Simplify 6 - 4[x + 1 - (y + 3)] + 5y.

A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 166 feet and a maximum height of 40 feet. Find

scoray [572]

Answer:

38.27775 feet

Step-by-step explanation:

The bridge has been shown in the figure.

Let the highest point of the parabolic bridge (i.e. vertex of the parabola) be at the origin, $O(0,0)$ in the cartesian coordinate system.

As the bridge have the shape of an inverted parabola, so the standard equation, which describes the shape of the bridge is

$x^2=4ay\;\cdots(i)$

where $a$ is an arbitrary constant (distance between focus and vertex of the parabola).

The span of the bridge = 166 feet and

Maximum height of the bridge= 40 feet.

The coordinate where the bridge meets the base is $A(83, -40)$ and $B(-83, -40).$

There is only one constant in the equation of the parabola, so, use either of one point to find the value of $a$ .

Putting $A(83,-40)$ in the equation (i) we have

$83^2=4a(-40)$

$\Rightarrow a=-43.05625$

So, on putting the value of $a$ in the equation (i), the equation of bridge is

$x^2=-172.225y$

From the figure, the distance from the center is measured along the x-axis, x coordinate at the distance of 10 feet is, $x=\pm 10$ feet, put this value in equation (i) to get the value of y.

$(\pm10)^2=-172.225y$

$\Rightarrow y=-1.72225$ feet.

The point $P_1(10,-1.72225)$ and $P_2(-10,-1.72225)$ represent the point on the bridge at a distance of 10 feet from its center.

The distance of these points from the x-axis is $d=1.72225$ feet and the distance of the base of the bridge from the x-axis is $h=40$ feet.

Hence, height from the base of the bridge at 10 feet from its center

$= h-d$

$=40-1.72225=38.27775$ feet.

8 0

3 years ago

a certain number of yams is sufficient to feed 20 students for 12 days for how many days would the same number of yams feed 15 s

Softa [21]

Step-by-step explanation:

no of days = (15×12)/20= 9 days

I hope this helps

6 0

2 years ago

Read 2 more answers

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 i

stich3 [128]

Complete question :

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 inches. Round answers to the nearest tenth of a percent. a) What percentage of years will have an annual rainfall of less than 43 inches? b) What percentage of years will have an annual rainfall of more than 39 inches? c) What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

Answer:

0.61053

0.66314

0.26647

Step-by-step explanation:

Given that :

Mean (m) = 41.4 inches

Standard deviation (s) = 5.7 inches

a) What percentage of years will have an annual rainfall of less than 43 inches?

P(x < 43)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (43 - 41.4) / 5.7 = 0.2807017

p(Z < 0.2807) = 0.61053 ( Z probability calculator)

b) What percentage of years will have an annual rainfall of more than 39 inches? c)

P(x > 39)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (39 - 41.4) / 5.7 = −0.421052

p(Z > −0.421052) = 0.66314 ( Z probability calculator)

What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

P(x < 38)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (38 - 41.4) / 5.7 = −0.596491

p(Z < −0.596491) = 0.27545 ( Z probability calculator)

P(x < 42)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (42 - 41.4) / 5.7 = 0.1052631

p(Z < 0.1052631) = 0.54192 ( Z probability calculator)

0.54192 - 0.27545 = 0.26647

6 0

3 years ago

Suppose that in a senior college class of 500500 students, it is found that 179179 smoke, 228228 drink alcoholic beverages, 1

olga2289 [7]

Answer: a) 0.16, b) 0.058, and c) 0.856.

Step-by-step explanation:

Since we have given that

Number of students = 500

Number of students smoke = 179

Number of students drink alcohol = 228

Number of students eat between meals = 119

Number of students eat between meals and drink alcohol = 59

Number of students eat between meals and smoke = 72

Number of students engage in all three = 30

a) Probability that the student smokes but does not drink alcohol is given by

$P(S-A)=P(S)-P(S\cap A)\\\\P(S-A)=\dfrac{179}{500}-\dfrac{99}{500}\\\\P(S-A)=\dfrac{179-99}{500}\\\\P(S-A)=\dfrac{80}{500}\\\\P(S-A)=0.16$

b) eats between meals and drink alcohol but does not smoke.

$P((M\cap A)-S)=P(M\cap A)-P(M\cap S\cap A)\\\\P((M\cap A)-S)=\dfrac{59}{500}-\dfrac{30}{500}\\\\P((M\cap A)-S)=\dfrac{59-30}{500}\\\\P((M\cap A)-S)=\dfrac{29}{500}\\\\P((M\cap A)-S)=0.058$