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iragen [17]
3 years ago
9

The formula s=12at ^2 is used to calculate the distance s travelled by a bike when a=3 and t=10, each correct to the nearest int

eger, calculate the lower bound of the distance s
Mathematics
1 answer:
seraphim [82]3 years ago
8 0

Answer:

Step-by-step explanation:

Given the formula for calculating the distance travelled expressed as;

s=1/2at^2

Given

a = 3

t = 10

Required

lower bound of s

Substitute the given values into the equation;

s=1/2at^2

S = 1/2(3)(10)^2

S = 1/2 * 3 * 100

S = 3 * 50

S = 150

Hence the lower bound of distance S is 150

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What is the discriminante of 9x^2+10x
Leni [432]
The discriminant of y=ax^2+bx+c is b^2-4ac

given
9x^2+10x+0
a=9
b=10
c=0

discriminant=10^2-4(9)(0)=100-0=100

the discriminant is 100
6 0
4 years ago
in 2011, an Action Comics no. 1, featuring first appearance of superman, was sold at auction for $2,161,000. The comic book was
Salsk061 [2.6K]

Price of book in 1938 = $ 0.10

Price of book in 2011 = $ 2,161,000

Number of years over 1938-2011 = 2011 - 1938 = 73

Annual increase in the price = ((\frac{\text{Price of book in 2011}}{\text{Price of book in 1938}}) ^ \frac{1}{\text{Number of years}} - 1 ) * 100

⇒ Annual increase in the price = ((\frac{2,161,000}{0.1}) ^ \frac{1}{73} - 1 ) * 100

⇒ Annual increase in the price = (1.260302283  - 1 ) × 100

⇒ Annual increase in the price = 0.260302283 × 100

⇒ Annual increase in the price = 26.0302283 %

⇒ Annual increase in the price = 26%

Hence, the comic book price increased by 26% annually

3 0
3 years ago
A fair-sided coin is tossed 18 times.
Lady bird [3.3K]
Most likely to be letter C
5 0
3 years ago
4. Peter participated in a 5 K race and finished the race in 17 min. A 5k race is
juin [17]

Answer: are you sure these are correct?

Step-by-step explanation:

You would calculate the speed be doing 5000 m / 17 min, which would be 294.117647059 m/min

17 minutes is extremely fast, but this is correct.

4 0
3 years ago
In a murder investigation, the temperature of a corpse was 32.5 degrees C at 1:30 pm and 30.3 degrees C an hour later. Normal bo
svp [43]
Let T(t) be the temperature of the body t hours after 1:30 PM. Then T(0) = 32.5 and T(1) = 30.3.

Using Newton's Law of Cooling, \dfrac{dT}{dt} = k(T - T_s), we have \dfrac{dT}{dt} = k(T-20). Now let y = T - 20, so
y(0) = T(0) - 20 = 32.5 - 20 = 12.5, so y is a solution to the initial value problem dy/dt = ky with y(0) = 12.5.
By separating and integrating, we have y(t) = y(0)e^{kt} = 12.5e^{kt}.

y(1) = 30.3 - 20\ \Rightarrow\  12.5e^{k(1)} = e^{k} = \frac{10.3}{12.5}\ \Rightarrow \\
k = \ln \frac{10.3}{12.5}


y(t) = 37 - 20\ \Rightarrow\ 12.5e^{kt} = 17\ \Rightarrow\ e^{kt} = \frac{10.3}{12.5} \ \Rightarrow \\ \\
kt = \ln \frac{10.3}{12.5}\ \Rightarrow\ t = \left( \ln \frac{17}{12.5}  \right)/ \frac{10.3}{12.5}  \approx -1.588 \mathrm{\ h}

≈ 95 minutes. Thus the murder took place about 95 minutes before 1:30 PM, or 11:55 AM.
6 0
3 years ago
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