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4 years ago

Please help me on problem 8 and 9 of my teacher created problem

Mathematics

1 answer:

Sunny_sXe [5.5K]4 years ago

4 0

Number 8 is C 251.2 formula multiply 40*3.14=125.6 125.6*2=251.2
and 9 is C get the area of the rectangle 15*5=75 and now of the circle 3.14*2.5^2=78.5 now divide 78.5/2=39.25 then add 39.25+75=114.3.

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The amphitheater has two types of tickets available, reserved seats and lawn seats. The maximum capacity of the venue is 20,000

Tema [17]

Let the number of reserved tickets = x

Let the number of lawn seats = y

Constraint functions:

Maximum capacity means $x+y\leq 20000$

For concert to be held $x+y\geq 5000$

$lawn seats\leq reserved$ means $y\leq x$

Objective functions :

Maximum profit equation p = 65x +40y

Intersection points :

(10000,10000) (20000,0)(2500,2500)(5000,0)

p at (10000,10000) = 65(10000) + 40(10000) = $1050000

p at (20000,0) = 65(20000) + 40(0) = $1300000

p at (2500,2500) = 65(2500) + 40(2500) = $262500

p at (5000,0) = 65(5000) + 40(0) = $325000

Hence maximum profit occurs when all 20000 reserved seats are sold and the profit is $1300000

Please find attached the graph of it.

8 0

4 years ago

Which equation models this problem?

OLEGan [10]

A 8p = 48.80
Since Susan paid for 8 dish towels, p. Then the 8 dish towels = $48.80

4 0

3 years ago

Read 2 more answers

Lines k and n intersect on the y-axis

avanturin [10]

a) The equation of line k is:

$y = -\frac{202}{167}x + \frac{598}{167}$

b) The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

The equation of a line, in slope-intercept formula, is given by:

$y = mx + b$

In which:

m is the slope, which is the rate of change.
b is the y-intercept, which is the value of y when x = 0.

Item a:

Line k intersects line m with an angle of 109º, thus:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

In which $m_1$ and $m_2$ are the slopes of k and m.

Line k goes through points (-3,-1) and (5,2), thus, it's slope is:

$m_1 = \frac{2 - (-1)}{5 - (-3)} = \frac{3}{8}$

The tangent of 109 degrees is $\tan{109^{\circ}} = -\frac{29}{10}$
Thus, the slope of line m is found solving the following equation:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

$-\frac{29}{10} = \frac{m_2 - \frac{3}{8}}{1 + \frac{3}{8}m_2}$

$m_2 - \frac{3}{8} = -\frac{29}{10} - \frac{87}{80}m_2$

$m_2 + \frac{87}{80}m_2 = -\frac{29}{10} + \frac{3}{8}$

$\frac{167m_2}{80} = \frac{-202}{80}$

$m_2 = -\frac{202}{167}$

Thus:

$y = -\frac{202}{167}x + b$

It goes through point (-2,6), that is, when $x = -2, y = 6$ , and this is used to find b.

$y = -\frac{202}{167}x + b$

$6 = -\frac{202}{167}(-2) + b$

$b = 6 - \frac{404}{167}$

$b = \frac{6(167)-404}{167}$

$b = \frac{598}{167}$

Thus. the equation of line k, in slope-intercept formula, is:

$y = -\frac{202}{167}x + \frac{598}{167}$

Item b:

Lines j and k intersect at an angle of 90º, thus they are perpendicular, which means that the multiplication of their slopes is -1.

Thus, the slope of line j is:

$-\frac{202}{167}m = -1$

$m = \frac{167}{202}$

Then

$y = \frac{167}{202}x + b$

Also goes through point (-2,6), thus:

$6 = \frac{167}{202}(-2) + b$

$b = \frac{(2)167 + 202(6)}{202}$

$b = \frac{1546}{202}$

The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

A similar problem is given at brainly.com/question/16302622

7 0

2 years ago

A rocket is launched from a tower what time will the rocket reach its max

Lena [83]

Answer:

Step-by-step explanation:

A science class designed a ball launcher and tested it by shooting a tennis ball up and off the top of a 15-story building. They determined that the motion of the ball could be described by the function: h(t) = -16t2 + 144t + 160, where ‘t’ represents the time the ball is in the air in seconds and h(t) represents the height, in feet, of the ball above the ground at time t.

a) Graph the function h(t) = -16t2 + 144t + 160 (see below)

b) What is the height of the building?

The height of the building is also the height of the tennis ball before it is launched into the air. This occurs when t=0 so substitute 0 for t and you get:

H(0) = -16(0)2 + 144(0) + 160

The height of the building is 160 feet.

c) At what time did the ball hit the ground?

The ball hits the ground when the height is 0. Therefore, we are looking for a solution to: -16t2 + 144t + 160 = 0

Use the quadratic formula or put this into a calculator. The solution is t=10 and -1, but only 10 makes sense. Therefore, the ball hits the ground at 10 seconds.

d) At what time did the ball reach its maximum height?

You can put this into the calculator or you can realize that the maximum height is also

− the vertex. The x-value (‘t’ in this case) is 2

−144

which is (2)(−16) = 4.5.

Therefore, the ball reached its maximum height at 4.5 seconds.

e) What is the maximum height of the ball?

We calculated the time of the maximum height (4.5 seconds). Therefore, substitute 4.5 into the function to find the maximum height.

-16(4.5)2 + 144(4.5) + 160

The maximum height is 484 feet.

5 0

3 years ago

With reference to the figure, sin X equals? 1.) 0.250 2.)0.447 3.)0.894 4.)1

Flura [38]

Answer:

3.) 0.894

Step-by-step explanation:

✔️First, find BD using Pythagorean Theorem:

BD² = BC² - DC²

BC = 17.89

DC = 16

Plug in the values

BD² = 17.89² - 16²

BD² = 64.0521

BD = √64.0521

BD = 8.0 (nearest tenth)

✔️Next, find AD using the right triangle altitude theorem:

BD = √(AD*DC)

Plug in the values into the equation

8 = √(AD*16)

Square both sides

8² = AD*16

64 = AD*16

Divide both sides by 16

4 = AD

AD = 4

✔️Find AB using Pythagorean Theorem:

AB = √(BD² + AD²)

AB = √(8² + 4²)

AB = √(64 + 16)

AB = √(80)

AB = 8.9 (nearest tenth)

✔️Find sin x using trigonometric ratio formula:

Reference angle = x

Opposite side = BD = 8

Hypotenuse = AB = 8.944

Thus:

$sin(x) = \frac{opp}{hyp} = \frac{8}{8.944} = 0.894$ (nearest thousandth)

6 0

3 years ago