on a horizontal line segment, point A is located at 21, point b is located at 66. point p is a point that divides segment ab in a ratio of 3:2 from a to b where is point p located
Answers
We have a one-dimensional horizontal line segment. Three points are indicated on the line as follows:
In the above sketch we have first denoted a reference point at the extreme left hand as ( Ref = 0 ). This is classified as the origin. The point ( A ) is located on the same line and is at a distance of ( 21 units ) from Reference ( Ref ). The point ( B ) is located on the same line and is at a distance of ( 66 units ) from Reference ( Ref ).
The point is located on the line segment ( AB ) in such a way that it given as ratio of length of line segment ( AB ). The ratio of point ( P ) from point ( A ) and from ( P ) to ( B ) is given as:
[tex]\textcolor{#FF7968}{\frac{AP}{PB}}\text{\textcolor{#FF7968}{ = }}\textcolor{#FF7968}{\frac{3}{2}\ldots}\text{\textcolor{#FF7968}{ Eq1}}[/tex]The length of line segment ( AB ) can be calculated as follows:
[tex]\begin{gathered} AB\text{ = OB - OA } \\ AB\text{ = ( 66 ) - ( 21 ) } \\ \textcolor{#FF7968}{AB}\text{\textcolor{#FF7968}{ = 45 units}} \end{gathered}[/tex]We can form a relation for the line segment ( AB ) in terms of segments related to point ( P ) as follows:
[tex]\begin{gathered} \textcolor{#FF7968}{AB}\text{\textcolor{#FF7968}{ = AP + PB }}\textcolor{#FF7968}{\ldots}\text{\textcolor{#FF7968}{ Eq2}} \\ \end{gathered}[/tex]We were given a ratio of line segments as ( Eq1 ) and we developed an equation relating the entire line segment ( AB ) in terms two smaller line segments as ( Eq2 ).
We have two equation that we can solve simultaneously:
[tex]\begin{gathered} \textcolor{#FF7968}{\frac{AP}{PB}}\text{\textcolor{#FF7968}{ = }}\textcolor{#FF7968}{\frac{3}{2\text{ }}\ldots}\text{\textcolor{#FF7968}{ Eq1}} \\ \textcolor{#FF7968}{AB}\text{\textcolor{#FF7968}{ = AP + PB }}\textcolor{#FF7968}{\ldots Eq2} \end{gathered}[/tex]Step 1: Use Eq1 and express AP in terms of PB.
[tex]AP\text{ = }\frac{3}{2}\cdot PB[/tex]Step 2: Substitute ( AP ) in terms of ( PB ) into Eq2
[tex]AB\text{ = }\frac{3}{2}\cdot PB\text{ + PB}[/tex]We already determined the length of the line segment ( AB ). Substitute the value in the above expression and solve for ( PB ).
Step 3: Solve for PB
[tex]\begin{gathered} 45\text{ = }\frac{5}{2}\cdot PB \\ \textcolor{#FF7968}{PB}\text{\textcolor{#FF7968}{ = 18 units}} \end{gathered}[/tex]Step 4: Solve for AP
[tex]\begin{gathered} AP\text{ = }\frac{3}{2}\cdot\text{ ( 18 )} \\ \textcolor{#FF7968}{AP}\text{\textcolor{#FF7968}{ = 27 units}} \end{gathered}[/tex]Step 5: Locate the point ( P )
All the points on the line segment are located with respect to the Reference of origin ( Ref = 0 ). We will also express the position of point ( P ).
Taking a look at point ( P ) in the diagram given initially we can augment two line segments ( OA and AP ) as follows:
[tex]\begin{gathered} OP\text{ = OA + AP} \\ OP\text{ = 21 + 27} \\ \textcolor{#FF7968}{OP}\text{\textcolor{#FF7968}{ = 48 units}} \end{gathered}[/tex]The point ( P ) is located at.
Answer:
[tex]\textcolor{#FF7968}{48}\text{\textcolor{#FF7968}{ }}[/tex]
Related Questions
On a particular day, the amount of untreated water coming into the plant can be modeled by f(t) = 100 + 30cos(t/6) where t is in hours since midnight and f(t) represents thousands of gallons of water. The amount of treated water at any given time, t, can be modeled by g(t) = 30e^cos(t/2)a) Define a new function, a′(t), that would represent the amount of untreated water inside the plant, at any given time, t.b) Find a′ (t).c) Determine the critical values of this function over the interval [0, 24).
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a)The amount of untreated water inside the plant will be the difference between the difference f(t) - g(t), then, a(t) can be defined as follows:
[tex]a(t)=100+30cos(\frac{t}{6})-30e^{cos(\frac{t}{2})}[/tex]b) the derivative of a(t) is the following:
[tex]a^{\prime}(t)=-5sin(\frac{t}{6})+15sin(\frac{t}{2})e^{cos(\frac{t}{2})}[/tex]c) the critical values of a(t) over the interval [0, 24) are:
[tex]\begin{gathered} t=0 \\ t=6\pi \end{gathered}[/tex]Can someone show me how to do this one correctly?
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ANSWER:
Juwan has 19 dimes and 6 quarters in his pocket.
STEP-BY-STEP EXPLANATION:
From the statement we can establish the following system of equations:
Taking into account that one dime is 10 cents and a quarter is 25 cents.
Let x be the number of dimes
Let y be the number of quarters
[tex]\begin{gathered} x+y=25\rightarrow x=25-y\text{ (1)} \\ 10x+25y=340\text{ (2)} \end{gathered}[/tex]We solve the system of equations by means of the substitution method, we substitute equation (1) in (2):
[tex]\begin{gathered} 10\cdot(25-y)+25y=340 \\ 250-10y+25y=340 \\ 15y=340-250 \\ y=\frac{90}{15} \\ y=6 \\ \\ \text{therefore, for x:} \\ x=25-6 \\ x=19 \end{gathered}[/tex]Therefore they are 19 dimes and 6 quarters
I need your help know
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Given the following data
Base area of the cone = 6cm
Height of the cone = 9 cm
The volume of a cone is given as
[tex]\begin{gathered} V\text{ = }\frac{1}{3}\cdot\text{ }\pi\cdot r^2\cdot\text{ h} \\ \text{where }\pi\text{ = 3.14},\text{ r = 6cm , and h = 9cm} \\ V\text{ = }\frac{1}{3}\cdot\text{ 3.14 }\cdot6^2\cdot\text{ 9} \\ V\text{ = }\frac{1}{3}\text{ x 3.14 x 36 x 9} \\ V\text{ = }\frac{3.14\text{ x 36 x 9}}{3} \\ V\text{ = }\frac{1017.36}{3} \\ V=339.1cm^3 \end{gathered}[/tex]Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.Solve the inequality and describe the solution set.y-6 > 1232, Math symbolsRelations► Geometry► Groups► Trgonometry3 of 3 AnsweredType here to searcho66F Mosty clou
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The problem gives the inequality:
[tex]y-6\ge12[/tex]solving for y we get:
[tex]\begin{gathered} y\ge12+6 \\ y\ge18 \end{gathered}[/tex]The solution set is all real numbers equal or greater than 18, i.e.,
[tex]\lbrack18,+\infty)[/tex]What is the sign of mlio Choose 1 answer: Positive Negative Neither positive nor negative the sum is zero.
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The sign will be positive.
Which values are solutions to the inequality below?Check all that apply.√x ≤ 5A. 1B. 18C. -5D. 25E. 24F. 625
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Given the inequality:
[tex]\sqrt[]{x}<5[/tex]We need to solve the inequality to get a range of values for x.
This we can do by finding the square of both sides:
[tex]\begin{gathered} (\sqrt[]{x})^2<5^2 \\ x<25 \end{gathered}[/tex]On checking the options given, we will pick the numbers that are strictly less than 25.
Therefore, the correct options are:
OPTION A
OPTION B
OPTION C
OPTION F
Write an expression to show how much Gretchen paid for drama,action, and comedy videos if she paid $4 for each at a sale. Evaluate the expressionGretchen’s video purchasesMystery 6Action 3Comedy 5Drama 2Romance 2
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Let d the number of drama videos, c the number of comedy videos and a the number of action videos.
If the cost per video (independently of the genre) is $4, then, for the total cost of the videos Gretchen payed, you can write:
total = 4d + 4c + 4a
Now, based on the given table, you have:
d = 2
c = 5
a = 3
By replacing the previous values into the expression for total, and by simplifying, you obtain:
total = 4(2) + 4(5) + 4(3)
total = 8 + 20 + 12
total = 40
Hence, Gretchen payed $40 for the videos
write an equation of each parabola in vertex form. Vertex (3,-2) Point (2,3)
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The equation of Parabola in the vertex form with vertex (3,-2) and point(2,3) is y = 5(x-3)² - 2 .
The equation of parabola with vertex (h,k) is denoted by the equation
y = a(x-h)² + k
In the question ,
it is given that
the vertex of the Parabola is (3,-2) and the point is (2,3)
So, the equation of the parabola with vertex (3,-2) will be
y = a(x-3)² - 2
Since the point (2,3) lies on the parabola ,
So, 3 = a(2-3)² - 2
3 + 2 = a*(-1)²
5 = a
Substituting a in the equation y = a(x-3)² - 2 ,
we get
y = 5(x-3)² - 2
Therefore , The equation of Parabola in the vertex form with vertex (3,-2) and point(2,3) is y = 5(x-3)² - 2 .
Learn more about Parabola here
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3-35. Fisher thinks that any two lines must have a point of intersection. Is he correct? If so, explain how youknow. If not, produce a counterexample. That is, find two lines that do not have a point of intersection and explainhow you know
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Any two lines must have a point of intersection, with ONLY ONE exception : when the lines are parallel. That means they have exactly the same inclination or Slope.
As an example of two lines that hasnt point of intersection they are
y = 5x + 16
y = 5x + 11
We know both lines have no point of intersection because
both have the same slope or inclination m. By comparation with slope intercept equation
y = mx + b
then we see both have same inclination, and b is 11 in one line and 16 in the other line. This means both lines are parallel with one line going over the other line, without touching it in any point.
(b) The area of a rectangular window is 3740 cmcm?If the length of the window is 68 cm, what is its width?Width of the windoow
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Step 1
The area of a rectangle = Length x width
Step 2
Parameters given include
Area of rectangular window= 3740 square cm
Length of window = 68cm
Step 3
Substitute and solve
[tex]\begin{gathered} 3740\text{ = 68 x width} \\ \text{width = }\frac{3740}{68}\text{ = 55cm} \end{gathered}[/tex]Therefore, the width of the window = 55 cm
identify the rate, base and portion.What percent of 126 is 44.1
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In this problem, we have that
Base=126 (represents the 100%)
Portion=44.1
Find out the percentage
100/126=x/44.1
solve for x
x=(100/126)*44.1
x=35%
The percentage is 35%
J(-6-2)3-*NWMark this and return2--9-8-7-6-5-4-3-2-3₁ 1 2 3 4 5 61-5737-2-cd-6--7--8-2 do-9--10--11--12--13-8 9 10 11 xK(8,-9)What is the x-coordinate of the point that divides thedirected line segment from J to K into a ratio of 2:5?X == (m²²7 m )(x₂ − ×₁) + X₁m+n0 000-22Save and ExitNextSubmit
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Use the given formula:
[tex]x=(\frac{m}{m+n})(x_2-x_1)+x_1[/tex]Being m: 2 and n: 5
x1: -6
x2: 8
[tex]\begin{gathered} x=(\frac{2}{2+5})(8-(-6))+(-6) \\ \\ x=\frac{2}{7}*(14)-6 \\ \\ x=4-6 \\ \\ x=-2 \end{gathered}[/tex]Then, the x-coordinate of th point that divides the directed line segment from J to K into a ratio 2:5 is -2Answer: -2State if the given binomial is a factor of the given polynomial.Question #9
Answers
We can use the Factor Theorem to state if the given binomial is a factor of the given polynomial.
The factor theorem states that when f(c)=0 that means the remainder is zero and (x-c) must be a factor of the polynomial.
The given polynomial is:
[tex]k^3+8k^2+6k-12[/tex]Then if (k+2) is a factor of the polynomial, k+2=0, k=-2, f(-2) must be equal to 0.
Let's check:
[tex]\begin{gathered} f(k)=k^3+8k^2+6k-12 \\ f(-2)=(-2)^3+8(-2)^2+6(-2)-12 \\ f(-2)=-8+8\cdot4-12-12 \\ f(-2)=-8+32-24 \\ f(-2)=32-32 \\ f(-2)=0 \end{gathered}[/tex]Thus, (k+2) is a factor of the given polynomial.
Dave and his brother. Theo, are selling cookies by the pound at the school bake sale Dave sold 14 84 pounds of cookies and Theo sold 21.45 pounds of cookies How many pounds did they sell altogether? A 35 29 OB 36 39 C36 25 0 D. 36 29
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For tis problem we have that Dave sold 14.84 pounds of cookies and Theo sold 21.45 pounds of cookies.
If we want to find the total of pounds that they sold together we just need to add the two values and we have:
[tex]14.84+21.45=36.29\text{pounds}[/tex]The reason is because 0.84+0.45=1.29
14+21=35. And finally 35+1.29=36.29
And the best answer for this case would be D. 36.29
22The value of the hypotenuse in the right triangle shown isinches.14 in48 inFigure not drawn to scale
Answers
SOL
Step 1 :
In this question, we are meant to find the value of
the hypotenuse in the right angle below:
Before, we proceed, we still need to remind ourselves of Pythagoras' theorem,
Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“.
Step 2 :
From the above theorem, we can see that that the two adjacent sides are :
14 inches and 48 inches.
From the principle of Pythagoras' Theorem,
[tex]\begin{gathered} c^2=a^2+b^2 \\ \text{where a = 14 inches} \\ b\text{ = 48 inches} \\ c^2=14^2+48^2 \\ c^2\text{ = ( 14 x 14 ) + ( 48 x 48 )} \\ c^2\text{ = 196 + 2304} \\ c^2\text{ = 2500} \\ \text{square root both sides, we have that :} \\ c\text{ = 50 inches} \end{gathered}[/tex]
CONCLUSION :
The value of the hypotenuse in the Right angle, c = 50 inches.
Which expression demonstrates how the Distributive Property could be used to find the product of 5 and 48?
A. 50 − 5 (2)
B. 5 (50) − 2
C. 5 (50) + 5 (2)
D. 5 (50 − 2) + 5 (2)
E. None of these
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Answer:
Step-by-step explanation:
C
Graph the line y = 5x – 1, then name the slope and y-intercept by looking at the graph.
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Answer:
m = 5
y-intercept = (0, -1)
Step-by-step explanation:
y = mx + b
y = 5x - 1
m = 5
y-intercept = b = (0, -1)
Point 1: (0, -1)
Point 2: (1, 4)
Point 3: (-1, -6)
I hope this helps!
Use the graph to answer the question.Which statement matches the vector operation shown on the coordinate grid?
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We have the correct statement about the vectors in the graph.
We can already see that it is a sum of vectors like:
[tex]v+w=u[/tex]As v has starting point at (-1,0) and ending point at (3,3), we can describe the vector as:
[tex]v=(3-(-1))\hat{i}+(3-0)\hat{j}=4\hat{i}+3\hat{j}[/tex]As w starts at (3, 3) and ends on (5, 2), we can describe it as:w
[tex]w=(5-3)\hat{i}+(2-3)\hat{j}=2\hat{i}-1\hat{j}[/tex]Finally, u starts at (-1,0) and ends at (5,2), so it can be described as:
[tex]u=(5-(-1))\hat{i}+(2-0)\hat{j}=6\hat{i}+2\hat{j}[/tex]Answer: v + w = u for v = 4i + 3j, u = 2i - j and u = 6i + 2j [Option C].
Solve. Your answer should be in simplest form. 2/5 (−3/7)
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Answer:
2/5 (-3/7) = -6/35 ≅ -0.1714286
Step-by-step explanation:
and that’s how you do it
Add: 2/5 + 3/7 = 2 · 7/5 · 7 + 3 · 5/7 · 5 = 14/35 + 15/35 = 14 + 15/35 = 29/35.
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(5, 7) = 35. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 5 × 7 = 35. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - two fifths plus three sevenths is twenty-nine thirty-fifths.
Compare the triangles and determjne whether they can be proven congruent, if possible by SSS, SAS, ASA, AAS or HL
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Since the triangles has a pair of congruent (equal) angles , and an equal side between the angles. It is congruent by ASA ( angle -side -angle)
Please help me with the question below(also please answer the question in a maximum of 5-10 minutes).
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Given that Tom's yard is in the shape of a trapezoid, you know that the formula for calculating the area of a trapezoid is:
[tex]A=\frac{(b_1+b_2)}{2}\cdot h[/tex]Where "h" is the height of the trapezoid and these are the bases:
[tex]\begin{gathered} b_1 \\ b_2 \end{gathered}[/tex]In this case, you can identify that:
[tex]\begin{gathered} b_1=65\text{ }ft \\ b_2=50\text{ }ft \\ h=30\text{ }ft \end{gathered}[/tex]Then, you can substitute values into the formula and evaluate:
[tex]A=\frac{(65\text{ }ft+50\text{ }ft)}{2}\cdot30\text{ }ft[/tex][tex]A=\frac{115\text{ }ft}{2}\cdot30\text{ }ft[/tex][tex]A=\frac{3450\text{ }ft^2}{2}[/tex][tex]A=\frac{3450\text{ }ft^2}{2}[/tex][tex]A=1725\text{ }ft^2[/tex]Hence, the answer is:
[tex]1725\text{ }ft^2[/tex]At a school on Monday, 3 out of every 4 students were wearing shirts. There were 600 students present in school on Monday. How many of the students were wearing shirts? A. 599, because 600 - (4 - 3) = 599 B. 450, because C. 50, because 600 - (4 x 3) = 50 600 - Student D. 800, because 450 4= Students 3=sludents 4 600 600 800 so
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3 out of 4 students mean
3/4th students were wearing shirts.
Total students = 600
So,
3/4th of 600 students were wearing shirt.
Let us calcualte (3/4)th of 600:
[tex]\begin{gathered} \frac{3}{4}\times600 \\ =\frac{3\times600}{4} \\ =\frac{1800}{4} \\ =450 \end{gathered}[/tex]Answer450 students
Find the solutions to the following quadratic equation negative 3X squared plus 4X plus one equals zero (-3x^2 + 4x + 1 = 0)
Answers
Answer:
Explanation:
Given:
[tex]-3x^2+4x+1=0[/tex]To find:
the value of x using the quadratic formula
The quadratic formula is given as:
[tex]$$x\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$$[/tex]where a = -3, b = 4, c = 1
[tex]\begin{gathered} x\text{ = }\frac{-4\pm\sqrt{(4)^2-4(-3)(1)}}{2(-3)} \\ \\ x\text{ = }\frac{-4\pm\sqrt{16+12}}{-6} \\ \\ x\text{ = }\frac{-4\pm\sqrt{28}}{-6} \end{gathered}[/tex][tex]undefined[/tex]What is a stem and leaf plot? How is it used and how exactly do i solve one? (an example would be great)
Answers
A stem and leaf plot is a table where each of the data is divided into two parts. The stem, that is the first digit and the leaf is the last digits. Let's say that we have the following set of data.
[tex]10,\text{ 12, 25, 28, 29, 35, 38, 40, 44}[/tex]If we want to make a stem and leaf plot of that data, we first write a column where we place the first digit of each number without repetition, like this:
[tex]\begin{gathered} 1 \\ 2 \\ 3 \\ 4 \end{gathered}[/tex]These are the stems. Now the leaves are the last digit of each number put in order next to the corresponding first digit, like this:
[tex]\begin{gathered} 1\parallel\text{ 0 2} \\ 2\parallel\text{ 5 8 9} \\ 3\text{ }\parallel\text{5 8} \\ 4\text{ }\parallel\text{0 4} \end{gathered}[/tex]Polynomial Functions:Find P(-1) and p(2) for each function.“P(x) = 4-3x”
Answers
P(-1):
[tex]\begin{gathered} P(-1)=4-3(-1) \\ P(-1)=4+3 \\ P(x)=7 \end{gathered}[/tex]P(2):
[tex]\begin{gathered} P(2)=4-3(2) \\ P(2)=4-6 \\ P(2)=-2 \end{gathered}[/tex]Which formula can be used to find the sum of the mesures of all the interior angles of a regular polygon with n sides?
A. S = (n-2)180 degrees
B. S = (n+2)180 degrees
C. S = (n-2)90 degrees
D. S = (n+2)90 degrees
Answers
Answer:180(n – 2),
Step-by-step explanation:
The mean height of women in a country (ages 20-29) is 63.6 inches. A random sample of 70 women in this age group are selected. What is the probability that the mean height for the sample is greater than 64 inches?Assuming sigma= 2.53 inches
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We have a mean of 63.6 inches and a standard deviation of 2.53 inches. We want to find the probability for our random sample to have a greater mean than 64 inches. We can do that by finding the probability of getting women higher than 64 inches in the original group. To do that, we're going to use a z-table.
First, let's convert 64 inches to a z-score:
[tex]\begin{gathered} z(64)=\frac{64-\mu}{\sigma/\sqrt[]{n}}=\frac{64-63.6}{2.53/\sqrt[]{70}}=\frac{0.4\sqrt[]{70}}{2.53}=1.32278265065\ldots \\ z(64)\approx1.32 \end{gathered}[/tex]Using a right z-table, we have
This z-table gives us the area between the middle of the bell curve and our desired value.
This means, the probability of getting a sample higher than our value, will be 0.5 minus the probability given by the z-table.
[tex]0.5-0.4066=0.0934[/tex]Then, we have our result.
[tex]P(\bar{x}>64)=0.0934[/tex]What is the remainder when 5x3 + 2x2 - 7 is divided by x + 9?-93,7503,800-3,490
Answers
Explanation
Given the expression
[tex]5x^3+2x^2-7[/tex]The remainder when it is divided by x+9 can be seen below;
[tex]r=5(-9)^3+2(-9)^2-7=-3645+162-7=-3490[/tex]Answer: -3490
The fraction models below represent two fractions of the same whole: How much of the8음을16
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So 4/5 times 5/8 is 1/2.
what are the coordinates of the focus of the conic section shown below (y+2)^2/16-(x-3)^2/9=1
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Given the function of the conic section:
[tex]\mleft(y+2\mright)^2/16-\mleft(x-3\mright)^2/9=1[/tex]This conic section is a hyperbola.
Use this form below to determine the values used to find vertices and asymptotes of the hyperbola:
[tex]\frac{(x-h)^2}{a^2}\text{ - }\frac{(y-k)^2}{b^2}\text{ = }1[/tex]Match the values in this hyperbola to those of the standard form.
The variable h represents the x-offset from the origin b, k represents the y-offset from origin a.
We get,
a = 4
b = 3
k = 3
h = -2
A. The first focus of a hyperbola can be found by adding the distance of the center to a focus or c to h.
But first, let's determine the value of c. We will be using the formula below:
[tex]\sqrt[]{a^2+b^2}[/tex]Let's now determine the value of c.
[tex]\sqrt[]{a^2+b^2}\text{ = }\sqrt[]{4^2+3^2}\text{ = }\sqrt[]{16\text{ + 9}}\text{ = }\sqrt[]{25}[/tex][tex]\text{ c = 5}[/tex]Let's now determine the coordinates of the first foci:
[tex]\text{Coordinates of 1st Foci: (}h\text{ + c, k) = (-2 + 5, 3) = 3,3}[/tex]B. The second focus of a hyperbola can be found by subtracting c from h.
[tex]\text{ Coordinates of 2nd Foci: (h - c, k) = (-2 - 5, 3) = -7,3}[/tex]Therefore, the conic section has two focus and their coordinates are 3,3 and -7,3.
In other forms, the foci of the hyperbola is:
[tex]\text{ }(h\text{ }\pm\text{ }\sqrt[]{a^2+b^2},\text{ k) or (-2 }\pm\text{ 5, 3)}[/tex]Therefore, the answer is letter B.
Answer :It's A lol
Step-by-step explanation: