After 4 years the percentage of never-married adults will be the same as the percentage of married adults.
The data can be modeled by the following system of linear equations.
-3x+10y = 160
x+2y=164
Multiply the second equation with 3
-3x + 10y = 160 .....equation 1
3x + 6y = 492........equation 2
adding equation 1 and 2
16y = 652
y = 40.75
x + 2y = 164
x = 164 - 2 (40.75)
x = 82.5
Let the number of years be t
-3x+10y x t = x+2y
t = 4x - 8y
t = 330 - 326
t = 4 years
Therefore, after 4 years the percentage of never-married adults will be the same as the percentage of married adults.
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Which of the following ordered pairs is a solution to the graph of the system of inequalities? Select all that apply(5.2)(-3,-4)(0.-3)(0.1)(-4,1)
For this type of question, we should draw a graph and find the area of the common solutions
[tex]\begin{gathered} \because-2x-3\leq y \\ \therefore y\ge-2x-3 \end{gathered}[/tex][tex]\begin{gathered} \because y-1<\frac{1}{2}x \\ \therefore y-1+1<\frac{1}{2}x+1 \\ \therefore y<\frac{1}{2}x+1 \end{gathered}[/tex]Now we can draw the graphs of them
The red line represents the first inequality
The blue line represents the second inequality
The area of the two colors represents the area of the solutions,
Let us check the given points which one lies in this area
Point (5, -2) lies on the area of the solutions
∴ (5, -2) is a solution
Point (-3, -4) lies in the blue area only
∴ (-3, -4) not a solution
Point (0, -3) lies in the red line and the red line is solid, which means any point on it will be on the area of the solutions
∴ (0, -3) is a solution
Point (0, 1) lies in the blue line and the blue line is dashed, which means any point that lies on it not belong to the area of the solutions
∴ (0, 1) is not a solution
Point (-4, 1) lies on the area of the solutions
∴ (-4, 1) is a solution
The solutions are (5, -2), (0, -3), and (-4, 1)
Find seco, coso, and coto, where is the angle shown in the figure.Give exact values, not decimal approximations.
step 1
Find out the value of cosθ
cosθ=8/17 ------> by CAH
step 2
Find out the value of secθ
secθ=1/cosθ
secθ=17/8
step 3
Find out the length of the vertical leg in the given right triangle
Applying the Pythagorean Theorem
17^2=8^2+y^2
y^2=17^2-8^2
y^2=225
y=15
step 4
Find out the value f cotθ
cotθ=8/15 -----> adjacent side divided by the opposite side
therefore
secθ=17/8cosθ=8/17cotθ=8/15Find the axis of symmetry, vertex and which direction the graph opens, and the y-int for each quadratic function
Solution
Part a
The axis of symmetry
Part b
The vertex
Vertex (2,3)
Part c
The graph opens downward
Part D
The y-intercept is the point where a graph crosses the y-axis. In other words, it is the value of y when x=0.
The y-intercept
[tex](0,-5)[/tex]Another
The y-intercept is the point where a graph crosses the y-axis. In other words, it is the value of y when x=0.
[tex]\begin{gathered} y=-2x^2+8x-5 \\ y=-2(0)+8(0)-5 \\ y=0+0-5 \\ y=-5 \end{gathered}[/tex]x=0 y=-5
[tex](0,-5)[/tex]find the value of x so that the function has the given value
j(x) = -4/3x + 7; j (x) = -5
Answer:
x = 13 [tex]\frac{2}{3}[/tex]
Step-by-step explanation:
j(x) = [tex]\frac{-4}{3}[/tex] x + 7 Substitute -5 for x
j(-5) = [tex]\frac{-4}{3 }[/tex] ( -5) + 7
or
j(-5) =[tex](\frac{-4}{3})[/tex] [tex](\frac{-5}{1})[/tex] + 7 A negative times a negative is a positive
j(-5) = [tex]\frac{20}{3}[/tex] + 7
j(-5) = [tex]\frac{20}{3}[/tex] + [tex]\frac{21}{3}[/tex] [tex]\frac{21}{3}[/tex] means the same thing as 7
j(-5) = [tex]\frac{41}{3}[/tex] = 13 [tex]\frac{2}{3}[/tex]
For a period of d days an account balance can be modeled by f(d) = d^ 3 -2d^2 -8d +3 when was the balance $38
Given a modelled account balance for the period of d days as shown below:
[tex]\begin{gathered} f(d)=d^3-2d^2-8d+3 \\ \text{where,} \\ f(d)\text{ is the account balance} \\ d\text{ is the number of days} \end{gathered}[/tex]Given that the account balance is $38, we would calculate the number of days by substituting for f(d) = 38 in the modelled equation as shown below:
[tex]\begin{gathered} 38=d^3-2d^2-8d+3 \\ d^3-2d^2-8d+3-38=0 \\ d^3-2d^2-8d-35=0 \end{gathered}[/tex]Since all coefficients of the variable d from degree 3 to 1 are integers, we would apply apply the Rational Zeros Theorem.
The trailing coefficient (coefficient of the constant term) is −35.
Find its factors (with plus and minus): ±1,±5,±7,±35. These are the possible values for dthat would give the zeros of the equation
Lets input x= 5
[tex]\begin{gathered} 5^3-2(5)^2-8(5)-35=0 \\ 125-2(25)-40-35=0 \\ 125-50-75=0 \\ 125-125=0 \\ 0=0 \end{gathered}[/tex]Since, x= 5 is a zero, then x-5 is a factor.
[tex]\begin{gathered} d^3-2d^2-8d-35=(d-5)(d^2+3d+7)=0 \\ (d-5)(d^2+3d+7)=0 \\ d-5=0,d^2+3d+7=0 \\ d=0, \end{gathered}[/tex][tex]\begin{gathered} \text{simplifying } \\ d^2+3d+7\text{ would give} \\ d=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=1,b=3,c=7 \end{gathered}[/tex][tex]\begin{gathered} d=\frac{-3\pm\sqrt[]{3^2-4\times1\times7}}{2\times1} \\ d=\frac{-3\pm\sqrt[]{9-28}}{2} \\ d=\frac{-3\pm\sqrt[]{-17}}{2} \end{gathered}[/tex]It can be observed that the roots of the equation would give one real root and two complex roots
Therefore,
[tex]d=5,d=\frac{-3\pm\sqrt[]{-17}}{2}[/tex]Since number of days cannot a complex number, hence, the number of days that would give a balance of $38 is 5 days
Fill In the proportion No explanation just need answer got disconnected from last tutor
Explanation
Since the given shapes are similar, which implies that they are proportional,
Therefore; we will have
Answer:
[tex]\frac{AB}{EF}=\frac{BC}{FG}[/tex]marie invested 10000 in a savings account that pays 2 interest quartarly 4 times a yesr. how much money will she have in her account in 7 years?
Okey, here we have the following:
Capital: 10000
Interest: 2%
Time: 7 Years= 7*4=28 quarters of year
Using the compound interest formula, we get:
[tex]C_f=10000(1+\frac{0.02}{4})^{4\cdot7}=1000(1+\frac{0.02}{4})^{28}[/tex]Working we get:
[tex]C_f\approx11.498.73[/tex]She will have aproximately $11,498.73 in her account after 7 years.
Use the given scale factor and the side lengths of the scale drawing to determine the side lengths of the real object. Scale factor. 4:1 10 in 10 in A C 12 in Scale drawing Object A. Side a is 6 inches long, side bis 6 inches long, and side cis 8 inches long. B. Side a is 14 inches long, side bis 14 inches long, and side cis 16 inches long. C. Side a is 40 inches long, side bis 40 inches long, and side c is 48 inches long D. Side a is 2.5 inches long, side bis 2.5 inches long, and side cis 3
As the scale factor is 4:1 it means that for each 4inches in scale drawing correspond to 1 inch in the object.
Then, to find the side lengths in the object you multiply the measure of each side in the scale drawing by 1/4:
[tex]\begin{gathered} 10in\cdot\frac{1}{4}=2.5in \\ \\ 10in\cdot\frac{1}{4}=2.5in \\ \\ 12in\cdot\frac{1}{4}=3in \end{gathered}[/tex]Then, side a is 2.5 inches, side b is 2.5in and side c is 3inchesSolve for x. Then find m
(8x+4)°
(10x-6)°
Both lines are intersecting and the two equations are vertical pairs
Answer:
x = 5
m∠QRT = 44
Step-by-step explanation:
8x + 4 = 10x - 6
-10x -10x
------------------------
-2x + 4 = -6
-4 -4
---------------------
-2x = -10
÷-2 ÷-2
--------------------
x = 5
m∠QRT
8x + 4
8(5) + 4
40 + 4 = 44
I hope this helps!
From the image, [tex](8x + 4 )° = (10x - 6)°[/tex]
Reason: Vertically opposite angles are equal.
Now, we solve for x
[tex]8x + 4 = 10x - 6[/tex]
collect like terms
[tex] 4 + 6 = 10x - 8x[/tex]
[tex]→ 10 = 2x[/tex]
Divide both sides by the coefficient of x to find the value of x
[tex] x = 5[/tex]
Now, substitute for the value of x in the expression for m∠QRT to find the degree
[tex](8x + 4 )° → (8(5) + 4)°[/tex]
Therefore: m∠QRT = 44°
I hope this helps7. The cylinder shown has a radius of 3inches. The height is three times the radiusFind the volume of the cylinder. Round yoursolution to the nearest tenth.
Answer:
250 cubic inches
Explanation:
Given that:
Radius of the cylinder, = 3 in.
Height of the cylinder = 3r
= 3(3)
=9 in.
The formula to find the volume of a cylinder is
[tex]V=\pi r^2h[/tex]Plug the given values into the formula.
[tex]\begin{gathered} V=\pi3^29 \\ =81\pi \\ =254.469 \end{gathered}[/tex]Rounding to nearest tenth gives 250 cubic inches, which is the required volume of the cylinder.
A bag of tokens contains 55 red, 44 green, and 55 blue tokens. What is the probability that a randomly selected token is not red? Enter your answer as a fraction.
Explanation
In the bag of tokens, we are told 55 red, 44 green, and 55 blue tokens. Therefore, the total number of tokens in the bag is
[tex]55+44+55=154[/tex]Hence to find the probability that a randomly selected token is not red becomes;
[tex]Pr(not\text{ red black})=\frac{n(green)+n(blue)}{n(tokens)}=\frac{44+55}{154}=\frac{99}{154}=\frac{9}{14}[/tex]Answer: 9/14
The elimination method is used in place over substitution when one equation is not easily solved for ______________ variable.A) a standardB) a dependentC) an independentD) a single
Given:
There are given the statement about the elimination method and substitution method.
Explanation:
According to the concept:
One equation cannot be easily solved for a single variable.
Final answer:
Hence, the correct option is D.
Where are all the tutors at??? Like it won’t even let me ask a tutor
The scatter plot is given and objective is to find the best line of fit for given scatter plot.
Let's take the few points of scatter plot,
(0,8) ,( -1,8) , (-4,10) ,( -8,12),(-10,14) (-12,14)
Take the line and check which graph contains most of the points of scatter plot.
[tex]1)\text{ f(x)=}\frac{-1}{2}x+8[/tex]The graph is ,
now take ,
[tex]2)\text{ f(x)=x+8}[/tex]The graph is,
This graph contains only one point of scatter plot.
Take,
[tex]3)\text{ f(x)= 10}[/tex]Now the take the last equation,
[tex]4)\text{ f(x)=-2x+14}[/tex]this graph contains no point of the scatter plot.
From all the four graph of the lines it is observed that option 1) is the best line of fit for given scatter plot. because it contains 3 points of scatter plotes . which is more than the other graph of line.
Answer: Option 1)
Fragment Company leased a portion of its store to another company for eight months beginning on October 1, at a monthly rate of $1,250. Fragment collected the entire $10,000 cash on October 1 and recorded it as unearned revenue. Assuming adjusting entries are only made at year-end, the adjusting entry made on December 31 would be:
Given:
Credit to rent earned for
Amount of total rent = $10,000
Amount unearned = amount of total rent ( 3 month / 8 month)
[tex]\begin{gathered} \text{Amount unearned=10000}\times\frac{3}{8} \\ =3750 \end{gathered}[/tex]Unearned rent is : $3750
identify the amplitude and period of the function then graph the function and describe the graph of G as a transformation of the graph of its parent function
Given the function:
[tex]g(x)=cos4x[/tex]Let's find the amplitude and period of the function.
Apply the general cosine function:
[tex]f(x)=Acos(bx+c)+d[/tex]Where A is the amplitude.
Comparing both functions, we have:
A = 1
b = 4
Hence, we have:
Amplitude, A = 1
To find the period, we have:
[tex]\frac{2\pi}{b}=\frac{2\pi}{4}=\frac{\pi}{2}[/tex]Therefore, the period is = π/2
The graph of the function is shown below:
The parent function of the given function is:
[tex]f(x)=cosx[/tex]Let's describe the transformation..
Apply the transformation rules for function.
We have:
The transformation that occured from f(x) = cosx to g(x) = cos4x using the rules of transformation can be said to be a horizontal compression.
ANSWER:
Amplitude = 1
Period = π/2
Transformation = horizontal compression.
Perform the indicated operation and write the answer in the form A+Bi
The Solution:
Given:
[tex](3+8i)(4-3i)[/tex]We are required to simplify the above expression in a+bi form.
Simplify by expanding:
[tex]\begin{gathered} (3+8i)(4-3i) \\ 3(4-3i)+8i(4-3i) \\ 12-9i+32i-24(-1) \end{gathered}[/tex]Collecting the like terms, we get:
[tex]\begin{gathered} 12-9i+32i+24 \\ 12+24-9i+32i \\ 36+23i \end{gathered}[/tex]Therefore, the correct answer is [option 3]
There is a bag filled with 5 blue and 4 red marbles.
A marble is taken at random from the bag, the colour is noted and then it is replaced.
Another marble is taken at random.
What is the probability of getting at least 1 blue?
The probability of getting exactly 1 blue marble from a bag which is filled with 5 blue and 4 red marbles is 40/81.
What is probability?Probability of an event is the ratio of number of favorable outcome to the total number of outcome of that event.
A bag is filled with 5 blue and 4 red marbles.
The total number of marble in the bag are,
5+4=9
One marble is taken at random from the bag, the color is noted and then it is replaced. The probability of getting blue marble is,
P(B)=5/9
probability of getting red marble is,
P(R)=4/9
The Probability of getting red marble in first pick and probability of getting blue marble in second pick
P1=5/9×4/9=20/81
The Probability of getting blue marble in first pick and probability of getting red marble in second pick is,
p2=4/9×5/9=20/81
The exactly 1 blue is taken out, when first marble is red and second is blue or the first one is blue and second one is red. Thus, the probability of getting exactly 1 blue is,
P=p1+p2
=20/81+20/81
40/81
Hence the probability of getting exactly 1 blue marble from a bag which is filled with 5 blue and 4 red marbles is 40/81.
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2. Axely says that 8is equivalent to –.125repeating. Without solving, evaluate her claim in thespace below.
we are asked about the claim that the fraction -12 / 8 is equivalent to the decimal expression -0.125... (repeating)
Without evaluating the expression, we can say that the clain is INCORRECT, since just the quotient 12/8 should give a number LARGER than "1" (one) in magnitude (the number 12 is larger than the number 8 in the denominator. We can also say that such division cannot ever give a repeating decimal at infinity, since divisions of integer numbers by 8 or 4 never render a repeating decimal, but a finite number of decimals.
PLEASE READ BEFORE ANSWERING: ITS ALL ONE QUESTION HENCE "QUESTION 6" THEY ARE NOT INDIVIDUALLY DIFFERENT QUESTIONS.
First, lets note that the given functions are polynomials of degree 2. Since the domain of a polynomial is the entire set of real numbers, the domain for all cases is:
[tex](-\infty,\infty)[/tex]Now, lets find the range for all cases. In this regard, we will use the first derivative criteria in order to obtain the minimum (or maximim) point.
case 1)
In the first case, we have
[tex]\begin{gathered} 1)\text{ }\frac{d}{dx}f(x)=6x+6=0 \\ which\text{ gives} \\ x=-1 \end{gathered}[/tex]which corresponds to the point (-1,-8). Then the minimum y-value is -8 because the leading coefficient is positive, which means that the curve opens upwards. So the range is
[tex]\lbrack-8,\infty)[/tex]On the other hand, the horizontal intercept (or x-intercept) is the value of the variable x when the function value is zero, that is,
[tex]3x^2+6x-5=0[/tex]which gives
[tex]\begin{gathered} x_1=-1+\frac{2\sqrt{6}}{3} \\ and \\ x_2=-1-\frac{2\sqrt{6}}{3} \end{gathered}[/tex]Case 2)
In this case, the first derivative criteria give us
[tex]\begin{gathered} \frac{d}{dx}g(x)=2x+2=0 \\ then \\ x=-1 \end{gathered}[/tex]Since the leading coefficient is positive, the curve opens upwards so the point (-1,5) is the minimum values. Then, the range is
[tex]\lbrack5,\infty)[/tex][tex]\lbrack5,\infty)[/tex]and the horizontal intercepts do not exists.
Case 3)
In this case, the first derivative criteris gives
[tex]\begin{gathered} \frac{d}{dx}f(x)=-2x=0 \\ then \\ x=0 \end{gathered}[/tex]Since the leading coeffcient is negative the curve opens downwards and the maximum point is (0,9). So the range is
[tex](-\infty,9\rbrack[/tex]and the horizontal intercepts occur at
[tex]\begin{gathered} -x^2+9=0 \\ then \\ x=\pm3 \end{gathered}[/tex]Case 4)
In this case, the first derivative yields
[tex]\begin{gathered} \frac{d}{dx}p(t)=6t-12=0 \\ so \\ t=2 \end{gathered}[/tex]since the leading coefficient is postive the curve opens upwards and the point (2,-12) is the minimum point. Then the range is
[tex]\lbrack-12,\infty)[/tex]and the horizontal intercetps ocurr when
[tex]\begin{gathered} 3x^2-12x=0 \\ which\text{ gives} \\ x=4 \\ and \\ x=0 \end{gathered}[/tex]Case 5)
In this case, the leading coefficient is positive so the curve opens upwards and the minimum point ocurrs at x=0. Therefore, the range is
[tex]\lbrack0,\infty)[/tex]and thehorizontal intercept is ('0,0).
In summary, by rounding to the nearest tenth, the answers are:
Write an equation of a line in SLOPE INTERCEPT FORM that goes through (-5,-3) and is parallel to the line y = x +5.
Since the slope of the line y=x+5 is m=1, then if the other line is parallel to y=x+5, then it must have the same slope, this is, m'=1.
Now we can use the point-slope formula to get the equation of the line:
[tex]\begin{gathered} m^{\prime}=1 \\ (x_0,y_0)=(-5,-3) \\ y-y_0=m(x-x_0) \\ \Rightarrow y-(-3)=1\cdot(x-(-5))=x+5 \\ \Rightarrow y+3=x+5 \\ \Rightarrow y=x+5-3=x+2 \\ y=x+2 \end{gathered}[/tex]therefore, the equation of the line in slope intercept form that goes through (-5,-3) and is parallel to the line y=x+5 is y=x+2
how do you find the x intercept for -(x-3)^2+12
This is the equation for the line.
Here the given equation is,
[tex]-(x-3)^2+12[/tex]We can calculate x intercept by substituting 0 for y, as there is no value of y,
the x intercept is none here. The graph is as follows,
What is an equation of the line that passes through the points (-3,3) and (3, — 7)?Put your answer in fully reduced form.
The equation of line passing through two points (x_1,y_1) and (x_2,y_2) is,
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]Substitute the points in the equation to obtain the equation of line.
[tex]\begin{gathered} y-3=\frac{-7-3}{3-(-3)}(x-(-3)) \\ y-3=\frac{-10}{6}(x+6) \\ 3(y-3)=-5(x+6) \\ 3y-9+5x+30=0 \\ 3y+5x+21=0 \end{gathered}[/tex]So equation of line is 3y+5x+21=0.
Use mental math to find all of the quotients equal to 50. Drag the correct division problems into the box.
4
,
500
÷
900
450
÷
90
45
,
000
÷
900
4
,
500
÷
90
450
÷
9
Quotients equal to 50
Answer: 45,000 ÷ 900=50
Step-by-step explanation:
Solve the equation3 x² - 12x +1 =0 by completing the
square.
By completing squares, we wll get that the solutions of the quadratic equation are:
x = 6 ± √35
How to complete squares?Here we have the quadratic equation:
x² - 12x + 1 = 0
We can rewrite this as:
x² - 2*6x + 1 = 0
So we can add and subtract 6² to get:
x² - 2*6x + 1 + 6² - 6² = 0
Now we rearrange the terms:
(x² - 2*6x + 6²) + 1 - 6² = 0
Now we can complete squares.
(x - 6)² + 1 - 36 = 0
(x - 6)² = 35
Now we solve for x:
x = 6 ± √35
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Your parents will retire in 25 years. They currently have $230,000 saved, and they think they will need $1,850,000 at retirement. What annual interest rate must they earn to reach their goal, assuming they don't save any additional funds? Round your answer to two decimal places.
6.62% is annual interest rate must they earn to reach their goal.
What exactly does "interest rate" mean?
An interest rate informs you of how much borrowing will cost you and how much saving will pay off. Therefore, the interest rate is the amount you pay for borrowing money and is expressed as a percentage of the entire loan amount if you are a borrower.N = 25
PV = - $230,000
FV = $1,850,000
PMT = 0
CPT Rate
Applying excel formula:
=RATE(25,0,-230,000,1,850,000)
= 6.62%
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A recent survey asked respondents how many hours they spent per week on the internet. Of the 15 respondents making$2,000,000 or more annually, the responses were: 0,0,0,0,0, 2, 3, 3, 4, 5, 6, 7, 10, 40 and 70. Find a point estimate of thepopulation mean number of hours spent on the internet for those making $2,000,000 or more.
Given
The total frequency is 15 respondents
The responses were: 0,0,0,0,0, 2, 3, 3, 4, 5, 6, 7, 10, 40 and 70
Solution
The population mean is the sum of all the values divided by the total frequency .
[tex]undefined[/tex]A) What is the perimeter of the regular hexagon shown above?B) What is the area of the regular hexagon shown above?(see attached image)
Remember that
A regular hexagon can be divided into 6 equilateral triangles
the measure of each interior angle in a regular hexagon is 120 degrees
so
see the attached figure to better undesrtand the problem
each equilateral triangle has three equal sides
the length of each side is given and is 12 units
Part A) Perimeter
the perimeter is equal to
P=6(12)=72 units
Part B
Find the area
Find the height of each equilateral triangle
we have
tan(60)=h/6
Remember that
[tex]\tan (60^o)=\sqrt[]{3}[/tex]therefore
[tex]h=6\sqrt[]{3}[/tex]the area of the polygon is
[tex]A=6\cdot\lbrack\frac{1}{2}\cdot(6\sqrt[]{3})\cdot(12)\rbrack[/tex][tex]A=216\sqrt[]{3}[/tex]alternate way to find out the value of happlying Pythagorean Theorem
12^2=6^2+h^2
h^2=12^2-6^2
h^2=108
h=6√3 units
In the given figure ABC is a triangle inscribed in a circle with center O. E is the midpoint of arc BC . The diameter ED is drawn . Prove that
Answer:
we can use two ways to write 180° along with the inscribed angle theorem to obtain the desired relation
Step-by-step explanation:
Given ∆ABC inscribed in a circle O where E is the midpoint of arc BC and ED is a diameter, you want to prove ∠DEA = 1/2(∠B -∠C).
SetupWe can add add arcs to make 180° in two different ways, then equate the sums.
arc EB +arc BA +arc AD = 180°
arc EC +arc CA -arc AD = 180°
Equating these expressions for 180°, we have ...
arc EB +arc BA +arc AD = arc EC +arc CA -arc AD
SolutionRecognizing that arc EB = arc EC, we can subtract (arc EB +arc BA -arc AD) from both sides to get ...
2·arc AD = arc CA -arc BA
The inscribed angle theorem tells us ...
arc AD = 2∠DEAarc CA = 2∠Barc BA = 2∠CMaking these substitutions into the above equation, we have ...
4∠DEA = 2∠B -2∠C
Dividing by 4 gives the relation we're trying to prove:
∠DEA = 1/2(∠B -∠C)
Which of the following steps were applied to ABC obtain A’BC’?
Given,
The diagram of the triangle ABC and A'B'C' is shown in the question.
Required:
The translation of triangle from ABC to A'B'C'.
Here,
The coordinates of the point A is (2,5).
The coordinates of the point A' is (5,7)
The translation of the triangle is,
[tex](x,y)\rightarrow(x+3,y+2)[/tex]Hence, shifted 3 units right and 2 units up.
Solve for t. If there are multiple solutions, enter them as a
we have the equation
[tex]\frac{12}{t}+\frac{18}{(t-2)}=\frac{9}{2}[/tex]Solve for t
step 1
Multiply both sides by 2t(t-2) to remove fractions
[tex]\frac{12\cdot2t(t-2)}{t}+\frac{18\cdot2t(t-2)}{(t-2)}=\frac{9\cdot2t(t-2)}{2}[/tex]simplify
[tex]12\cdot2(t-2)+18\cdot2t=9\cdot t(t-2)[/tex][tex]24t-48+36t=9t^2-18t[/tex][tex]\begin{gathered} 60t-48=9t^2-18t \\ 9t^2-18t-60t+48=0 \\ 9t^2-78t+48=0 \end{gathered}[/tex]Solve the quadratic equation
Using the formula
a=9
b=-78
c=48
substitute
[tex]t=\frac{-(-78)\pm\sqrt[]{-78^2-4(9)(48)}}{2(9)}[/tex][tex]t=\frac{78\pm66}{18}[/tex]The solutions for t are
t=8 and t=2/3
therefore
the answer is
t=2/3,8