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2 years ago

ALGEBRA 2 word PROBLEM HELP ME PLEASE !!!

Mathematics

1 answer:

gizmo_the_mogwai [7]2 years ago

3 0

Answer:

l = 231 ft (nearest ft)

Step-by-step explanation:

sin60 = 200/x

xsin60 = 200

x = 200/sin60

= 231 ft (rounded to nearest ft)

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What does the underlined conjunction connect in the sentence? The family enjoyed fishing, but they loved hiking during their tri

pychu [463]

The answer is c objects

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3 years ago

Quadrilateral JKLM is a rhombus. The diagonals intersect at N. If angle JNK equals 5x - 15, find the value of x.

VLD [36.1K]

One property of a rhombus is that the diagonals are perpendicular. This means that the diagonals intersect at 90 degree angles (right angles).

So we know that angle JNK is 90 degrees

angle JNK = 90 degrees
5x - 15 = 90
5x - 15 + 15 = 90+15
5x = 105
5x/5 = 105/5
x = 21

Answer: x = 21

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3 years ago

brandy is playing a game where she wins 50 points per round for 8 rounds. However, during the last round she gets penalized and

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50*8=400
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3 years ago

A swimming pool at a park measures 9.75 meters by 7.2 meters.

neonofarm [45]

Answer:

Part a) The area of the swimming pool is $70.2\ m^2$

Part b) The total area of the swimming pool and the playground is $105.3\ m^2$

Step-by-step explanation:

Part a) Find the area of the swimming pool

we know that

The area of the swimming pool is

$A=LW$

where

L is the length side

W is the width side

we have

$L=9.75\ m\\W=7.2\ m$

substitute the values

$A=(9.75)(7.2)$

$A=70.2\ m^2$

therefore

The area of the swimming pool is $70.2\ m^2$

Part b) The area of the playground is one and a half times that of the swimming pool. Find the total area of the swimming pool and the playground

we know that

To obtain the area of the playground multiply the area of the swimming pool by one and a half

$\frac{1}{2}(70.2)=35.1\ m^2$

To obtain the total area of the swimming pool and the playground, adds the area of the swimming pool and the area of the playground

$70.2\ m^2+35.1\ m^2=105.3\ m^2$

therefore

The total area of the swimming pool and the playground is $105.3\ m^2$

6 0

3 years ago

The force F (in newtons) of the hydraulic cylinder in a press is proportional to the square of sec x where x is the distance (in

DanielleElmas [232]

Answer:

Part A) F(x) =500 sec(x)

Part B) Average force exerted by the press = 750 N

Step-by-step explanation:

Given:

F(x) ∝ sec(x)

Range of x = [0,Pi/3] or x = [0 , 60°]

F(0) = 500 N i.e. F is 500 N at x = 0.

As we now that:

Force is proportional to sec(x) therefore we may write an equation by introducing Proportionality Constant (k) as under:

F(x) = k * sec(x) - Say it Equation 1

Then using given information of F=500 N at x = 0 we have:

500 = k * sec(0)

As sec(0) = 1, therefore we get:

k = 500

By putting value of k in equation 1 we have:

F(x) = 500 * sec(x)

Now by putting max value of x from the given range that is Pi/3 in the above equation we get:

F(Pi/3) = 500 * sec(Pi/3)

As Pi = 180 there we simplify the above equation as:

F(60°) = 500 * sec(60°)

F(60°) = 500 * 2 ; By putting sec(60°) = 2

F(60°) = 1000 N

Now the avg. of force exerted by the hydraulic press is given by:

Avg. Force = Minimum Force + [(Maximum Force - Minimum Force) / 2] - Say it equation 2

Considering sec(x) is minimum at x = 0 and maximum at x = 60° within the given range [0 , 60°] therefore we have:

The minimum force being put at x = 0 i.e. F = 500 N and max. force at x = 60° which is F = 1000 N.

Finally we get the average force using equation 2 as under:

Average force = 500 + [(1000 - 500) / 2]

Average Force = 500 + 250

Average Force = 750 N

7 0

3 years ago