Solve the quadratic equation by the square root method and write the solutions in radical forms simplify the solutions
Answers
Given the following quadratic equation:
[tex](8x+20)^2=50[/tex]We will solve the equation by the square root method.
Taking the square root of both sides:
[tex]\begin{gathered} \sqrt{(8x+20)^2}=\pm\sqrt{50} \\ \\ 8x+20=\pm\sqrt{50} \\ Note:\sqrt{50}=\sqrt{25*2}=5\sqrt{2} \\ \\ 8x+20=\pm5\sqrt{2} \end{gathered}[/tex]Subtract 20 from both sides
[tex]\begin{gathered} 8x+20-20=-20\pm5\sqrt{2} \\ 8x=-20\pm5\sqrt{2} \end{gathered}[/tex]Divide both sides by 8
[tex]x=\frac{-20\pm5\sqrt{2}}{8}[/tex]So, the answer will be:
[tex]x=\frac{-20+5\sqrt{2}}{8};or;\frac{-20-5\sqrt{2}}{8}[/tex]Related Questions
the perimeter of a rectangle is 222 millimeters. if the length is 60 millimeters .what is the width?
Answers
Given :
The perimeter of the rectangle = p = 222 mm
And the length of the rectangle = l = 60 mm
Let the width = w
The perimeter of the rectangle = p = 2 ( l + w )
so,
[tex]\begin{gathered} p=2\cdot(l+w) \\ 222=2(60+w) \end{gathered}[/tex]solve the equation to find w:
[tex]\begin{gathered} \frac{222}{2}=60+w \\ 111=60+w \\ 111-60=w \\ \\ w=51 \end{gathered}[/tex]The answer is : the width of the rectangle = 51 mm
Find the six trigonometric functions of 0 in simplest radical form. Rationalize all fractions.
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step 1
Find the hypotenuse of the right triangle
applying Pythagorean theorem
c^2=2^2+3^2
c^2=4+9
[tex]c=\sqrt[]{13}[/tex]step 2
Find sin(theta)
we have
[tex]\sin (\theta)=\frac{2}{\sqrt[]{13}}[/tex]simplify
[tex]\sin (\theta)=\frac{2}{\sqrt[]{13}}=\frac{2\sqrt[\square]{13}}{13}[/tex]opposite side divided by the hypotenuse
step 3
Find cos(theta)
[tex]\cos (\theta)=\frac{3}{\sqrt[\square]{13}}[/tex]adjacent side divided by the hypotenuse
simplify
[tex]\cos (\theta)=\frac{3}{\sqrt[\square]{13}}=\frac{3\sqrt[]{13}}{13}[/tex]step 4
find tan(theta)
[tex]\tan (\theta)=\frac{2}{3}[/tex]opposite side divided by the adjacent side
step 5
find cot(theta)
[tex]\cot (\theta)=\frac{1}{\tan (\theta)}=\frac{3}{2}[/tex]adjacent side divided by the opposite side
step 6
Find sec(theta)
[tex]\sec (\theta)=\frac{1}{\cos (\theta)}=\frac{\sqrt[]{13}}{3}[/tex]hypotenuse divided by the adjacent side
step 7
Find csc(theta)
[tex]\csc (\theta)=\frac{1}{\sin (\theta)}=\frac{\sqrt[]{13}}{2}[/tex]hypotenuse divided by the opposite side
12+Which statement is true regarding the functions on thegraph?10-86.fo42Of(-3) = g(-4)Of(-4) = 9(-3)O f(-3) = g(-3)Of(-4) = g(4)-5-4-3-1145 6-46--91Mark this and returnSave and ExitSubmit
Answers
Both functions intersect at the point (-3,-4)
This means that when x=-3 f(x)=-4 and g(x)=-4
You can express this as
[tex]\begin{gathered} f(-3)=-4 \\ g(-3)=-4 \\ So\text{ that} \\ f(-3)=g(-3) \end{gathered}[/tex]The correct statement is the third one.
Consider the expressions 3x(x - 2) + 2 and 2x2 + 3x – 16.
Answers
For x = 3, each expression has a value of 11. For x = 6, each expression has a value of 74. These results suggest that the expressions are equivalent
Explanation:The given expressions are:
3x(x - 2) + 2 and 2x² + 3x - 16
Substitute x = 3 into 3x(x - 2) + 2
3(3)(3 - 2) + 2
= 9(1) + 2
= 9 + 2
= 11
Substitute x = 3 into 2x² + 3x - 16
2(3)² + 3(3) - 16
2(9) + 9 - 16
18 + 9 - 16
= 11
Substitute x = 6 into 3x(x - 2) + 2
3(6)(6 - 2) + 2
18(4) + 2= 74
Substitute x = 6 into 2x² + 3x - 16
2(6)² + 3(6) - 16
2(36) + 18 - 16
72 + 2
= 74
For x = 3, each expression has a value of 11. For x = 6, each expression has a value of 74. These results suggest that the expressions are equivalent
[tex]7.5 + 5f + 16.2 + 2f[/tex]simplified expression
Answers
given :
7.5 + 5f + 16.2 + 2f =
combine like terms
So,
(7.5 + 16.2) + ( 5f + 2f ) = 23.7 + 7f
So,
the simplified expression = 23.7 + 7f
Find the area of aninscribed circle in a squarehaving 5 inch sides.
Answers
Explanation
Given: An inscribed circle in a square of side 5 inches.
We are required to determine the area of the inscribed circle.
This is achieved thus:
We know that the side of the square equals the diameter of the circle. Therefore, we have:
[tex]\begin{gathered} D=5in \\ \\ Area\text{ }of\text{ }circle=\frac{\pi D^2}{4} \\ Area=\frac{\pi\times5^2}{4} \\ Area=\frac{25}{4}\pi in^2 \end{gathered}[/tex]Hence, the answer is:
[tex]\begin{equation*} \frac{25}{4}\pi in^2 \end{equation*}[/tex]The graph represents two linear relations. What is the point of intersection of the two lines?
Answers
Solution
The graph below shows the point of intersection
Thus, the point of intersection is (20, 32)
Option C
if m<10=77, m<7=47 and m<16=139, find the measure of the missing angle m<8=?
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Line a and line b are two parallel lines and a cut by the transverse c and d.
But to obtain what angle 8, we will only consider the transverse c.
Below is a plot:
The measure of angle 10 = The measure of angle 8
(Reason: They are alternate angles and alternate angles are equal)
Where does the line y = 6x - 13 cross the Y axis?
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A function cross the y-axis when the value of x is 0.
[tex]y=6x-13[/tex]Find the value of y, when x is 0:
[tex]\begin{gathered} y=6(0)-13 \\ y=-13 \end{gathered}[/tex]Then, the line y=6x-13 crosses the y-axis at y= -13i was hoping someone could help me find the answers
Answers
A) We will determine the annual rate of change between 1991 and 2006 as follows:
We replace the values in the expression and solve for r, thatt is:
[tex]15000=34000(1+r)^{15}\Rightarrow\frac{15}{34}=(1+r)^{15}[/tex][tex]\Rightarrow\sqrt[15]{\frac{15}{34}}=1+r\Rightarrow\Rightarrow r=\sqrt[15]{\frac{15}{34}}-1[/tex][tex]\Rightarrow r=-0.0530926\ldots\Rightarrow r\approx-0.0531[/tex]So, the annual rate of change between 1991 and 2006 was of approximately -5%.
B) The correct answer in percentage form is -5.31%.
C) If we assume that te car value continues to drop by the same percentage, for the year 210 would have been:
[tex]S\approx34000(1-0.0531)^{19}\Rightarrow S\approx12057.49614[/tex]So, the value would end up being approximately $12057.5.
Select the correct answer. Consider these functions: $($)=3.13 + 2 g(t)=V Which statements, if any, are true about these functions? 1. The function fg(x)) = x for all real x. II. The function & x)) = x for all real x. III. Functions fand gare inverse functions. O A. I only B. ll only C.I, II, and III D. None of the statements are true.
Answers
Let's verify every statement:
I.
[tex]f(g(x))=3(\sqrt[3]{\frac{x-2}{3}})^3+2=3(\frac{x-2}{3})+2=x-2+2=x[/tex]So, the first one is true.
II:
[tex]g(f(x))=\sqrt[3]{\frac{(3x^3+2)-2}{3}}=\sqrt[3]{x^3}=x[/tex]So, this one is true
III.
Since:
[tex](fog)(x)=(gof)(x)=x[/tex]We can conclude that f(x) and g(x) are inverse functions. So, this statements is also true
Answer:
C.I, II, and III
A number is decreased by 50%, and the result is increased by 50% to get99. What is the original number? *
Answers
Let the original number be
[tex]=x[/tex]The number decreased by 50% means that we will first calculate 50% of the original number and then subtract it from the original number x
[tex]\begin{gathered} 50\text{ \% of the original number will give} \\ =50\text{ \% of x} \\ =\frac{50}{100}\times x=0.5x \end{gathered}[/tex]Then, let's subtract the decreased value from the original value
The remaining value will be
[tex]\begin{gathered} \text{remaining value} \\ =x-0.5x \\ =0.5x \end{gathered}[/tex]The result is increased by 50% to get
[tex]\begin{gathered} increasedvalue=50\text{ \% of the remaining value} \\ =\frac{50}{100}\times0.5x \\ =0.5\times0.5x \\ =0.25x \end{gathered}[/tex]The new value after increasing by 50 % can be gotten by adding the increased value to the remaining value
[tex]\begin{gathered} \text{the new value will be} \\ =0.5x+0.25x \\ =0.75x \end{gathered}[/tex]From the question, the new value =99
Therefore, the original value will be
[tex]\begin{gathered} 0.75x=99 \\ \text{divide both sides by 0.75} \\ \frac{0.75x}{0,75}=\frac{99}{0.75} \\ x=132 \end{gathered}[/tex]Hence,
The original number = 132
5-10c + 8c = 45 - 30What is the value of c?
Answers
Collecting like terms, we have:
-10c + 8c = 45 - 30 - 5
-2c = 15 -5
-2c = 10
Dividing both side by -2
[tex]undefined[/tex]Solve and graph the solutions to the inequalities below using any of the method from class: a table, a graph,"undoing," or algebraic operations.(You may need to use a separate sheet of notebook paper or graph paper.)15x < 45
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Step 1
Given;
[tex]15x<45[/tex]Required to solve algebraically or by graph paper.
Step 2
Solve algebraically
[tex]\begin{gathered} 15x<45 \\ \text{Divide bot sides by 15} \\ \frac{15x}{15}<\frac{45}{15} \end{gathered}[/tex]Simplify;
[tex]x<3[/tex]Step 3
Graph the inequality
The times, in seconds, for 8 athletes who ran a 100 m sprint on the sametrack are shown below. Find the mean time (seconds).10.2, 10.8, 10.9, 10.3, 10.2, 10.4, 10.1, and 10.4
Answers
Given: Time records for 8 athletes who ran 100 m sprint is 10.2, 10.8, 10.9, 10.3, 10.2, 10.4, 10.1, and 10.4 (in seconds).
Required: To determine the mean time.
Explanation: The mean of n number of observations is the ratio of the sum of all the observations to the number of observations.
Thus,
[tex]Mean=\frac{10.2+10.8+10.9+10.3+10.2+10.4+10.1+10.4}{8}[/tex]Further solving-
[tex]\begin{gathered} Mean=\frac{83.3}{8} \\ =10.4125 \end{gathered}[/tex]Final Answer: The mean time is 10.4125 seconds.
Arbuckle County has an area of 1,424 square miles with a population of 854,786 people.a) Determine the population density. Round to the nearest tenth.b.) What is the population density in people per square kilometers (Recall 1 km = 0.62 miles). Round to the nearest tenth.______people per sqaure kilo c.) Baxter County has an area of 2,608 square miles. How many people would be in Baxter County, if the population density were the same as Arbuckle County? Round to the nearest person._____people.
Answers
Given that Arbuckle County has an area of:
[tex]1,424\text{ }square\text{ }miles[/tex]And has a population of:
[tex]854,786\text{ }people[/tex]a) You need to use the formula for calculating the Population Density:
[tex]Population\text{ }Density=\frac{Number\text{ }of\text{ }people}{Land\text{ }area}[/tex]Therefore, by substituting values into the formula and evaluating, you get:
[tex]Population\text{ }Density=\frac{854786\text{ }people}{1424\text{ }square\text{ }miles}[/tex][tex]Population\text{ }Density\approx600.3\frac{people}{square\text{ }mile}[/tex]b) You need to convert the area from square miles to square kilometers.
You know that:
[tex]1\text{ }km=0.62\text{ }mi[/tex][tex](1\text{ }km)^2=(0.62\text{ }mi)^2\Rightarrow1km^2=0.3844\text{ }mi^2[/tex]Now you can set up the conversion:
[tex](1424\text{ }mi^2)(\frac{1\text{ }km^2}{0.3844mi^2})[/tex]Evaluating, you get:
[tex]\approx3704.5\text{ }km^2[/tex]Using the formula for Population Density, you get:
[tex]Population\text{ }Density=\frac{854786\text{ }people}{3704.5\text{ }square\text{ }kilometers}[/tex][tex]Population\text{ }Density\approx230.7\frac{people}{square\text{ }kilometer}[/tex]c) You know that Baxter County has an area of 2,608 square miles, if the population density were the same as Arbuckle County, you can substitute values into the formula used before and solve for the number of people:
[tex]600.3=\frac{People}{2608}[/tex][tex]\begin{gathered} 600.3\cdot2608=People \\ People\approx1,565,580 \end{gathered}[/tex]Hence, the answers are:
a)
[tex]600.3\text{ }people\text{ }per\text{ }square\text{ }mile[/tex]b)
[tex]230.7\text{ }people\text{ }per\text{ }square\text{ }kilometer[/tex]c)
[tex]1,565,580\text{ }people[/tex]In this diagram, circle A has radius = 4.9 and DC = 8.1. BC is tangent line, calculate the distance of BC.
Answers
We have a tangent line that means the m
we have
AB= radius=4.9
AC=AD+DC=4.9+8.1=13
Because we have that m
[tex]c^2=a^2+b^2[/tex]where
a=AB=4.9
b=?=x
c=AC=13
we substitute the values
[tex]\begin{gathered} b=\sqrt[]{13^2-4.9^2} \\ b=\sqrt[]{169-24.01} \\ b=\sqrt[]{144.99} \\ b=12.04 \end{gathered}[/tex]For the functions f(x)=6x−5 and g(x)=6x2−3x, find (f∘g)(x).Question 6 options:(f∘g)(x) = 36x2−18x−5(f∘g)(x) = 24x2−18x−5(f∘g)(x) = 36x−18x−5(f∘g)(x) = 36x2−18x−2
Answers
36x²-18x- 5 is the (f∘g)(x) for the functions
How to determine the (f∘g)(x) for the functions?A function is a relationship between inputs where each input is related to exactly one output. Every function has a domain and codomain or range. A function is generally denoted by f(x) where x is the input
Given: f(x)= 6x−5 and g(x)=6x²−3x
In order to determine (f∘g)(x), first, g(x) = 6x²−3x into (f∘g)(x):
(f∘g) = f(6x²−3x)
Then substitute x = 6x²−3x into f(x) = 6x−5:
(f∘g) = 6(6x²−3x)−5 = 36x²-18x- 5
Therefore, (f∘g) of the functions is 36x²-18x- 5. So the 1st option is the answer
Learn more about function on:
brainly.com/question/29233399
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Convert 205° to radians. Radians = degrees timesPie/180
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The number in degree is 205º
The expression to convert degree into radians is by multiply the given angle degree by π/180.
[tex]\begin{gathered} \text{Radians}=\frac{180}{\Pi}\times205 \\ \text{Radians }=3.578 \end{gathered}[/tex]205º is written as 3.578 radians
I need to know the steps and the answer please and thank you.
Answers
Solution:
Let the following function:
[tex]g(n)=n^2+2[/tex]if we evaluate n=3 in the previous equation, we get:
[tex]g(3)=3^2+2[/tex]this is equivalent to
[tex]g(3)=\text{ (3x3) +2 = 9+2 = 11}[/tex]that is:
[tex]g(3)=11[/tex]so that, we can conclude that the correct answer is:
[tex]g(3)=11[/tex]hey tutor do you know this? it says convert 6/20 to a decimal
Answers
To convert a fraction to decimal,we have to do the long division of the numerator 6 by the denominator 20 to get the following:
therefore, the decimal form is 0.3
Jose solved the following problem:2x +4< 98x +4< 49x < 45What is Jose's mistake?
Answers
In the first step, Joe divided both sides of the inequality by 2, but forgot to divide the constant 4. The correct approach would be this:
[tex]\begin{gathered} \frac{1}{2}(2x+4<98) \\ \Rightarrow x+2<49 \end{gathered}[/tex]A bicyclist rides 11.2 kilometerseast and then 5.3 kilometers southWhat is the angle of thebicyclist's resultant vector?
Answers
Given
A = 11.2 km East
B = 5.3 km south
Find
Angle of the resultant vector
Explanation
Angle of the resultant vector is given by
[tex]tan\emptyset=\frac{Bsin\alpha}{A+Bcos\alpha}[/tex]Here angle between both the vectors is 90 degree
[tex]\begin{gathered} tan\emptyset=\frac{Bsin\alpha}{A+Bcos\alpha}=\frac{11.2(sin90)}{5.3+11.2(cos90)} \\ tan\emptyset=\frac{11.2}{5.3} \\ tan\emptyset=2.11320755 \\ \emptyset=64.67degrees \end{gathered}[/tex]So Angle made from origin = 270+64.67
= 334.67 degree = 334.7 degree(approx)
Final Answer
Angle of resultant Vector = 334.7 degree
x = y +10x = 2y + 3(17,7)is it consistent and independent
Answers
Simplify the equation to obtain the y coordinate.
[tex]\begin{gathered} y+10=2y+3 \\ 2y-y=10-3 \\ y=7 \end{gathered}[/tex]Substitute 7 for y in the equaation x = y +10 to obtain the value of x.
[tex]\begin{gathered} x=7+10 \\ =17 \end{gathered}[/tex]So solution of equations is (17,7).
The equations consist of the solution, so equation is consistent and equation has only one solution those equation is independent.
Thus it is consistent and independent.
Hello, can you please help me solve question 3 !!
Answers
Use the next trigonometric rules:
[tex]\begin{gathered} \cos 2t=\cos ^2t-\sin ^2t \\ \\ \sin ^2t=1-\cos ^2t \end{gathered}[/tex]Use Cos2t
[tex]\cos ^2t-\sin ^2t-2\sin ^2t=0[/tex]Combine similar terms:
[tex]\cos ^2t-3\sin ^2t=0[/tex]Use sin²t:
[tex]\begin{gathered} \cos ^2t-3(1-\cos ^2t)=0 \\ \\ \cos ^2t-3+3\cos ^2t=0 \end{gathered}[/tex]Combine similar terms:
[tex]4\cos ^2t-3=0[/tex]Add 3 in both sides of the equation:
[tex]\begin{gathered} 4\cos ^2t-3+3=0+3 \\ 4\cos ^2t=3 \end{gathered}[/tex]Divide both sides of the equation into 4:
[tex]\begin{gathered} \frac{4\cos ^2t}{4}=\frac{3}{4} \\ \\ \cos ^2t=\frac{3}{4} \end{gathered}[/tex]Find the square root of both sides of the equation:
[tex]\begin{gathered} \sqrt[]{\cos^2t}=\sqrt[]{\frac{3}{4}} \\ \\ \cos t=\pm\frac{\sqrt[]{3}}{2} \\ \\ \cos t=+\frac{\sqrt[]{3}}{2} \\ \\ \cos t=-\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]Use the unit circle to find wich angles in the given interval have a cos equal to:
[tex]\cos t=\pm\frac{\sqrt[]{3}}{2}[/tex]Solution:
[tex]t=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}[/tex]state the direction of opening for f(x)=-(1)/(2)x^(2)+3
Answers
The given expression is a quadratic function. If the leading coefficient is greater than zero in a quadratic function, the parabola opens upward, and if the leading coefficient is less than zero, the parabola opens downward. Graphically,
In this case, the leading coefficient is less than zero, the direction of opening is downward.
5 Which of the following functions are an example of continuous growth?
Answers
In order to have a function that represents continuous growth, the base value that has a variable as exponent must be the constant value "e":
[tex]\begin{gathered} f(x)=a\cdot b^{cx}\\ \\ b=e \end{gathered}[/tex]Looking at the options, the functions that have this base value are options I and II.
Therefore the correct option is A.
Find the radius of the circle containing 60° arc of a circle whose length is 14 m.
Answers
Given:
It is given that
[tex]\begin{gathered} \theta\text{ = 60}^0 \\ Arc\text{ length = 14}\pi \end{gathered}[/tex]Required:
The radius of the circle
Explanation:
The length of an arc is given by the formula,
[tex]\begin{gathered} Arc\text{ length = }\frac{\theta}{360}\text{ }\times\text{ 2}\pi r \\ \end{gathered}[/tex]Substituting the values in the formula,
[tex]\begin{gathered} 14\pi\text{ = }\frac{60}{360}\text{ }\times\text{ 2}\times\pi\times r \\ r\text{ = }\frac{14\times360}{60\times2} \\ r\text{ = }\frac{5040}{120} \\ r\text{ = 42} \end{gathered}[/tex]Answer:
Thus the radius of the circle is 42 m.
How do I solve this problem? I need to graph the segment
Answers
Step 1:
Draw a table of time with inches of snow.
Sketch the graph of the line whose equation , in point-slope form , is y-3 =9/5 (×+1 ). Also write the equation of this line in slope-intercept form . (y=mx+b)
Answers
To sketch this line equation, it would be a good idea to write the equation of the line in the slope-intercept form first.
Finding the equation of the line in slope-intercept formThe equation of the line in the slope-intercept form is given by:
[tex]y=mx+b[/tex]Where:
• m is the slope of the line.
,• b is the y-intercept of the line (the point where the line passes through the y-axis. At this point, x = 0.
Now, we have that the line equation is given in point-slope form as follows:
[tex]y-3=\frac{9}{5}(x+1)[/tex]We can multiply both sides of the equation by 5:
[tex]\begin{gathered} 5(y-3)=5\cdot\frac{9}{5}(x+1) \\ 5(y-3)=\frac{5}{5}\cdot9(x+1)\Rightarrow\frac{a}{a}=1,\frac{5}{5}=1 \\ 5(y-3)=9(x+1) \end{gathered}[/tex]Now, we have to apply the distributive property to both sides of the equation:
[tex]\begin{gathered} 5(y-3)=9(x+1) \\ 5y-15=9x+9 \end{gathered}[/tex]Add 15 to both sides of the equation, and then divide by 5:
[tex]\begin{gathered} 5y-15+15=9x+9+15 \\ 5y=9x+24 \\ \frac{5y}{5}=\frac{1}{5}(9x+24) \\ y=\frac{9}{5}x+\frac{24}{5} \end{gathered}[/tex]Therefore, the equation is slope-intercept form is:
[tex]y=\frac{9}{5}x+\frac{24}{5}[/tex]Sketching the graph for the lineSince we have that the original equation of the line was:
[tex]y-3=\frac{9}{5}(x+1)[/tex]We already know that one of the points of the line is (-1, 3) since the point-slope form of the line is given by:
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-(3)=\frac{9}{5}(x-(-1)) \\ y-3=\frac{9}{5}(x+1) \end{gathered}[/tex]We need another point to graph the line. We can use the y-intercept obtained before:
[tex]\frac{24}{5}=4.8[/tex]And since we know it is the y-intercept, we have that this point is (0, 4.8). Therefore, we can graph this equation using the following points:
(0, 4.8) and (-1, 3). Then we can sketch the line as follows:
To have a more precise graph for the line, we can use a graphing calculator:
We can see that the line passes through the x-axis at the point:
[tex]\begin{gathered} y=0\Rightarrow y=\frac{9}{5}x+\frac{24}{5} \\ 0=\frac{9}{5}x+\frac{24}{5} \\ -\frac{24}{5}=\frac{9}{5}x \\ \frac{5}{9}\cdot(-\frac{24}{5})=\frac{5}{9}\cdot(\frac{9}{5})x \\ -\frac{24}{9}=x \\ x=-\frac{8}{3}\approx-2.66666666667 \end{gathered}[/tex]