For this question, you need to know that the average velocity of a 100m Olympic athlete is 10 m/s before you start solving the equation. The equation is K (energy in Jules) = 1/2 mv^2 (where m is the mass in kg and v the speed in meters per second).
The correct answer is 4000J. To see how I got it, take a look at the attachment.
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The correct answer is that their estimate is longer than the time it will actually take.
Explanation:
Since Shandra can mow the entire lawn in 45 minutes, she can mow 1/45 of it in 1 minute.
Kelly mows the entire lawn in 1 hour 15 minutes, which is 75 minutes; this means she mows 1/75 of it in 1 minute.
Using these rates, and letting x be the amount of time it takes, we have the equation
1/45x + 1/75x = 1 (1 represents the entire lawn).
Rather than try to find a common denominator, we can “cancel” the denominators by multiplying everything by them. First multiply everything by 45:
45(1/45x)+45(1/75x)=1*45
1x+45/75x=45.
Now multiply everything by 75:
75(1x)+75(45/75x)=45*75
75x+45x=3375.
Combine like terms:
120x=3375.
Divide both sides by 120:
120x/120=3375/120
x=28.125.
It will take them a little over 28 minutes together, which means their estimate was longer than the actual time.
The first step is to solve the horizontal component using the given values from the problem and the illustration,
horizontal component = 6.4sin(40°) + 5.1sin(35°) – 4.8sin(23°)Then the vertical component is determined,
vertical component = 6.4cos(40°) – 5.1cos(35°) – 4.8cos(23°)
horizontal component = 6.4sin(40°) + 5.1sin(35°) – 4.8sin(23°)Then the vertical component is determined,
vertical component = 6.4cos(40°) – 5.1cos(35°) – 4.8cos(23°)
The distance of the fourth leg is calculated using the Pythagorean Theorem,
distance = (h²+v²)^1/2 = 6.35 km
Finally, the angle is computed by applying the inverse tangent function,
theta = tan^-1(v/-h) = 54.42°
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This is just simple. For example you have a plane of the form x=a, then you just substitute x with a, and you’ll get an equation with y and z only, hence you have a 2-d trace of the intersection. It is just similar for y=b and z=c.
(1) At z=1.5, 2x^2 + 5y^2 + 1.5^2 = 4
2x^2 + 5y^2 = 1.75
Now you have an ellipse in the z=1.5 plane as your trace.
(2) At x=1, 2(1)^2 + 5y^2 + z^2 = 4
5y^2 + z^2 = 2
Now you have an ellipse in the x=1 plane as your trace.
(3) At z=0, 2x^2 + 5y^2 + (0)^2 = 4
2x^2 + 5y^2 = 4
Now you have an ellipse in the z=0 plane as your trace.
(4) At y=0, 2x^2 + 5(0)^2 + z^2 = 4
2x^2 + z^2 = 4
Now you have an ellipse in the y=0 plane as your trace.
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Answer:
Option 4 – She will have $16,126.80
Step-by-step explanation:
Given : Gracie Shay wants to buy a new Hummer in 5 years. Gracie estimates the cost of the Hummer will be $28,000. If she invests $12,000 now, at a rate of 6 percent compounded semiannually.
To find : Which option she will have ?
Solution :
Using compound interest formula,
Where, A is the amount
P is the principle P=12,000
Compounded semiannually,
r is the rate 6%=0.06
t is the time 5 years
Substitute the value,
The amount she will have $16126.99.
Therefore, Option 4 is correct.
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Use linear approximation to estimate 15^.25 …?
A linear approximation of f(x) in a neighborhood of a
f(x) ~ f(a) + f'(a) (x-a)
f(x) = x^1/4
2^4 = 16 i.e. a = 16
f(15) ~ f(16) + f'(16) (15-16)
f'(x) = 1/4 x^(-3/4)
f(15) ~ 2 + 1/4 (16)^-3/4 (-1)
~ 2 – 1/(32) = 1.96875
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
In living matter, the proportion of carbon 14, an isotope of normal carbon 12, is constant. When the
matter dies, the carbon 14 atoms decay into nitrogen 14 with an emission of radiation such that there
will only be half as many carbon 14 atoms every 5570 years. Thus the half-life of carbon 14 is 5570
years. By measuring the emission of radiation, scientists can infer the population of carbon 14 atoms in
similar living matter and estimate when it died.
(a) In charcoal found in an ancient fire pit on Java, the amount of carbon 14 was found to be 1/16 of
that of a living sample of the same size. Estimate the age of the charcoal.
(b) A refuse deposit near the Strait of Magellan was excavated in the 1930’s and dated using carbon
dating methods. Emissions from carbon 14 disintegrations were found to be occurring 60% of that
of a contemporary living sample. Estimate the age of the sample.
(c) If ten percent of a radioactive substance decays in 10 years, what is its half-life?
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