What is the best method to solve the equation below

What’s the answer for number 9

-BARSIC- [3]

The answer is C because complimentary is equal to 90 so 24+66 equals 90 and 34+56 equal 90

3 0

3 years ago

Order the following numbers from least to greatest -9/2 0 1 3/5 9/7 2/3

marysya [2.9K]

$\frac{-9}{2}$

0

$\frac{2}{3}$

$\frac{9}{7}$

1 $\frac{3}{5}$

3 0

3 years ago

Read 2 more answers

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

The Weight On Earth Compared To Some Rando Planet... pls help

earnstyle [38]

Answer:

if your talking about the earth i dont know because no one have ever compared some random planet and earths weight

Step-by-step explanation:

That’s because the planets weigh different amounts, and therefore the force of gravity is different from planet to planet. For example, if you weigh 100 pounds on Earth, you would weigh only 38 pounds on Mercury. That’s because Mercury weighs less than Earth, and therefore its gravity would pull less on your body.

3 0

2 years ago

the area of a rectangle is 90 in2^. the ratio of the length to the width is 5:2. find the length and the width

-BARSIC- [3]

Length and width of rectangle is 15 inches and 6 inches respectively

<h3>Solution:</h3>

Given that area of a rectangle is 90 square inch

Ratio of length to the width = 5: 2.

Need to determine length and width of rectangle.

As ratio of length to the width is 5 : 2

Lets assume length of rectangle = 5x inches and width of rectangle = 2x inches.

The formula for area of rectangle is given as:

$\text { Area of rectangle }=\text { length of rectangle } \times \text { width of rectangle}$

Substituting the given value of area of rectangle and assumed value of length and width of rectangle we get:

$\begin{array}{l}{90=5 x \times 2 x} \\\\ {=>90=10 x^{2}}\end{array}$

On solving the above expression for x we get

$\begin{array}{l}{=>\frac{90}{10}=x^{2}} \\\\ {=>x^{2}=9} \\\\ {=>x=\sqrt{9}=3}\end{array}$

$\begin{array}{l}{\text { Length of rectangle }=5 \times x=5 \times 3=15 \text { inches }} \\\\ {\text { Width of rectangle }=2 \times} x=2 \times} 3=6 \text { inches }}\end{array}$

Hence length and width of rectangle is 15 inches and 6 inches.

4 0

3 years ago