9514 1404 393
Answer:
vertical shift: 7 (up)horizontal shift: 2 (right)vertical asymptote: x=2domain: x > 2range: all real numbersStep-by-step explanation:
For any function f(x), the transformation f(x -h) +k represents a horizontal shift of h units to the right and k units upward.
Here, the parent function is log₂(x) and the transformation to log₂(x -2) +7 represents translation 2 units right and 7 units upward.
The parent function has a vertical asymptote at x=0, so the shifted function will have a vertical asymptote at x-2=0, or x = 2.
The parent function has a domain of x > 0, so the shifted function will have a domain of x-2 > 0, or x > 2.
The parent function has a range of "all real numbers." Shifting the function vertically does not change that range. The range of the shifted function is still "all real numbers."
The graph is shown below. The vertical asymptote is the dashed orange line.
_____
The "work" is in matching the pattern f(x -h) +k to the function log₂(x -2) +7.
Simplify cube root of 7 over fifth root of 7. 7 to the power of one fifth 7 to the power of eight fifteenths 7 to the power of five thirds 7 to the power of two fifteenths
Answer:
[tex]\huge\boxed{7^{\frac{2}{15}}}[/tex]
Step-by-step explanation:
[tex]\dfrac{\sqrt[3]7}{\sqrt[5]7}\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\dfrac{7^\frac{1}{3}}{7^\frac{1}{5}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=7^{\frac{1}{3}-\frac{1}{5}}\qquad\text{find the common denominator (15)}\\\\=7^{\frac{(1)(5)}{(3)(5)}-\frac{(1)(3)}{(5)(3)}}=7^{\frac{5-3}{15}}=7^{\frac{2}{15}}[/tex]
Answer:
D. 7 to the power of two fifteenths
Step-by-step explanation:
50q + 43 > −11q + 70
Answer:
q > 27/61
Step-by-step explanation:
50q + 43 > −11q + 70
Add 11 q to each side
50q+11q + 43 > −11q+11q + 70
61q+43> 70
Subtract 43 from each side
61q> 27
Divide each side by 61
61q/61> 27/61
q > 27/61
Write "six and thirty-four thousandths" as a decimal
Answer:
6.034
Step-by-step explanation:
6 is a whole number.
.034 because it is 34 thousandths, not 34 hundredths.
5/7 minus 2/9 please
Answer:
[tex]\large \boxed{31/63}[/tex]
Step-by-step explanation:
5/7 - 2/9
Make denominators equal by LCM.
(5 × 9)/(7 × 9) - (2 × 7)/(9 × 7)
45/63 - 14/63
Subtract fractions since denominators are equal.
(45 - 14)/63
31/63
Answer:
[tex]\frac{31}{63}[/tex]
Step-by-step explanation:
Find the LCM of 7 and 9: 63Find how much we increased each number to get to 63: we increased 7 by 9, and we increased 9 by 7Multiply the numerators by the corresponding increase numbers: 5 × 9 = 45, and 2 × 7 = 14Put the new numerators over the new denominators, so it looks like this: [tex]\frac{45}{63}[/tex] and [tex]\frac{14}{63}[/tex] Finally, subtract one from the other and here's what you get: [tex]\frac{31}{63}[/tex]Therefore, the answer is [tex]\frac{31}{63}[/tex].
Can someone please help !
In functions; transformations are used to move lines, points, curves or shapes across the graph (i.e. up, down, right or left). After the transformation of [tex]f(x) = \log\ x[/tex], the resulting function is [tex]g(x) = 2\log\ (-x + 4) - 6[/tex]
Given that:
[tex]f(x) = \log\ x[/tex]
When translated 4 units left, the rule of translation is:
[tex](x,y) \to (x + 4,y)[/tex]
So, we have:
[tex]f'(x) = \log\ (x + 4)[/tex]
When translated 3 units down, the rule of translation is:
[tex](x,y) \to (x,y-3)[/tex]
So, we have:
[tex]f"(x) = \log\ (x + 4) - 3[/tex]
When reflected about the y-axis, the rule of reflection is:
[tex](x,y) \to (-x,y)[/tex]
So, we have:
[tex]f'"(x) = \log\ (-x + 4) - 3\\[/tex]
Lastly, when it is vertically stretched by 2; the rule is:
[tex](x,y) \to (x,2y)[/tex]
So, we have:
[tex]g(x) = 2[\log\ (-x + 4) - 3][/tex]
Open bracket
[tex]g(x) = 2\log\ (-x + 4) - 6[/tex]
See attachment for the graph of function f(x) and the new function g(x)
Read more about function transformations at:
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Solve -9 < 4x + 3 5 19.
Answer:
C -3 < x ≤ 4
Step-by-step explanation:
-9 < 4x + 3 ≤ 19.
Subtract 3 from all sides
-9-3 < 4x + 3-3 ≤ 19-3
-12 < 4x ≤ 16
Divide by 4
-12/4 < 4x/4 ≤ 16/4
-3 < x ≤ 4
Tom is afraid of heights above 9 feet. He is asked to repair a side of a high deck. The bottom of the ladder must be placed 6 feet from a deck. The ladder is 10 feet long. How far above the ground does the ladder touch the deck? Is Tom afraid of the height?
Answer:
8 ftnoStep-by-step explanation:
The height on the side of the deck (h) can be found using the Pythagorean theorem. It tells you ...
6^2 + h^2 = 10^2
h = √(10^2 -6^2) = √64 = 8
The ladder touches the deck 8 feet above the ground. Tom is not afraid of that height.
Find m∠DFC.
A. 23
B. 167
C. 46
D. 69
Answer:
A.23
maaf kalosalah !(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!
identify(describe) each part of the ellipse as labeled by a letter
Answer: see below
Step-by-step explanation:
A) y has the smaller radius so this is the Minor Axis
B) y has the smaller radius so these are the CoVertices
C) x has the bigger radius so these are the Vertices
D) This is the Center of the ellipse.
F & G) These are the Foci (plural for Focus)
H) x has the bigger radius so this is the Major Axis
HELP ME PLEASE!!!!!!!!! WORTH 100 POINTS WILL FOLLOW AND RATE BRAINLIEST ANSWER!!!!!!!!!!!! If the function f(x) has a domain of (a,b] and a range of [c,d), then what is the domain and range of g(x)=m×f(x)+n?(1 point) A.The domain of g(x) is (ma+n,mb+n], and the range is [c,d). B.The domain of g(x) is (a,b], and the range is [mc+n,md+n). C.The domain of g(x) is (a,b], and the range is [c,d). D.The domain of g(x) is (ma+n,mb+n], and the range is [mc+n,md+n).
Greetings from Brasil...
If domain is (a,b] and the range is [c,d), so:
f(a) = c
f(b) = d
so
g(x) = m × f(x) + n
g(a) = m × f(a) + n how f(a) = c, then
g(a) = mc + n
g(b) = m × f(b) + n
g(b) = md + n
So
Domain = (a; b]Range = [mc + n; md + n)graph the linear equation. Find three points that solve the equation, the plot them on the graph. -2y= 5x +11
Answer:
Three points are (0,-5.5), (-1,-3), (-2.2,0) and graph is shown below.
Step-by-step explanation:
The given equation is
[tex]-2y=5x+11[/tex]
We need to find three points that solve the equation.
Put x=0,
[tex]-2y=5(0)+11[/tex]
[tex]-2y=11[/tex]
[tex]y=-5.5[/tex]
Put x=-1,
[tex]-2y=5(-1)+11[/tex]
[tex]-2y=6[/tex]
[tex]y=-3[/tex]
Put y=0,
[tex]-2(0)=5x+11[/tex]
[tex]5x=-11[/tex]
[tex]x=-2.2[/tex]
So, three points (0,-5.5), (-1,-3) and (-2.2,0) are the solutions of the given equation.
Plot these points on a coordinate plane and connect them by a straight line as shown below.
Consider the surface f(x,y) = 21 - 4x² - 16y² (a plane) and the point P(1,1,1) on the surface.
Required:
a. Find the gradient of f.
b. Let C' be the path of steepest descent on the surface beginning at P, and let C be the projection of C' on the xy-plane. Find an equation of C in the xy-plane.
c. Find parametric equations for the path C' on the surface.
Answer:
A) ( -8, -32 )
Step-by-step explanation:
Given function : f (x,y) = 21 - 4x^2 - 16y^2
point p( 1,1,1 ) on surface
Gradient of F
attached below is the detailed solution
find the response of the function at * (t=4) using Laplace transform (y() + 2y" + y = sint) y(0)=1, y (0)=-2, y"(0)=3 , y"(0)=0
Considering you have four initial conditions (the last of which should probably read [tex]y'''(0)=0[/tex]), I'm assuming the ODE is
[tex]y^{(4)}(t)+2y''(t)+y(t)=\sin t[/tex]
with [tex]y(0)=1[/tex], [tex]y'(0)=-2[/tex], [tex]y''(0)=3[/tex], and [tex]y'''(0)=0[/tex].
Take the Laplace transform of both sides, denoting the transform of [tex]y(t)[/tex] by [tex]Y(s)[/tex]:
[tex](s^4Y(s)-s^3y(0)-s^2y'(0)-sy''(0)-y'''(0))+2(s^2Y(s)-sy(0)-y'(0))+Y(s)=\dfrac1{s^2+1}[/tex]
Solve for [tex]Y(s)[/tex]:
[tex](s^4+2s^2+1)Y(s)-s^3+2s^2-5s+4=\dfrac1{s^2+1}[/tex]
[tex]Y(s)=\dfrac{1+(s^3-2s^2+5s-4)(s^2+1)}{(s^2+1)(s^4+2s^2+1)}[/tex]
Notice that
[tex]s^4+2s^2+1=(s^2+1)^2[/tex]
[tex]\implies Y(s)=\dfrac{1+(s^3-2s^2+5-4)(s^2+1)}{(s^2+1)^3}[/tex]
and simplify a bit to get
[tex]Y(s)=\dfrac{s^5-2s^4+6s^3-6s^2+5s-3}{(s^2+1)^3}[/tex]
Decompose [tex]Y(s)[/tex] into partial fractions:
[tex]\dfrac{s^5-2s^4+6s^3-6s^2+5s-3}{(s^2+1)^3}=\dfrac{a_0+a_1s}{s^2+1}+\dfrac{b_0+b_1s}{(s^2+1)^2}+\dfrac{c_0+c_1s}{(s^2+1)^3}[/tex]
[tex]s^5-2s^4+6s^3-6s^2+5s-3=(a_0+a_1s)(s^2+1)^2+(b_0+b_1s)(s^2+1)+(c_0+c_1s)[/tex]
[tex]s^5-2s^4+6s^3-6s^2+5s-3=a_1s^5+a_0s^4+(2a_1+b_1)s^3+(2a_0+b_0)s^2+(a_1+b_1+c_1)s+(a_0+b_0+c_0)[/tex]
[tex]\implies\begin{cases}a_1=1\\a_0=-2\\2a_1+b_1=6\\2a_0+b_0=-6\\a_1+b_1+c_1=5\\a_0+b_0+c_0=-3\end{cases}[/tex]
[tex]\implies a_0=-2,a_1=1,b_0=-2,b_1=4,c_0=1,c_1=0[/tex]
So we have
[tex]Y(s)=\dfrac{s-2}{s^2+1}+\dfrac{4s-2}{(s^2+1)^2}+\dfrac1{(s^2+1)^3}[/tex]
Split up the first term to get two easy inverse transforms:
[tex]L^{-1}\left[\dfrac s{s^2+1}\right]=\cos t[/tex]
[tex]L^{-1}\left[-\dfrac2{s^2+1}\right]=-2\sin t[/tex]
Also split up the second term, but use the convolution theorem, which says
[tex]L\left[(\alpha \ast \beta)(t)\right]=A(s)\cdot B(s)[/tex]
where [tex]A(s)[/tex] and [tex]B(s)[/tex] are the Laplace transforms of [tex]\alpha(t)[/tex] and [tex]\beta(t)[/tex], respectively, and the convolution is defined by
[tex](\alpha \ast \beta)(t)=\displaystyle\int_0^t\alpha(\tau)\beta(t-\tau)\,\mathrm d\tau[/tex]
Take
[tex]A(s)=\dfrac{4s}{s^2+1}\text{ and }B(s)=\dfrac1{s^2+1}[/tex]
so that
[tex]\alpha(t)=4\cos t\text{ and }\beta(t)=\sin t[/tex]
and their convolution is
[tex]L^{-1}\left[\dfrac{4s}{(s^2+1)^2}\right]=(\alpha \ast \beta)(t)=2t\sin t[/tex]
Next, take
[tex]A(s)=-\dfrac2{s^2+1}\text{ and }B(s)=\dfrac1{s^2+1}[/tex]
[tex]\implies \alpha(t)=-2\sin t\text{ and }\beta(t)=\sin t[/tex]
[tex]\implies L^{-1}\left[-\dfrac2{(s^2+1)^2}\right]=t\cos t-\sin t[/tex]
You can treat the third term similarly, but with an extra step. First compute
[tex]L^{-1}\left[\dfrac1{(s^2+1)^2}\right][/tex]
by taking
[tex]A(s)=B(s)=\dfrac1{s^2+1}[/tex]
[tex]\implies \alpha(t)=\beta(t)=\sin t[/tex]
Then
[tex]L^{-1}\left[\dfrac1{(s^2+1)^2}\right]=\dfrac{\sin t-t\cos t}2[/tex]
Next, take
[tex]A(s)=\dfrac1{(s^2+1)^2}\text{ and }B(s)=\dfrac1{s^2+1}[/tex]
[tex]\implies \alpha(t)=\dfrac{\sin t-t\cos t}2\text{ and }\beta(t)=\sin t[/tex]
[tex]\implies L^{-1}\left[\dfrac1{(s^2+1)^3}\right]=\dfrac{(3-t^2)\sin t-3t\cos t}8[/tex]
Thus we end up with the solution,
[tex]y(t)=(\cos t-2\sin t)+(2t\sin t+t\cos t-\sin t)+\dfrac{(3-t^2)\sin t-3t\cos t}8[/tex]
[tex]\boxed{y(t)=\dfrac{(8+5t)\cos t+(-21+16t-t^2)\sin t}8}[/tex]
Estimate the mean exam score for the 50 students in Prof. Burke's class.
Score
f
40 but less than 50
21
50 but less than 60
39
60 but less than 70
40
70 but less than 80
34
80 but less than 90
28
Total
162
Group of answer choices
63.78
62.34
64.89
65.56
==========================================================
Explanation:
The interval [tex]40 \le x < 50[/tex] has the midpoint (40+50)/2 = 90/2 = 45 which represents the average value from this interval. We don't know what the exact 21 values are from this interval, but the best guess we can make is each value is on average 45.
Similarly, the interval [tex]50 \le x < 60[/tex] has the midpoint (50+60)/2 = 110/2 = 55 which represents the average value in this interval.
The set of all midpoints is: {45, 55, 65, 75, 85}. You start at 45 and add 10 to each term to get the next term. Let's say x represents the midpoint. We'll multiply each x value with its corresponding frequency (f) to get a new column of values you see in the table below.
For example, in the first row, we have 45*21 = 945
Add up everything in the x*f column and we'll get this sum:
945+2145+2600+2550+2380 = 10,620
We'll divide this over the total frequency, which is the sum of the frequency column (21+39+40+34+28 = 162)
We then arrive at this estimated final answer: (10,620)/(162) = 65.55555... where the '5's go on forever. This rounds to 65.56
What is the error in this problem
Answer:
The error is the use of wrong trigonometric ratio formula.
Sine was used instead of tangent.
It should be: [tex] tan(A) = \frac{36}{84} [/tex]
Step-by-step explanation:
Side length, 36, is opposite to <A. Side length, 84, is the adjacent side. Therefore, the right trigonometric ratio formula to use is:
[tex] tan(A) = \frac{opposite}{adjacent} [/tex]
[tex] tan(A) = \frac{36}{84} [/tex]
[tex] A = tan^{-1}(\frac{36}{84}) [/tex]
m<A ≈ 23°
The error made was the use of wrong trigonometric ratio formula.
Margaret drives 188 miles
with 8 gallons of gas. Find the unit rate
The unit rate will be "23.5 miles/gallon". In the below segment, a further solution to the given question is provided.
Given values in the question are:
Total distance,
= 188 miles
Total gas used,
= 8
Now,
⇒ The rate of gas consumption will be:
= [tex]\frac{Total \ distance}{Total \ gas \ used}[/tex]
By putting the given values in the above formula, we get
= [tex]\frac{188}{8}[/tex]
= [tex]23.5 \ miles/gallon[/tex]
Thus the above is the appropriate solution.
Learn more about gas consumption here:
https://brainly.com/question/17321062
the point p(-3,4) is reflected in the line x +2=0. find the coordinate of the image x
Answer:
(- 1, 4 )
Step-by-step explanation:
The line x + 2 = 0 can be expressed as
x + 2 = 0 ( subtract 2 from both sides )
x = - 2
This is the equation of a vertical line parallel to the y- axis and passing through all points with an x- coordinate of - 2
Thus (- 3, 4 ) is 1 unit to the left of - 2
Under a reflection in the line x = - 2
The x- coordinate will be the same distance from x = - 2 but on the other side while the y- coordinate remains unchanged.
Thus
(- 3, 4 ) → (- 1, 4 )
Please help!! find the value of the expression
Answer:
7
Step-by-step explanation:
First plug in the variable amounts so the expression now looks like this:
(3 × 4 - 12) + 1/2(4 × 6 - 10)
Now, start by solving the multiplication parts first, so it now looks like this:
(12 - 12) + 1/2(24 - 10)
Now, apply the rules of order of operation, so start by solving what's in parenthesis. It should now look like this: (0) + 1/2(14)
Next, solve the multiplication part, so it now looks like this: 0 + 7
Solve that and the answer is 7.
How do I solve? Show with steps.
Step-by-step explanation:
or,[(√-x-1)+(√x+9)]^2=4^2
or,(√-x-1)^2+2√(-x-1)(x+9)+(√x+9)^2=16
or,-x-1+2√-x^2-10x-9 +x+9=16
or,2√-x^2-10x-9=8
or,√-x^2-10x-9=4
squaring on both sides
or,-x^2-10x-9=16
or,-x^2-10x=25
or,-x(x+10)=25
Either,
x=-25 or, x=15.
Which equation represents the data shown in the table provided in the image?
A. y = 2x + 1
B. y = 3x — 1
C. y = 2.5x
D. y = 2.5x + 1
Please include ALL work! <3
The correct answer is A. y = 2x +1
Explanation:
An equation is a statement that shows equality. In this context, the equation should lead to two equal numbers even if the values of y and x change. In this context, the correct equation is y= 2X + 1 because this is the only one, in which, the value of Y is always equivalent to 2x + 1. To prove this, let's replace y and x for the values of the table.
First column
5 = 2 · 2 + 1
5 = 4 + 1
5 = 5
Second Column
9 = 2 · 4 + 1
9 = 8 + 1
9 = 9
Third column
13 = 6 · 2 + 1
13 = 12 + 1
13 = 13
Fourth column
17 = 8 · 2 + 1
17 = 16 + 1
17 = 17
I have a circle that has a radius of 8 in. What is the circumference of the circle? What is the area of the circle? ( use 3.14 for pi).Explain your steps. Please Give A clear explanation The best answer gets brainliest.
Answer:
The circumference is 50.24 in. and the area is 200.96 in².
Step-by-step explanation:
The circumference formula is C = 2πr where C = Circumference, π = pi and r = radius. We know that r = 8 and π = 3.14 and that we're solving for C, so we can substitute those values into the equation to get C = 2 * 3.14 * 8 = 50.24 in.
The area formula is A = πr² where A = Area, π = pi and r = radius. Again, we're solving for A and we know that r = 8 and π = 3.14 so A = 3.14 * 8² = 3.14 * 64 = 200.96 in².
Answer:
The circumference is 50.24 in. and the area is 200.96 in².
Step-by-step explanation:
MARK SNOG AS BRAINLIEST
What is the congruence correspondence, if any, that will prove the given triangles congruent?
A. ASA
B. SSS
C. SAS
B. HL
Answer:
HLStep-by-step explanation:
The congruence correspondence between the two given triangles is HL as one leg and hypotenuse are given equal.
----------------------
OAmalOHopeO
----------------------
A line passes through (-5, -3) and is parallel to -3x - 7y = 10. The equation of the line in slope-intercept form is _____
Answer:
-3x - 7y = 36
Step-by-step explanation:
The given line -3x - 7y = 10 has an infinite number of parallel lines, all of the form -3x - 7y = C.
If we want the equation of a line parallel to -3x - 7y = 10 that passes through (-5, -3), we substitute -5 for x in -3x - 7y = 10 and substitute -3 for y in -3x - 7y = 10:
-3(-5) - 7(-3) = C, or
15 + 21 = C, or C = 36
Then the desired equation is -3x - 7y = 36.
The balances in two separate bank accounts that grow each month at different rales are represented by the functions f(x) and gix) In what month do the funds in the f(x) bank account exceed those in the glx)
bank account?
Month (x) f(x) = 2* g(x) = 4x + 12
1
2
16
2.
4
20
O Month 3
O Month 4
O Month 5
O Month 6
Answer:
The balance in two separate bank accounts grows each month at different rates. the growth rates for both accounts are represented by the functions f(x) = 2x and g(x) = 4x 12. in what month is the f(x) balance greater than the g(x) balance?
Answer:
6 months
A function is a relationship between inputs where each input is related to exactly one output.
x = 5,
f(5) = [tex]2^5\\[/tex] = 32
g(5) = 4 x 5 + 12 = 20 + 12 = 32
x = 6,
f(6) = [tex]2^6[/tex] = 64
g(6) = 4 x 6 + 12 = 24 + 12 = 36
At month 6 the funds in the f(x) bank account exceed those in the g(x) bank account.
What is a function?A function is a relationship between inputs where each input is related to exactly one output.
Example:
f(x) = 2x + 1
f(1) = 2 + 1 = 3
f(2) = 2 x 2 + 1 = 4 + 1 = 5
The outputs of the functions are 3 and 5
The inputs of the function are 1 and 2.
We have,
f(x) = [tex]2^{x}[/tex]
g(x) = 4x + 12
x = number of months
Now,
x = 3,
f(3) = 2³ = 8
g(3) = 4 x 3 + 12 = 12 + 12 = 24
x = 4,
f(4) = [tex]2^4[/tex] = 16
g(4) = 4 x 4 + 12 = 16 + 12 = 28
x = 5,
f(5) = [tex]2^5\\[/tex] = 32
g(5) = 4 x 5 + 12 = 20 + 12 = 32
x = 6,
f(6) = [tex]2^6[/tex] = 64
g(6) = 4 x 6 + 12 = 24 + 12 = 36
We see that,
At x = 6,
f(5) = 64
g(5) = 36
Thus,
At month 6 the funds in the f(x) bank account exceed those in the g(x) bank account.
Learn more about functions here:
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I will mark u brainleiest if u help me and 5 stars
Answer:
[tex]\boxed{50}[/tex]
Step-by-step explanation:
Because the initial temperature is 40 degrees and it increases by 10, add the two values together to get the final temperature.
40 + 10 = 50
Therefore, the final answer is 50 degrees.
Answer:
50
Step-by-step explanation:
If it starts at 40 degrees and increases 10 degrees, it is going to be 50 degrees. Increases means adding, so it is asking you to add 10 to 40 which is 50. If it asks decreases in the future you will have to subtract.
find the area of the triangle
9514 1404 393
Answer:
108 cm²
Step-by-step explanation:
The area of a triangle is given by the formula ...
A = 1/2bh
where b represents the base length, and h represents the height--the perpendicular distance from the base to the opposite vertex. The area of this triangle is ...
A = 1/2(12 cm)(18 cm) = 108 cm²
find the value of x and y
Answer:
x=50 and y=80
Step-by-step explanation:
ATQ, x+50+y=180 and y+2x=180. x=50 and y=80
solve for x in the equation:log6-2logx=1.4
9514 1404 393
Answer:
x = (√6)/10^0.7 ≈ 0.488737
Step-by-step explanation:
It can work reasonably well to take antilogs first.
[tex]\log(6)-2\log(x)=1.4\\\\\dfrac{6}{x^2}=10^{1.4}\\\\\dfrac{6}{10^{1.4}}=x^2\\\\x=\dfrac{\sqrt{6}}{10^{0.7}}\approx0.488737[/tex]
HELP MEEEE FAST PLZ
MY MOM THINKS IM SMART BUT I DON'T GET THIS AHHHH HELP
Answer:
10
Step-by-step explanation:
-3 = 7 - 10
-3 = -3
determine the coordinator of the point
of intersection of lines
3x-2y=13 and 2y+x+1=0
Answer:
(3,-2)
Step-by-step explanation:
Given equations of line
3x-2y=13
2y+x+1=0
=> x = -1 -2y
Point of intersection will coordinates where both equation have same value of (x,y)
top get that we have to solve the both equations by using method of substitution of simultaneous equation.
using this value of x in 3x-2y=13, we have
3(-1-2y) -2y = 13
=> -3 -6y-2y = 13
=> -8y = 13+3 = 16
=> y = 16/-8 = -2
x = -1 - 2y = -1 -2(-2) = -1+4= 3
Thus, point of intersection of line is (3,-2)