Cost = 200 + 50x
The relationship between time and cost of the plan is not proportional. By definition, proportional ;relationships between two variables have equivalent ratio; one variable is always a constant value times the other which in this case is not .
To illustrate,
Month 1 Cost = 200 + 50(1) = 250 Ratio ( time:cost) = 1:250
Month2 Cost = 200 + 50(2) = 300 Ratio ( time:cost) = 2:300 = 1:150
Month 3 Cost = 200 + 50(3) = 350 Ratio ( time:cost) = 3:350
Month 4 Cost = 200 + 50(4)= 400 Ratio ( time:cost) = 4 :400 = 1:100
The ratios are not equivalent,thus the relationship is not proportional.
The ordered pairs are graphed from the table in question 27 which of the following lines show the correct relationship
To find the price per pound of oranges, we need to divide the cost by the number of pounds:
[tex]\text{ Price / pound=}\frac{12}{3}=4\text{ \$/pound}[/tex]Thus, each pound costs $4.
The cost of 5 pounds is:
[tex]5*4=20[/tex]The answer is C. $20
o from Mobile. How Tar IS Il IIUMI U 3117 12. Cincinnati is 488 miles from New York City and along the way to Brownsville, Texas which is 2007 miles from New York. How far, then, is it from Cincinnati to Brownsville?
ANSWER:
2495 miles.
STEP-BY-STEP EXPLANATION:
To answer the question we make the following scheme:
Therefore the distance between Cincinnati and Brownsville is the sum of the distance between Cincinnati to New York and New York to Brownsville, like this
[tex]\begin{gathered} d=488+2007 \\ d=2495 \end{gathered}[/tex]Therefore the distance between Cincinnati and Brownsville is 2495 miles.
Find the missing side length and angles of ABC given that m B = 137º, a = 15, and c = 17. Round to the nearest tenth.(Find angles A and C and side b)
Given the triangle ABC and knowing that:
[tex]\begin{gathered} a=BC=15 \\ c=AB=17 \\ m\angle B=137º \end{gathered}[/tex]You need to apply:
• The Law of Sines in order to solve the exercise:
[tex]\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}[/tex]That can also be written as:
[tex]\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}[/tex]Where A, B, and C are angles and "a", "b" and "c" are sides of the triangle.
• The Law of Cosines:
[tex]b=\sqrt[]{a^2+c^2-2ac\cdot cos(B)}[/tex]Where "a", "b", and "c" are the sides of the triangle and "B" is the angle opposite side B.
Therefore, to find the length "b" you only need to substitute values into the formula of Law of Cosines and evaluate:
[tex]b=\sqrt[]{(15)^2+(17)^2-2(15)(17)\cdot cos(137\degree)}[/tex][tex]b\approx29.8[/tex]• To find the measure of angle A, you need to set up the following equation:
[tex]\begin{gathered} \frac{a}{sinA}=\frac{b}{sinB} \\ \\ \frac{15}{sinA}=\frac{29.8}{sin(137\degree)} \end{gathered}[/tex]Now you can solve for angle A. Remember to use the Inverse Trigonometric Function "Arcsine". Then:
[tex]\frac{15}{sinA}\cdot sin(137\degree)=29.8[/tex][tex]\begin{gathered} 15\cdot sin(137\degree)=29.8\cdot sinA \\ \\ \frac{15\cdot sin(137\degree)}{29.8}=sinA \end{gathered}[/tex][tex]A=\sin ^{-1}(\frac{15\cdot sin(137\degree)}{29.8})[/tex][tex]m\angle A=20.1\degree[/tex]• In order to find the measure of Angle C, you need to remember that the sum of the interior angles of a triangle is 180 degrees. Therefore:
[tex]m\angle C=180º-137º-20.1\degree[/tex]Solving the Addition, you get:
[tex]\begin{gathered} m\angle C=180º-137º-20.1\degree \\ m\angle C=22.9\degree \end{gathered}[/tex]Therefore, the answer is:
[tex]undefined[/tex]Barry Bonds holds the major league home run record with 73 in one season. If Pete Alonso wants to break his record, how many homeruns would he have to hit on average over 162 games to break Bonds' record?
73 homeruns in one season
162 games Pete Alonso must do at least 74 homeruns
He must do 74 homeruns and 88 will not be homeruns.
On average he must do 74/162 = 0.45 homeruns per game
Convert the following Quadratic Equations from Vertex Form to Standard Form.
4)
Given:
The vertex form is given as,
[tex]y=-\frac{1}{3}(x+6)^2-1[/tex]The objective is to convert the vertex form to standard form.
The standard form can be obtained as,
[tex]\begin{gathered} y=-\frac{1}{3}(x+6)^2-1 \\ =-\frac{1}{3}(x^2+6^2+2(x)(6))-1 \\ =-\frac{1}{3}(x^2+36+12x)-1 \\ =-\frac{x^2}{3}-\frac{36}{3}+\frac{12x}{3}-1 \\ =-\frac{x^2}{3}-12+4x-1 \\ =-\frac{x^2}{3}+4x-13 \end{gathered}[/tex]Here,
[tex]\begin{gathered} a=-\frac{1}{3} \\ b=4 \\ c=-13 \end{gathered}[/tex]Hence, the required standard form of the equation is obtained.
Find the area of the triangle below.Be sure to include the correct unit in your answer.025 ft24 ft7 ft
To calculate the area of the triangle you have to multiply its base by its height and divide the result by 2, following the formula
[tex]A=\frac{bh}{2}[/tex]Considering the given right triangle, the base and the height of the triangle are its legs:
The area can be calculated as:
[tex]\begin{gathered} A=\frac{bh}{2} \\ A=\frac{7\cdot24}{2} \\ A=\frac{168}{2} \\ A=84ft^2 \end{gathered}[/tex]The area of the triangle is 84ft²
Writing Equations Is As Easy As 1, 2, 3 Digital Write the equation of the line that has the indicated slope and y-intercept. Slope = 2; y-intercept is (0,5)
The general structure of a linear function is "slope-intercept" form is
y=mx+b
Where
m is the slope
b is the y-intercept
To write the equation for a slope 2 and y-intercept (0,5) you have to replace said values in the formula:
m=2
b=5
y=2x+5
Similar PcIn this session, you will apply your knowledge of similar polygons to real-lifesituations.An artist plans to paint a picture. He wants to use a canvas that is similar to aphotograph with a height of 8 in, and a width of 10 in. If the longer horizontalsides of the canvas are 30 in. wide, how high should the canvas be?
24 in
Explanation / Steps:
Since the aim of the artists is to duplicate the photograph on a different scale canvas:
8 * 30 / 10 = 24 in
More details:
Since for a side of 8 in is requires a side of 10 in => a side of 1 in requires 8/10 ~ 0.8 in
then a side of 30 in requires 30 *0.8 = 24 in
Of the following real functions of real variable it calculates: a) the domain, b) the intersections with the axes c) the limits at the ends of the domain. d) Draw the graph e) calculates the first derivative and identifies the minimum or maximum points of the function
Given:
[tex]g(x)=\frac{x+\sqrt{x}}{x+1}[/tex]Find-: Domain, intersection with axis, limit at end the domain, draw the graph, first derivative the minimum and maximum of the point is:
Sol:
Domain:
The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined. In the function machine metaphor, the domain is the set of objects that the machine will accept as inputs.
[tex]\begin{gathered} g(x)=\frac{x+\sqrt{x}}{x+1} \\ \\ Domain:x>0 \end{gathered}[/tex]The domain is greater than the zero "0" because the negative value of "x" is create undefine form inside the square root.
Intersection with axis:
For x-intersection value of "y" is zero so,
[tex]\begin{gathered} g(x)=\frac{x+\sqrt{x}}{x+1} \\ \\ 0=\frac{x+\sqrt{x}}{x+1} \\ \\ 0=x+\sqrt{x} \\ \\ \sqrt{x}(\sqrt{x}+1)=0 \\ \\ x=0,\sqrt{x}=-1 \\ \\ \sqrt{x}=-1\text{ Not possible } \\ x=0 \end{gathered}[/tex]For y intersection value of "x" is zero then,
[tex]\begin{gathered} y=\frac{x+\sqrt{x}}{x+1} \\ \\ y=\frac{0+\sqrt{0}}{0+1} \\ \\ y=0 \end{gathered}[/tex]Graph of function is:
The first derivative is:
[tex]\begin{gathered} g(x)=\frac{x+\sqrt{x}}{x+1} \\ \\ g^{\prime}(x)=\frac{(x+1)(1+\frac{1}{2\sqrt{x}})-(x+\sqrt{x)}}{(x+1)^2} \\ \\ g^{\prime}(x)=\frac{x+\frac{\sqrt{x}}{2}+1+\frac{1}{2\sqrt{x}}-x-\sqrt{x}}{(x+1)^2} \\ \\ g^{\prime}(x)=\frac{1+\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}}{(x+1)^2} \\ \\ \end{gathered}[/tex]For maximum is:
[tex]\begin{gathered} g^{\prime}(x)=0 \\ \\ 1+\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}=0 \\ \\ \frac{2\sqrt{x}+1-x}{2\sqrt{x}}=0 \\ \\ x-1=2\sqrt{x} \end{gathered}[/tex]Square both side then:
[tex]\begin{gathered} (x-1)^2=2\sqrt{x} \\ \\ x^2+1-2x=4x \\ \\ x^2-6x+1=0 \\ \\ x=\frac{6\pm\sqrt{36-4}}{2} \\ \\ x=\frac{6\pm\sqrt{32}}{2} \\ \\ x=5.83,x=0.17 \end{gathered}[/tex]For the maximum or minimum point are
(5.83) and 0.17
20. Monroe's teacher wants each student to draw a sketch of the longest specimen. Which specimen is the longest? 21. Seen through the microscope, a specimen is 0.75 cm long. What is its actual length?
20)
The sketch of the given data is
By observing the sketch, the longest line is A.
Hence the specimen A is the longest.
21)
In the microscope, the size of the specimen is 0.75cm long.
It is given that the microscope enlarges the actual length 100 times.
[tex]\text{Size =100}\times Actual\text{ length}[/tex][tex]\frac{\text{Size }}{100}\text{=}Actual\text{ length}[/tex]Substitute Size=0.75, we get
[tex]\frac{0.75}{100}=\text{Actual length}[/tex][tex]\text{Actual length=0.0075 cm}[/tex]Hence the actual length of the specimen is 0.0075cm.
Question: Jack has an old scooter. He wants to sell it for 60% off the current price. The market price is $130.
What should his asking price be? Explain your reasoning.
The asking price for the scooter is $78.
How to calculate the price?From the information, Jack has an old scooter. He wants to sell it for 60% off the current price and the market price is $130.
In this case, the asking price will be:
= Percentage × Market price
= 60% × $130
= 0.6 × $130
= $78
The price is $78.
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From a point on the North Rim of the Grand Canyon, the angle of depression to a pointon the South Rim is 2. From an aerial photo, it can be determined that the horizontaldistance between the two points is 10 miles. How many vertical feet is the South Rimbelow the North Rim (nearest whole foot).
Question:
Solution:
The situation of the given problem can be drawn in the following right triangle:
This problem can be solved by applying the trigonometric identities as this:
[tex]\tan(2^{\circ})=\frac{y}{10}[/tex]solving for y, this is equivalent to:
[tex]y\text{ = 10 tan\lparen2}^{\circ}\text{\rparen=0.349 miles}[/tex]now, 0.349 miles is equivalent to 1842.72 feet. Now, this number rounded to the nearest whole foot is equivalent to 1843.
Then, the correct answer is:
1843 feet.
Estimate the fraction 3/8 by rounding to the nearest whole or one-half
SOLUTION
The fraction given is
[tex]\frac{3}{8}[/tex]To round-off the fraction, we need to convert it to decimal number
[tex]\frac{3}{8}\text{ to decimal is }[/tex]Hence
The estimated fraction in decimal will be
[tex]0.375=0.4=\frac{4}{10}=\frac{2}{5}[/tex]Answer: 1/2 is the answer
hope this helped :)
Step-by-step explanation:
Joaquin has been working on homework for 3 1/2 hours. If each assignment takes him of 1/4 an hour, how many assignments has he completed? Select one OA. 13 O B. 14 OC. 15 OD. 16
The total amount of time he has spent doing homework:
[tex]3\frac{1}{2}[/tex]And the time it takes to complete each assignment:
[tex]\frac{1}{4}[/tex]To find the number of assignments he has completed in that time, we divide the total time 3 1/2 by the time for each assignment 1/4:
[tex]3\frac{1}{2}\div\frac{1}{4}[/tex]To make this division we need to convert 3 1/2 to a fraction as follows:
[tex]3\frac{1}{2}=\frac{3\times2+1}{2}=\frac{6+1}{2}=\frac{7}{2}[/tex]We multiply the whole number by the denominator 3x2 and we add to that the numerator 1, and divide that by the original denominator 2.
Now instead of 3 1/2 we use 7/2 for our division:
[tex]\frac{7}{2}\div\frac{1}{4}[/tex]And we use the formula to divide two fractions, which is:
[tex]\frac{a}{b}\div\frac{c}{d}=\frac{a\times d}{b\times c}[/tex]Applying this to our division:
[tex]\frac{7}{2}\div\frac{1}{4}=\frac{7\times4}{2\times1}[/tex]Solving the operations:
[tex]\frac{7}{2}\div\frac{1}{4}=\frac{28}{2}=14[/tex]Answer: 14 assignments
use the first derivative test to classify the relative extrema. Write all relative extrema as ordered pairs
The given function is
[tex]f(x)=-10x^2-120x-5[/tex]First, find the first derivative of the function f(x). Use the power rule.
[tex]\begin{gathered} f^{\prime}(x)=-10\cdot2x^{2-1}-120x^{1-1}+0 \\ f^{\prime}(x)=-20x-120 \end{gathered}[/tex]Then, make it equal to zero.
[tex]-20x-120=0[/tex]Solve for x.
[tex]\begin{gathered} -20x=120 \\ x=\frac{120}{-20} \\ x=-6 \end{gathered}[/tex]This means the function has one critical value that creates two intervals.
We have to evaluate the function using two values for each interval.
Let's evaluate first for x = -7, which is inside the first interval.
[tex]f^{\prime}(-7)=-20(-7)-120=140-120=20\to+[/tex]Now evaluate for x = -5, which is inside the second interval.
[tex]f^{\prime}(7)=-20(-5)-120=100-120=-20\to-[/tex]As you can observe, the function is increasing in the first interval but decreases in the second interval. This means when x = -6, there's a maximum point.
At last, evaluate the function when x = -6 to find the y-coordinate and form the point.
[tex]\begin{gathered} f(-6)=-10(-6)^2-120(-6)-5=-10(36)+720-6 \\ f(-6)=-360+720-5=355 \end{gathered}[/tex]Therefore, we have a relative maximum point at (-6, 355).In a right triangle, cos (2x) = sin (8x + 5)'. Find the smaller of the triangle's two
acute angles.
According to the given problem,
[tex]\cos (2x)^{\circ}=\sin (8x+5)^{\circ}[/tex]Consider the formula,
[tex]\sin (90-\theta)=\cos \theta[/tex]Apply the formula,
[tex]\sin (90-2x)=\sin (8x+5)[/tex]Comparing both sides,
[tex]\begin{gathered} 90-2x=8x+5 \\ 8x+2x=90-5 \\ 10x=85 \\ x=\frac{85}{10} \\ x=8.5 \end{gathered}[/tex]Obtain the value of the two angles,
[tex]\begin{gathered} 2x=2(8.5)=17 \\ 8x+5=8(8.5)+5=73 \end{gathered}[/tex]It is evident that the smaller angle is 17 degrees, and the larger angle is 73 degrees.
Thus, the required value of the smaller acute angle of the triangle is 17 degrees.
Are the ratios 1:6 and 14:18 equivalent
I don't know I need points you are loved tho I don't know I need points you are loved tho
Express the fraction as a percentage use the bar notation if necessary
We have to express the fraction as a percentage.
We can think of the percentage as a fraction with denominator 100.
This means that 25% is equivalent to 25/100.
This is because 100% is the unit, then 100/100 = 1.
We can use this to convert a fraction to a percentage by transforming the denominator into 100.
In this case, the fraction is 2/5.
The denominator is 5, so to convert it to 100 we have to multiply it by 100/5 = 20.
Then, we apply this both to the numerator and denominator:
[tex]\frac{2}{5}*\frac{20}{20}=\frac{2*20}{5*20}=\frac{40}{100}=40\%[/tex]Answer: 2/5 = 40%
1. According to a recent poll, 4,060 out of 14,500 people in the United States were clas-sified as obese based on their body mass index. What’s the relative frequency of obe-sity according to this poll?
The relative frequency of obesity is the ratio between obese people in the sample and total people in the sample:
[tex]\frac{4060}{14500}=0.28[/tex]The relative frequency is 0.28
c(t)=2(t-4)(t+1)(t-6)
Question: find the x- or t intercepts of the polynomial function:
c(t)=2(t-4)(t+1)(t-6).
Solution:
the t-intercept (zeros of the function) of the given polynomial function occurs when c (t) = 0, that is when:
[tex]c(t)\text{ = 0 = }2\mleft(t-4\mright)\mleft(t+1\mright)\mleft(t-6\mright)[/tex]this can only happen when any of the factors of the polynomial are zero:
t-4 = 0, that is when t = 4
t + 1 = 0 , that is when t = -1
and
t-6 = 0, that is when t = 6.
then, we can conclude that the t-intercept (zeros of the function) of the given polynomial are
t = 4, t = -1 and t = 6.
Round to the nearest tenthRound to the nearest hundredthRound to the nearest whole number
Round to the nearest tenth
15. 7.953 is equal to 8.0 rounded to the nearest tenth.
16. 4.438 is equal to 4.4 rounded to the nearest tenth.
17. 5.299 is equal to 5.3 rounded to the nearest tenth.
18. 8.171 is equal to 8.2 rounded to the nearest tenth.
Round to the nearest hundredth
19. 5.849 is equal to 5.85 rounded to the nearest hundredth.
20. 4.484 is equal to 4.48 rounded to the nearest hundredth.
21. 0.987 is equal to 0.99 rounded to the nearest hundredth.
22. 0.155 is equal to 0.16 rounded to the nearest hundredth.
Round to the nearest whole number
23. 98.55 is equal to 99 rounded to the nearest whole number.
24. 269.57 is equal to 270 rounded to the nearest whole number.
25. 14.369 is equal to 14 rounded to the nearest whole number.
26. 23.09 is equal to 23 rounded to the nearest whole number.
What is the inmage of A(2, 1) after reflecting it across x =4 and then across the x-axis? a.16. 1)b. (6,-1) c(-6,-1) d. -6, 1)
If we reflect the point A (2, 1) across x=4, the point will be at (6, 1). Then if we reflect it again across the x axis, the point will be at (6, -1)
Answer: (6, -1)
(G.12, 1 point) Which point lies on the circle represented by the equation (x-4)2 + (y - 2)2 = 72? O A. (-1,4) B. (8,3) O C. (9,0) O D. (-2, 2)
To know if a point lies on a circle you use the (x,y) of each point in the equation and prove it that correspond to a mathematical congruence.
A. (- 1 , 4)
[tex]\begin{gathered} (-1-4)^2+(4-2)^2=7^2 \\ -5^2+2^2=49 \\ 25+4=49 \\ 29=49 \end{gathered}[/tex]As 29 is not equal to 49, this point doesn't lie in the circle
B. ( 8 , 3)
[tex]\begin{gathered} (8-4)^2+(3-2)^2=7^2 \\ 4^2+1^2=49 \\ 16+1=49 \\ 17=49 \end{gathered}[/tex]As 17 is not equal to 49, this point doesn't lie in the circle
C. (9 , 0)
[tex]\begin{gathered} (9-4)^2+(0-2)^2=7^2 \\ 5^2+(-2)^2=49 \\ 25+4=49 \\ 29=49 \end{gathered}[/tex]As 29 is not equal to 49, this point doesn't lie in the circle
D. (-2 , 2)
[tex]\begin{gathered} (-2-4)^2+(2-2)^2=7^2 \\ -6^2+0^2=49 \\ 36=49 \end{gathered}[/tex]As 36 is not equal to 49, this point doesn't lie in the circle
None of the points lies on the circle.The next graph represents the circle and the 4 given points:
what is the slope of y=x-7
The given equation is in slope-intercept form, which means the coefficient of x is the slope.
Therefore, the slope is 1.How to know if you have the slope-intercept form?
The slope intercept-form is y = mx + b, as you can observe, the variable y is completely isolated, and the other side of the equation has two terms. Whenever you have a linearGrea equation like this, it means you have a slope-intercept form where the slope is m, and the y-intercept is b.
Which of the following lines is parallel to the line y= -3/2x-4?
ANSWER:
1st option: y = -3/2x + 5
STEP-BY-STEP EXPLANATION:
We have that the equation in its slope-intercept form is the following:
[tex]\begin{gathered} y=mx+b \\ \\ \text{ where m is the slope and y-intercept is b} \end{gathered}[/tex]Two lines are parallel when the slope is the same, therefore, the line parallel to this line must have a slope equal to -3/2.
We can see that option 2 is the same line, therefore, they cannot be parallel, so the correct answer is 1st option: y = -3/2x + 5
on which of the following lines does the point (1,6) lie
It is required to determine which line the point (1,6) lies on.
To do this, substitute the point into each equation and check which of the equations it satisfies.
Check the first equation:
[tex]\begin{gathered} y=x+5 \\ \text{ Substitute }(x,y)=(1,6): \\ \Rightarrow6=1+5 \\ \Rightarrow6=6 \end{gathered}[/tex]Since the equation is true, it follows that the point lies on the line.
Check the second equation:
[tex]\begin{gathered} y=-x+7 \\ \text{ Substitute }(x,y)=(1,6): \\ \Rightarrow6=-1+7 \\ \Rightarrow6=6 \end{gathered}[/tex]Since the equation is true, it follows that the point also lies on the line.
Check the third equation:
[tex]\begin{gathered} y=2x-1 \\ \text{ Substitute }(x,y)=(1,6): \\ \Rightarrow6=2(1)-1 \\ \Rightarrow6=1 \end{gathered}[/tex]Notice that the equation is not true. Hence, the point does not lie on the line.
So the given point only lies on lines a and b.
The answer is option D.
the post office and city hall are marked on a coordinate plane. write the equation of the line in slope intercept form that passes through these two points.
The given points are (1,6) and (-5,-3).
First, we have to find the slope
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Replacing given points, we have
[tex]m=\frac{-3-6}{-5-1}=\frac{-9}{-6}=\frac{3}{2}[/tex]Now, we use the point-slope formula
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-6=\frac{3}{2}(x-1) \\ y=\frac{3}{2}x-\frac{3}{2}+6 \\ y=\frac{3}{2}x+\frac{-3+12}{2} \\ y=\frac{3}{2}x+\frac{9}{2} \end{gathered}[/tex]Therefore, the equation would be y = 3/2x + 9/2Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.At a candy store, Carla bought 6 pounds of jelly beans and 6 pounds of gummy worms for $108. Meanwhile, Rachel bought 6 pounds of jelly beans and 1 pound of gummy worms for $63. How much does the candy cost?
Given:
Let x and y be the cost of 1 pound of Jelly beans and 1 pound of gummy worms.
[tex]\begin{gathered} 6x+6y=108 \\ x+y=18\ldots\text{ (1)} \end{gathered}[/tex][tex]6x+y=63\ldots\text{ (2)}[/tex]Subtract equation (1) from equation(2)
[tex]\begin{gathered} 6x+y-x-y=63-18 \\ 5x=45 \\ x=9 \end{gathered}[/tex]Substitute x=9 in equation(1)
[tex]\begin{gathered} 9+y=18 \\ y=18-9 \\ y=9 \end{gathered}[/tex]Cost of 1 pound of Jelly beans is $9
Cost of 1 pound of gummy worms is $9
Evaluate x^1 - x^-1 + x^0 for x = 2.
The value of x^1 - x^-1 + x^0 for x = 2 is 5
we need to evaluate x¹ - x⁻¹ + x⁰
A positive exponent tells us how many times to multiply a base number, while a negative exponent tells us how many times to divide a certain base number. Negative exponents can be re written. x⁻ⁿ as 1 / x
x¹ - 1/x + x⁰
(any number or variable which is raised to the power zero is considered as 1)
x¹ - 1/x + 1
x² - 1 + x
x² + x - 1
The value of this expression for x = 2
2² + 2 - 1
4 + 2 - 1
5
Therefore,the value of x^1 - x^-1 + x^0 for x = 2 is 5
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You are hiking and are trying to determine how far away the nearest cabin is, which happens to be due north from your current position. Your friendr walks 220 yards due west from your position and takes a bearing on the cabin of N 21.2°E. How far are you from the cabin?A. 638 yardsB. 608 yardsC. 567 yardsD. 589 yards
In the given right triangle
we have that
[tex]\begin{gathered} tan(21.2^o)=\frac{220}{x}\text{ -----> by TOA} \\ \\ solve\text{ for x} \\ \\ x=\frac{220}{tan(21.2^o)} \\ \\ x=567\text{ yrads} \end{gathered}[/tex]The answer is option C