r = 0.15/100 = 0.0015. Replacing the values in our function:

f(t) = a (1-b)^t

f(t) = 300 (1-0.0015)^t

f(t) = 300 (0.9985)^t equation (1)

For Substance B:

Since we have 500 grams of the substance, a= 500. To convert the decay rate to decimal form, we are going to divide the rate by 100%:

r=0.37/100= 0.0037. Replacing the values in our function:

f(t) = a (1-b)^t

f(t)= 500 (1-0.0037)^t

f(t)=500(0.9963)^t equation (2)

Since they are trying to determine how many years it will be before the substances have an equal mass M, we can replace f(t) with M in both equations:

M=300(0.9985)^t equation (1)

M=500(0.9963)^t equation (2)

We can conclude that the system of equations that can be used to determine how long it will be before the substances have an equal mass, M, is :

{M=300(0.9985)^t

{M=500(0.9963)^t

7 0
3 years ago
For how many integers $n$ is it true that $\sqrt{n} \le \sqrt{4n - 6} < \sqrt{2n + 5}$?
ale4655 [162]

\bf \sqrt{n}\leqslant \sqrt{4n-6}

\bf \sqrt{n}< \sqrt{2n+5}\implies \stackrel{\textit{squaring both sides}}{n< 2n+5}\implies 0\leqslant 2n - n + 5 \\\\\\ 0 < n+5\implies \boxed{-5 < n} \\\\\\ \stackrel{-5\leqslant n < 2}{\boxed{-5}\rule[0.35em]{10em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}2}


namely, -5, -4, -3, -2, -1, 0, 1.  Excluding "2" because n < 2.

6 0
3 years ago
What is one third plus one seventh
Kryger [21]
To do this, change into common denominators.  The common denominator is 21 so change the problem from 1/3 + 1/7 to 7/21 + 3/21.  This gives you 10/21.  It is already in simplest form.
8 0
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3.
nasty-shy [4]

Answer:

y    

Step-by-step explanation:

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2 years ago
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Those are the answers
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