-11. Given points (x, y) and (x2, y2), derive the two-point form of a line. , , , 10. 13. Given that a line is parallel to the x-axis through (x, y), derive the parallel to x-axis form a line.
Answers
11. Given the two points (x1, y1) & (x2, y2) we will have the following line and we derivate it:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex][tex]y-y_1=m(x-x_1)\Rightarrow y=mx-mx_1+y_1[/tex]It's derivative is:
[tex]\frac{\delta y}{\delta x}=m=\frac{y_2-y_1}{x_2-x_1}[/tex]This is since the derivative of constants is 0 and the only variable accompanied m. This is proof that the derivative of a function can be interpreted as the slope of the function at that point.
13. If we have that the line is parallel to the x-axis and passes through the point (x1, y1), we will have that the line is a constant function, so when we derivate no matter the point, it will be equal to 0.
That is:
[tex]y=x_1[/tex][tex]\frac{\delta y}{\delta x}=0[/tex]***Explanation:
point 11:
Since we are given two points (x1, y1) & (x2, y2), we will always have that the slope of the line that passes through those points will always have the form:
Related Questions
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?
Answers
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
what is the slope of y=x-7
Answers
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.
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?
Answers
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
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?
Answers
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.
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)
Answers
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)
Answers
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:
6.[–/1 Points]DETAILSALEXGEOM7 8.3.006.MY NOTESASK YOUR TEACHERIn a regular polygon, each interior angle measures 120°. If each side of the regular polygon measures 4.6 cm, find the perimeter of the polygon in centimeters. cm
Answers
Given:
Measure of each interior angle = 120 degrees
Length of each side = 4.6 cm
Let's find the perimeter of the polygon.
Since the measure of each interior angle is 120 degrees, let's find the number of sides of the polygon using the formula below:
[tex]120=\frac{(n-2)*180}{n}[/tex]Let's solve for n.
We have:
[tex]\begin{gathered} 120n=(n-2)*180 \\ \\ 120n=180n-180(2) \\ \\ 120n=180n-360 \\ \\ 180n-120n=360 \\ \\ 60n=360 \end{gathered}[/tex]Divide both sides by 60:
[tex]\begin{gathered} \frac{60n}{60}=\frac{360}{60} \\ \\ n=6 \end{gathered}[/tex]Therefore, the polygon has 6 sides.
To find the perimeter, apply the formula:
Perimeter = number of sides x length of each side
Perimeter = 6 x 4.6
Perimeter = 27.6 cm
Therefore, the perimeter of the polygon is 27.6 cm
ANSWER:
27.6 cm
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.
Answers
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 + 9/2the problem was sent in a picture
Answers
The vertex of the given parabola is (h, k)=(0,0).
(x, y)=(2, -4) is a point on the parabola.
The vertex form of a parbola is,
[tex]y=a(x-h)^2+k\text{ ------(1)}[/tex]Here, (h, k) is the vertex of parabola.
Put h=0, k=0, x=2 and y=-4 in the above equation.
[tex]\begin{gathered} -4=a(2-0)+0 \\ \frac{-4}{2}=a \\ -2=a \end{gathered}[/tex]Put a=-2, h=0, k=0 in equation (1) to find the function.
[tex]y=-2x^2[/tex]Put y=0 to obtain a quadratic function and solve for x.
[tex]\begin{gathered} 0=-2x^2 \\ x=0 \end{gathered}[/tex]So, there is only one solution to the graph.
Short cut:
Since the parabola touches the x axis when the x intercept is zero, the solution of the quadratic function of the parabola is x=0. So, there is only one solution to the graph.
The ordered pairs are graphed from the table in question 27 which of the following lines show the correct relationship
Answers
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
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)
Answers
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]use the first derivative test to classify the relative extrema. Write all relative extrema as ordered pairs
Answers
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).Evaluate x^1 - x^-1 + x^0 for x = 2.
Answers
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
Answers
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 CRound to the nearest tenthRound to the nearest hundredthRound to the nearest whole number
Answers
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.
A line that includes the point (10,5)and has a slope of 1.What is its equation in slope intercept form?
Answers
The line equation is y = x - 5
EXPLANATION
Given:
Point (10, 5)
x=10 and y=5
slope (m)=1
We need to first find the intercept(b).
Substitute x=10 , y=5 and m=1 into y=mx + b and solve for intercept(b).
That is;
5 = 1(10) + b
5 = 10 + b
5 - 10 = b
-5 = b
Form the equation by substituting m=1 and b=-5.
Hence, the line equation is y = x - 5
Convert the following Quadratic Equations from Vertex Form to Standard Form.
Answers
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
Answers
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
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
Answers
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
on which of the following lines does the point (1,6) lie
Answers
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.
Write 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?
Answers
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
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.
Answers
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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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?
Answers
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
Express the fraction as a percentage use the bar notation if necessary
Answers
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%
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).
Answers
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.
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
Answers
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
c(t)=2(t-4)(t+1)(t-6)
Answers
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.
Marshall has a rectangular garden that he wants to enclose with a fence. To calculate the perimeter, he used the expression below; where w represents the width and I represents the length of the garden. 2w + 2L Which other expression could Marshall use to calculate the perimeter?A. wLB.2wLC.2(w+2L)D.2(w+L)
Answers
by factoring the expression, you can also write
[tex]\begin{gathered} 2w+2l \\ 2(w+l) \end{gathered}[/tex]hi how do you find the area of this figuer
Answers
we can divide the figure in two parts, a triangule and a square. So the area is sum of the area fo each figure
[tex]A=3\cdot3+\frac{2\cdot3}{2}=9+3=12[/tex]therefore the area is 12m^2
In a right triangle, cos (2x) = sin (8x + 5)'. Find the smaller of the triangle's two
acute angles.
Answers
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.