PLEASE HELP!!! 50 POINTS!! WILL GIVE BRAINLIEST!!!A person standing at the top of Mountain Everest would be approximately 5.5 mi high. The radius of earth is 3959 mi.

What is the distance to the horizon from this point?

Enter your answer as a decimal in the box. Round only your final answer to the nearest tenth.
PLZ SHOW WORK!!!

Answers

Answer 1

Answer:

Answer:

208.8 mi

Step-by-step explanation:

Form a triangle with the centre of the Earth (C) at one point, the horizon (H) as a second point, and the top of Mt. Everest (O) as the third (see diagram).

Let

r = Earth’s radius

h = height of Mt Everest

d = distance to horizon

∆CHO is a right triangle.

d² + r² = (r+h)²

d² + 3959² = (3959+5.5)²

d² + 15 673 681 = 3964.5²

d² + 15 673 681 = 15 717 260 Subtract 15 673 681 from each side

d² = 43 579 Take the square root of each side

d = 208.8 mi

The distance from the top of Mt Everest to the horizon is 208.8 mi.

Answer 2

Answer:

Answer:

I JUST TOOK THE TEST THE ANSWER IS CORRECT.

Step-by-step explanation:

208.8

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A triangle has two sides of lengths 10 and 14 .value could the length of the thrid side be
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A sum of 17 and a difference of 5?

Which inequality has no solution?6(x+2)> x-3

3+4x<2(1+2x)

-2(x+6)< x-20

x-9 < 3 (x-3)

Answers

we will proceed to solve all cases to determine the solution of the problem

case A) $6(x+2)> x-3$

Eliminate the parenthesis of the left side

$6x+12> x-3$

Group terms that contain the same variable, and move the constants to the opposite side of the inequality

$6x-x> -3-12$

Combine like terms

$5x> -15$

$x> -3$

The solution is the interval--------> (-3,∞)

All real numbers greater than $-3$

therefore

The inequality A has solution

case B) $3+4x <2(1+2x)$

Eliminate the parenthesis of the right side

$3+4x <2+4x$

Group terms that contain the same variable, and move the constants to the opposite side of the inequality

$4x-4x <2-3$

Combine like terms

$0 <-1$ ---------> is not true

therefore

The inequality B has no solution

case C) $-2(x+6)< x-20$

Eliminate the parenthesis of the left side

$-2x-12< x-20$

Group terms that contain the same variable, and move the constants to the opposite side of the inequality

$-2x-x< -20+12$

Combine like terms

$-3x< -8$

Multiply by $-1$ both sides

$3x> 8$

$x> (8/3)$

The solution is the interval--------> (8/3,∞)

All real numbers greater than $8/3$

therefore

The inequality C has solution

case D) $x-9<3(x-3)$

Eliminate the parenthesis of the right side

$x-9<3x-9$

Group terms that contain the same variable, and move the constants to the opposite side of the inequality

$x-3x< -9+9$

Combine like terms

$-2x< 0$

Multiply by $-1$ both sides

$2x> 0$

$x>0$

The solution is the interval--------> (0,∞)

All real numbers greater than $0$

therefore

The inequality D has solution

therefore

the answer is

$3+4x <2(1+2x)$

$\boxed{3 + 4x < 2\left( {1 + 2x} \right)}$ inequality has no solution. Option (b) is correct.

Further explanation:

The linear equation with slope m and intercept c is given as follows.

$\boxed{y = mx + c}$

The formula for slope of line with points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ can be expressed as,

$\boxed{m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}}$

Given:

The inequalities are as follows,

(a). $6\left( {x + 2} \right) > x - 3$

(b). $3 + 4x < 2\left( {1 + 2x} \right)$

(c). $- 2\left( {x + 6} \right) < x - 20$

(d). $x - 9 < 3\left( {x - 3} \right)$

Explanation:

Solve the inequality $6\left( {x + 2} \right) > x - 3$ to check whether the inequality has solution.

$\begin{aligned}6\left( {x + 2} \right) >& x - 3\n6x + 12 >& x - 3\n6x - x >& - 3 - 12\n 5x >&- 15\nx >& \frac{{ - 15}}{5}\nx >&- 3\n\end{aligned}$

Option (a) has solution.

Solve the inequality $3 + 4x < 2\left( {1 + 2x} \right)$ to check whether the inequality has solution.

$\begin{aligned}3+ 4x &< 2\left( {1 + 2x} \right)\n3 + 4x &< 2 + 4x\n3&< 2\n\end{aligned}$

Option (b) has no solution.

Solve the inequality $- 2\left( {x + 6} \right) < x - 20$ to check whether the inequality has solution.

$\begin{aligned}- 2\left( {x + 6}\right) &< x - 20\n- 2x - 12 &< x - 20\n- 12 + 20 &< x + 2x\n8&< 3x\n(8)/(3) &< x\n\end{aligned}$

Option (c) has solution.

Solve the inequality $x - 9 < 3\left( {x - 3} \right)$ to check whether the inequality has solution.

$\begin{aligned}x - 9&< 3\left( {x - 3} \right)\nx - 9&< 3x - 9\nx&< 3x \n1&< 3\n\end{aligned}$

Option (d) has solution.

$\boxed{3 + 4x < 2\left( {1 + 2x} \right)}$ inequality has no solution. Option (b) is correct.

Learn more:

Learn more about inverse of the functionbrainly.com/question/1632445.
Learn more about equation of circle brainly.com/question/1506955.
Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Linear equation

Keywords: numbers,slope, slope intercept, inequality, equation, linear inequality, shaded region, y-intercept, graph, representation, origin.

I really need help on geometry

Answers

For example, exc. 15.

x + y = 10 and 6/x = 9/y; You need y!

=> x = 10 -y and 9x = 6y => 9(10-y) = 6y => 90 - 9y = 6y => 15y = 90 => y = 6.

The perimeter of a rectangle is 70 cm. If its length is decreased by 5 cm and its width is increased by 5 cm, its area will increase by 50 cm2. Find the length and the width of the original rectangle.

Answers

x=length of the original rectangle. (in cm)
y=width o f the original rectangle. (in cm)

Perimeter of a rectangle=sum of the all sides.
Perimeter of the original rectangle=x+x+y+y=2x+2y

Area of a rectangle=length x width
Area of the original rectangle=xy

x-5=length decreased by 5 cm
y+5=width increase by 5 cm.

We can suggest this system of equations:
2x+2y=70
(x-5)(y+5)=xy+50

We solve this system of equations by susbstitution method:
2x+2y=70 ⇒x+y=35 ⇒ y=35-x

(x-5)(35-x+5)=x(35-x)+50
(x-5)(40-x)=35x-x²+50
40x-x²-200+5x=35x-x²+50
40x-35x+5x=200+50
10x=250
x=250/10
x=25

y=35-x
y=35-25
y=10

Answer: the lenght and the width of the original rectangle is :
lenght=25 cm
width=10 cm.

The length of the rectangle is 25cm and the width of the rectangle is 45cm and this can be determine by forming the linear equations.

Given :

Perimeter of a rectangle is 70 cm.

Rectangle length is decreased by 5 cm and its width is increased by 5 cm, its area will increase by 50 $\rm cm^2$ .

Let 'a' be the length of the rectangle and 'b' be the width of the rectangle. Than the perimrter of the rectangle will be:

a + b = 70 ---- (1)

Now, the area of the rectangle will be:

$\rm (a-5)(35-a+5)=a(35-a)+50$

$\rm 35a -a^2+5a-175+5a-25=35a-a^2+50$

$\rm10a -200=50$

a = 25

Now, put the value of 'a' in equation (1).

b = 70 - 25 = 45

b = 45

Therefore, the length of the rectangle is 25cm and the width of the rectangle is 45cm.

For more information, refer the link given below:

brainly.com/question/919810

The square represents a scale model that was created by using a factor of 4. 8 ft 8 ft [Not drawn to scale] Which is true of the measures of the sides of the original square? Each side of the original square is 7 the length of each side of the scale model. O Each side of the original square is 4 times the length of each side of the scale m O Each side of the original salvare is the same length as each side of the scale ma

Answers

Answer:

Step-by-step explanation:

Simplify the expression (4x - 3)(x + 5).

Answers

=>(4x - 3) (x + 5)=>(4x × x )+( 4x × 5) - (3 × x ) -( 3 × 5)=>4x^2 + 20x -3x - 15=>4x^2 + 17x -15Hope it helps!!!

Wait I think it's 9x-3.. Sorry...

Mrs. Bergstedt teaches six classes. In her first period class she has 19 students, her third period she has 13 students, her fourth period 19 students, her fifth period 9 students, her sixth period 14 students, and her seventh period 15 students. What is her average class size, to the nearest tenth

Answers

19 + 13 + 19 + 9 + 14 + 15

19 + 13 = 32
32 + 19 = 51
51 + 9 = 60
60 + 14 = 74
74 + 15 = 89

89 / 6 (the number of classes she has)
= 14.8

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