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3 years ago

Angles a and b are supplementary how would you fix value of x

Mathematics

1 answer:

kotykmax [81]3 years ago

5 0

Answer: 180 = 30 + (x + 60)

Step-by-step explanation: Supplementary means that the angles add together to 180 degrees. This equations shows that angle A (30 degrees) plus angle B (x + 60) equal 180. If you solve the equation you will find that angle B is equal to 150 degrees, and that x = 90.

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4) The path of a satellite orbiting the earth causes it to pass directly over two

Naily [24]

Answers:

Satellite is approximately 2446.43 km from station A.
Satellite is approximately 2441.61 km above the ground.

=========================================================

Explanation:

I'm assuming tracking stations A and B are at the same elevation and are on flat ground. In reality, this is likely not the case; however, for the sake of simplicity, we'll assume this is the case.

The diagram is shown below. Points A and B describe the two stations, while point C is the satellite's location. Point D is on the ground directly below the satellite. We have these lengths

AB = 60 km
AD = x
CD = h

Focusing on triangle ACD, we can apply the tangent rule to isolate h.

tan(angle) = opposite/adjacent

tan(A) = CD/AD

tan(86.4) = h/x

x*tan(86.4) = h

h = x*tan(86.4)

We'll use this later in the substitution below.

--------------------

Now move onto triangle BCD. For the reference angle B = 85, we can use the tangent rule to say

tan(angle) = opposite/adjacent

tan(B) = CD/DB

tan(B) = CD/(DA+AB)

tan(85) = h/(x+60)

tan(85)*(x+60) = h

tan(85)*(x+60) = x*tan(86.4) ............. apply substitution; isolate x

x*tan(85)+60*tan(85) = x*tan(86.4)

60*tan(85) = x*tan(86.4)-x*tan(85)

60*tan(85) = x*(tan(86.4)-tan(85))

x*(tan(86.4)-tan(85)) = 60*tan(85)

x = 60*tan(85)/(tan(86.4)-tan(85))

x = 153.612786190499

--------------------

We'll use this approximate x value to find h

h = x*tan(86.4)

h = 153.612786190499*tan(86.4)

h = 2441.60531869599

h = 2441.61 km is how high the satellite is above the ground.

Return to triangle ACD. We'll use the cosine rule to determine the length of the hypotenuse AC

cos(angle) = adjacent/hypotenuse

cos(A) = AD/AC

cos(86.4) = x/AC

cos(86.4) = 153.612786190499/AC

AC*cos(86.4) = 153.612786190499

AC = 153.612786190499/cos(86.4)

AC = 2446.43279498247

AC = 2446.43 km is the distance from the satellite to station A.

6 0

3 years ago

Which is the same as 10-2? A) 0.0001 B) 0.001 C) 0.01 D) 100

IceJOKER [234]

The correct answer is C. This is because
10^-2 = 1/10^2 which is 0.01

3 0

3 years ago

I need to know the answer asap

Doss [256]

subtract lowest time from highest time for each kind

129-29 = 100

320-210 = 110

the ranges are about the same but the times are higher for the dark chocolate

7 0

3 years ago

A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0

3 years ago

There are 16 ounces in one pound. If 3.4 pounds of beef cost $4.95, what is the cost per ounce, to the nearest cent?

Alina [70]

$0.09 per ounce Because there are 16 ounces in a pound, 3.4 pounds is 3.4(16) = 54.4 ounces. To find cost per ounce, divide the cost in dollars by the number of ounces

4 0

3 years ago