Two bicyclists leave the center of town at the same time. One heads due north and the other heads due west. Later, the two cyclists are exactly 25 mi apart. The cyclist headed north has traveled 5 mi farther than the cyclist going west.How far has the cyclist going west traveled?

Answers

Answer 1

Answer:

Answer:

Distance traveled by bicyclist traveling west = 15 miles

Step-by-step explanation:

Two bicyclists leave the center of town at the same time. One heads due north and the other heads due west. Later, the two cyclists are exactly 25 mi apart. The cyclist headed north has traveled 5 mi farther than the cyclist going west.

These two cyclists travel at angle 90°

Relative displacement can be calculated using Pythagoras theorem.

Let d be the distance traveled by bicyclist traveling west

Distance traveled by bicyclist traveling north = d + 5

$25^2=d^2+(d+5)^2\n\nd^2+d^2+10d+25=625\n\n2d^2+5d-600=0\n\nd^2+5d-300=0\n\nd=(-5\pm √(5^2-4* 1* (-300)))/(2* 1)\n\nd=(-5\pm √(25+1200))/(2)\n\nd=(-5\pm √(1225))/(2)\n\nd=(-5\pm 35)/(2)\n\nd=15miles\texttt{ or }d=-20miles$

Negative displacement is not possible.

Hence d = 15 miles

Distance traveled by bicyclist traveling west = d = 15 miles

Answer 2

Answer: This situation is represented in annex.
From Pythagorean theorem you've got equation:
$x^2+(x+5)^2=25 \n \n \hbox{From formula} \ \ \ (a+b)^2 = a^2+2ab+b^2: \n \n x^2+x^2+10x+25=25 \n 2x^2+10x=0 \n \n \hbox{Factor out} \ \ 2x: \n 2x(x+5)=0 \n \n \hbox{So you've got:} \n \n x_1=0 \n x_2=-5$

There isn't any solutions where x>0, so this situation is IMPOSSIBLE.

Did you write correctly this question....?

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Keshawn is asked to compare and contrast the domain and range for the two functions.f(x) = 5x
g(x) = 5

Answers

For the function 5x, the range is the value of f(x) which is five times the x and the domain is the value of x which can be the ratio of f(x) and 5. For the second function which is g(x) = 5, the range is 5 all through out the graph while the domain is infinity.

Function: y = 2(x + 3)² +1

Answers

Answer:

10x

Step-by-step explanation:

How would you write y = –3x2 + 12x – 21 in vertex form?

Answers

The vertex form of that expression would be y = 3 ( x +2 )squared - 25. This is done by transporting first the -21 to the other side then simplifying the 3(x2 + 4x). Then making the "x2 + 4x" a perfect binomial we need to add 4 and that makes it "x2 + 4x + 4". Then after that we now have the vertex form of y = 3 ( x +2 )squared - 25

Which functions are symmetric with respect to the origin?y = arcsinx and y = arccosx
y = arccosx and y = arctanx
y = arctanx and y = arccotx
y = arcsinx and y = arctanx

Answers

The function are symmetric with respect to the origin is y = arcsinx and y = arctanx and this can be determined by using the trigonometric property.

The following steps can be used in order to determine which functions are symmetric with respect to the origin:

Step 1 - Check all the options given in order to determine which functions are symmetric with respect to the origin.

Step 2 - Substitute (x = 0) in option a), that is:

$\rm y = sin^(-1)(0)=0$

$\rm y = cos^(-1)(0)=1.57$

Step 3 - Substitute (x = 0) in option b), that is:

$\rm y = cos^(-1)(0)=1.57$

$\rm y = tan^(-1)(0)=0$

Step 4 - Substitute (x = 0) in option c), that is:

$\rm y = tan^(-1)(0)=0$

$\rm y = cot^(-1)(0)=1.57$

Step 5 - Substitute (x = 0) in option d), that is:

$\rm y = tan^(-1)(0)=0$

$\rm y = sin^(-1)(0)=0$

For more information, refer to the link given below:

brainly.com/question/13710437

Answer:

The correct option is;

y = arcsinx and y = arctanx

Step-by-step explanation:

The given options are;

1) y = arcsinx and y = arccosx

Here, we have at the origin, where x = 0, arccosx ≈ 1.57 while arcsinx = 0

Therefore arccosx does not intersect arcsinx at the origin for it to be symmetrical to arcsinx or the origin

2) y = arccosxy and y = arctanx

Here arctanx = 0 when x = 0 and arcos x = 1.57 when x = 0 therefore, they are not symmetrical

3) y = arctanx and y = arccotx

Similarly, At x = 0, arccotx = 1.57 therefore, they are not symmetrical

4) y = arcsinx and y = arctanx

Both functions arcsinx and arctanx pass through the origin and their shapes are similar but inverted as they go from negative to positive.

Write the equations of two lines whose graphs are perpendicular to one another

Answers

Write 2 equations in slope - intercept form: y = mx + b
The slopes of perpendicular lines are negative reciprocal of each other. That is, if you have a slope of 2/3, the perpendicular slope will be -3/2.
y = 2/3x + 3
is perpendicular to
y = -3/2x - 9
Only the slope matters in the equations. The y-intercept can be anything. I randomly chose +3 and -9.

Final answer:

Perpendicular lines have slopes that are negative reciprocals of each other. By using the slope-intercept form of a linear equation, you can find two equations for lines that are perpendicular to each other. An example pair of equations is y = 2x + 1 and y = (-1/2)x + 3.

Explanation:

In order for two lines to be perpendicular, their slopes must be negative reciprocals of each other. Let's say the equation of the first line is y = mx + b, where m is the slope. The equation of the second line that is perpendicular to the first line would have a slope of -1/m. So, you can choose any values for m and b for the first line and then find the slope of the second line using -1/m.

For example, let's say the equation of the first line is y = 2x + 1. The slope of this line is 2. The equation of the second line that is perpendicular to the first line would have a slope of -1/2. So, an example equation for the second line could be y = (-1/2)x + 3.

Therefore, two example equations of lines that are perpendicular to each other are y = 2x + 1 and y = (-1/2)x + 3.

Learn more about Perpendicular Lines here:

brainly.com/question/18271653

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D= 5/6 (F+G) solve for G

Answers

$D = 5/6(F + G)$
$(6 * D)/5 = F + G$
$(6 * D - F)/5 = G$

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