1 yards equals 3 feet in one mile equals 5280 feet how many yards are in 5 MI

Metal strips 11 feet long and 2 feet wide . of the stripping costs $6 per foot , find the total costs

adoni [48]

The area of the metal strips:
A = L * W , where L stays for length and W stays for wirth.
A = 11 ft * 2 ft ;
A = 22 ft^2.
If it costs $6 per foot:
Total cost = 22 * 6 = $132.
Answer: The total cost of a metal strip is $132.

5 0

3 years ago

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Need help ASAP!!

Ilia_Sergeevich [38]

An object that is in motion as a projectile follows a path or trajectory of a parabola

The function and values are;

a) The equation of the quadratic function is; $\underline{y = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}$

b) The maximum height of the ball is approximately 7.334 m

c) Horizontal distance at maximum height 18.8 meters

Reason:

a) Known parameters are;

Let f(x) = a·x² + b·x + c represent the equation of the parabola modelling the path of the ball, we have;

Points on the path of the parabola = (0, 0), (35, 1.5), 37, 0)

Plugging the values gives;

0 = a·0² + b·0 + c

Therefore, c = 0

1.5 = 35²·a + 35·b

0 = 37²·a + 37·b

Solving gives;

a = -3/140, b = 111/140

The equation of the quadratic function is therefore;

$\underline{y = f(x) = \dfrac{111}{140} \cdot x - \dfrac{3}{140} \cdot x^2}$

b) The maximum height is given by the vertex of the parabola

The x-coordinate at the vertex is the point $-\dfrac{b}{2 \cdot a}$

Which gives;

$x-coordinate = \dfrac{\frac{111}{140} }{2 \times \dfrac{3}{140} } = 17.5$

The maximum height is therefore;

$f(x)_{max} = \dfrac{111}{140} \times 17.5 - \dfrac{3}{140} \cdot 17.5^2 \approx 7.334$

The maximum height of the ball is approximately 7.334 m

c) The distance the ball has travelled to horizontally is given by half of the range, R as follows;

The range of the motion, R = 37 meters

$Horizontal \ distance \ to \ maximum \ height = \dfrac{R}{2}$

Therefore;

$Horizontal \ distance \ to \ maximum \ height = \dfrac{37}{2} = 18.5$

The distance the ball has travelled horizontally to reach the maximum height horizontally 18.5 meters

Learn more about the trajectory of a projectile here:

brainly.com/question/13646224

8 0

3 years ago

The mean score of a competency test is 82, with a standard deviation of 2. Between what two values do about 99.7% of the values

goldenfox [79]

99.7% encompasses about 3 standard deviations either side of the mean.

82 ±3*2 = (76, 88)

About 99.7% of the values lie between 76 and 88.

5 0

3 years ago

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Given that a line has a slope

zheka24 [161]

Answer: the last option (:

Step-by-step explanation:

You always put the slope in front of the x intercept in the equation.

5 0

4 years ago

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A cell of some bacteria divides into two cells every 10 minutes.The initial population is 3 bacteria. (a) Find the size of the p

mars1129 [50]

Answer:

(a) $P_{t}=3(2)^{6t}$

(b) $3(2)^{42}$

(c) 28.07 minutes

Step-by-step explanation:

A cell of some bacteria divides itself into 2 cells in every 10 minutes and initial population of the bacteria was 3.

That means sequence formed will be 3, 6, 12, 24............

We can easily say that this sequence is a geometric sequence having common ratio (r) = $\frac{T_{2}}{T_{1}}=\frac{6}{3}$

r = 2

Now we know the explicit formula of a geometric sequence is given by

$P_{t}=P_{0}(r)^{\frac{60t}{10}}=P_{0}(r)^{6t}$

Where a = Initial population = 3 bacteria

r = common ratio = 2

and t = time in hours

So explicit formula will be $P_{t}=3(2)^{6t}$

(a) Now we have to calculate the size of population after t hours

$P_{t}=3(2)^{6t}$

(b) We have to find the size of population after 7 hours or 420 minutes

$P_{t}=3(2)^{6\times7}$

= $3(2)^{42}$

After 7 hours bacteria population will be $3(2)^{42}$

(c) Time to reach population as 21

By the explicit formula

$21=3(2)^{6t}$

$2^{6t}=\frac{21}{3}=7$

Now we take log on both the sides of the equation

$log(2^{6t})=log(7)$

6t log2 = log 7

6t(0.301) = 0.845

t(1.806) = 0.845

t = $\frac{0.845}{1.806}=0.468$ hours

Or t = 0.468×60 = 28.07 minutes

Therefore, after 28.07 minutes bacterial population will be 21

5 0

3 years ago