Given:
The initial amount of substance, No=10 g.
The amount of substance left after 9 years, N=5 g.
Since 10 g of substance is present initially, and it became 5 g(half of the initial amount) in 9 years, the half life of the substance is, t =9 years.
Hence, the expression for the amount remaining after T years is,
[tex]N(t)=N_0(\frac{1}{2})^{\frac{T}{t_{}}}[/tex]To find the amount of substance remaining after 18 years, put T=18, N0=10 and t=9 in the above equation.
[tex]\begin{gathered} N(18)=10\times(\frac{1}{2})^{\frac{18}{9}} \\ N(18)=10(\frac{1}{2})^2 \\ =\frac{10}{4} \\ =2.5\text{ g} \end{gathered}[/tex]Therefore, after 18 years 2.5 g of the radioactive substance will remain.
Instructions: Fill in the table of values for the exponential function. Insert all answers as fractions, when applicable.
Given,
The expression is:
[tex]y=-2(\frac{1}{2})^x[/tex]Required:
The value of y at x = -2, -1, 0, 1, 2.
The value of y at x = -2.
[tex]y=-2(\frac{1}{2})^{-2}=-2\times(2)^2=-2\times4=-8[/tex]The value of y at x = -1.
[tex]y=-2(\frac{1}{2})^{-1}=-2\times(2)^1=-2\times2=-4[/tex]The value of y at x = 0.
[tex]y=-2(\frac{1}{2})^0=-2\times(2)^0=-2\times1=-2[/tex]The value of y at x = 1.
[tex]y=-2(\frac{1}{2})^1=-2\times\frac{1}{2}=-1[/tex]The value of y at x = 2.
[tex]y=-2(\frac{1}{2})^2=-2\times\frac{1}{4}=-\frac{1}{2}=-0.5[/tex]The table for the different value of the function:
x y
-2
help meeeeeeeeee pleaseee !!!!!
The function 2x + 3x^2 represents the result of adding the two provided functions, f(x) and g(x).
Composite performance.An operation known as "function composition" takes two functions, f and g, and produces a new function, h, that is equal to both g and f and has the property that h(x) = g.
Given the f(x) = 2x and g(x) = 3x^2 functions
The sum of the two functions must be calculated as illustrated;
f(x) + g = (f+g)(x)
Put the provided functions in place of (f+g)(x) to have:
(f+g)(x) = 2x + 3x^2
Standard version of the expression is (f+g)(x) = 2x + 3x^2
Consequently, the sum of the functions f(x) and g(x) is2x + 3x^2
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Write a value that will make the relation not represent a function
Given:
There are given that the data for x and y are in the form of a table.
Explanation:
According to the concept of function:
The function is not defined when the value of x will be repeated.
That means if the input value is repeated again and again then the given relation will not function.
In the given relation, we can put 7 into the input box.
Final answer:
Hence, the value is 7.
Kiran is solving 2x-3/x-1=2/x(x-1) for x, and he uses these steps.He checks his answer and finds that it isn’t a solution to the original equation, so he writes “no solutions.” Unfortunately, Kiran made a mistake while solving. Find his error and calculate the actual solution(s).
Solution:
Given:
[tex]\begin{gathered} To\text{ solve,} \\ \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \end{gathered}[/tex]Kiran multiplied the left-hand side of the equation by (x-1) and multiplied the right-hand side of the equation by x(x-1).
That was where he made the mistake. He ought to have multiplied both sides with the same quantity (Lowest Common Denominator) so as not to change the actual value of the question.
Multiplying both sides by the same quantity does not change the real magnitude of the question.
The actual solution goes thus,
[tex]\begin{gathered} \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \\ \text{Multiplying both sides of the equation by the LCD,} \\ \text{The LCD is x(x-1)} \\ x(x-1)(\frac{2x-3}{x-1})=x(x-1)(\frac{2}{x(x-1)}) \\ x(2x-3)=2 \\ \text{Expanding the bracket,} \\ 2x^2-3x=2 \\ \text{Collecting all the terms to one side to make it a quadratic equation,} \\ 2x^2-3x-2=0 \end{gathered}[/tex]Solving the quadratic equation;
[tex]\begin{gathered} 2x^2-3x-2=0 \\ 2x^2-4x+x-2=0 \\ \text{Factorizing the equation,} \\ 2x(x-2)+1(x-2)=0 \\ (2x+1)(x-2)=0 \\ 2x+1=0 \\ 2x=0-1 \\ 2x=-1 \\ \text{Dividing both sides by 2,} \\ x=-\frac{1}{2} \\ \\ \\ OR \\ x-2=0 \\ x=0+2 \\ x=2 \end{gathered}[/tex]Therefore, the actual solutions to the expression are;
[tex]\begin{gathered} x=-\frac{1}{2} \\ \\ OR \\ \\ x=2 \end{gathered}[/tex]Hello, can you please help me solve this question ASAP!!!
SOLUTION:
Step 1:
In this question, we have that:
Step 2:
Part A:
We are meant to show that the equation:
[tex]5sinx=1+2cos^2x[/tex]can be written in the form
[tex]2sin^2\text{x + 5 sin x - 3=0}[/tex]Proof:
[tex]\begin{gathered} \text{5 sin x = 1 + 2 cos }^2x\text{ } \\ \text{But cos}^2x+sin^2x\text{ = 1} \\ \text{Then,} \\ \cos ^2x=1-sin^2x\text{ } \\ \text{Hence,} \\ 5sinx=1+2(1-sin^2x_{}) \\ 5sinx=1+2-2sin^2x \\ 5sinx=3-2sin^2x \end{gathered}[/tex]Re-arranging, we have that:
[tex]2sin^2x\text{ + 5 sin x - 3 = 0 }[/tex]Part B:
b) Hence, solve for x in the interval:
[tex]0\text{ }\leq\text{ x }\leq\text{ 2}\pi[/tex]7+[9÷(9x1 to the second power)]
The value of the expression 7+[9÷(9x1 to the second power)] is 64/9
What is a fraction?A fraction can be described as the part of a whole set or element.
There are several types of fractions, which includes;
Simple fractionsComplex fractionsMixed fractionsProper fractionsImproper fractionsSome examples of these fractions are given as;
Simple fractions: 1/5, 1/6
Mixed fractions: 2 1/8, 3 1/4
Proper fractions: 2/3, 4/5
Improper fractions; 4/1, 6/3
Given the expression;
7+[9÷(9x1 to the second power)]
This is expressed as;
7 + ( 9 ÷ (9)^2
Find the square
7 + ( 9 ÷ 81)
find the ratio
7 + 1/9
Find the common multiple
63 + 1 /9
64/9
Hence, the value is 64/9
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find the volume or missing value 3ft, 2.5ft, 6ft
The formula to find the volume of a rectangular prism is
[tex]\begin{gathered} V=l\cdot w\cdot h \\ \text{ Where V is the volume}, \\ l\text{ is the length,} \\ w\text{ is the width and} \\ \text{h is the height of the rectangular prism} \end{gathered}[/tex]Graphically,
So, in this case, you have
[tex]\begin{gathered} l=3ft \\ w=2.5ft \\ h=6ft \\ V=l\cdot w\cdot h \\ V=3ft\cdot2.5ft\cdot6ft \\ V=45ft^3 \end{gathered}[/tex]Therefore, the volume of the rectangular prism is 45 cubic feet.
4/7 X 1/2 = in fraction
Consider the given expression,
[tex]P=\frac{4}{7}\times\frac{1}{2}[/tex]The product of fractions is obtained in the form of a fraction whose numberator is the product of numerators of fractions, and the denominator of the product is the product of denominators of the given fractions,
[tex]\begin{gathered} P=\frac{4\times1}{7\times2} \\ P=\frac{4}{14} \end{gathered}[/tex]Thus, the product of the given fractions is 4/14 .
laws exponents multiplication band power to a power simplifymake it small steps please the smallest you canbare minimum of steps
Answer:
[tex](4r^4s^{-2})(-3rs^{-3})(rs)=-12r^6s^{-4}[/tex]Explanation:
Given the expression:
[tex](4r^4s^{-2})(-3rs^{-3})(rs)[/tex]This can be rearranged using law of multiplication (That multiplication is cummutative) to become:
[tex](4)(-3)(r^4rr)(s^{-2}s^{-3}s)[/tex]This becomes, using the law of exponents:
[tex]-12r^{4+1+1}s^{-2-3+1}[/tex]and finally, we have:
[tex]-12r^6s^{-4}[/tex]Sally started on the 12th floor. She walked up 4 flights. Then she went down 2 flights. Then she ran up 8 flights of stairs. a) Write an ADDITION expression b) What floor did she end up on? SHOW ALL WORK!
1) Gathering the data
Initial point 12th floor
2) She started on 12th floor and walked up 4 flights of stairs, assuming from each floor to another we have just 1 flight of stair. And we're using an addition expression, Hence, we can say:
12 +4-2+8=
16 +6
22
She ended up on the 22th floor
Solve for x using the quadratic formula.3x^2 +10x+8=3
The quadartic equation is 3x^2+10x+8=3.
Simplify the quadratic equation to obtain the equation in standard form ax^2+bx+c=0.
[tex]\begin{gathered} 3x^2+10x+8=3 \\ 3x^2+10x+5=0 \end{gathered}[/tex]The coefficent of x^2 is a=3, coefficient of x is b=10 and constant term is c=5.
The quadartic formula for the values of x is,
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Substitute the values in the formula to obtain the value of x.
[tex]\begin{gathered} x=\frac{-10\pm\sqrt[]{(10)^2-4\cdot3\cdot5}}{2\cdot3} \\ =\frac{-10\pm\sqrt[]{100-60}}{6} \\ =\frac{-10\pm\sqrt[]{40}}{6} \\ =\frac{-10\pm2\sqrt[]{10}}{6} \\ =\frac{-5\pm\sqrt[]{10}}{3} \end{gathered}[/tex]The value of x is,
[tex]\frac{-5\pm\sqrt[]{10}}{3}[/tex]2) Katie and Jacob are enlarging pictures in a school yearbook on the copy machine. The ratio of the width to the length of the enlarged photo will be the same as the ratio of the width to the length of the original photo. 25 points One of the photographs that they want to enlarge is a 3" x 4"photo. katie says that she can enlarge the photo to a 9" x 12", but Jacob disagrees. He says it will be 11" x 12". Who is correct? Explain your reasoning in words. * Enlarged Photo Original Photo 3 inches 4 inches
The original picture Katie and Jacob want to enlarge is 3 by 4 photographs
This means that the initial length of the photograph is 3 and the intial width of the photographs is 4
If both of them want to enlarge the photograph, then the scaling factor must be the same for both the width and length
Katie enlarge the photo to a 9 x 12
The ratio of the original photograph is 3 to 4
That is, 3 : 4
Katie enlarge the photo to a 9 x 12
Ratio of the enlarged photo by katie is 9 to 12
That is, 9 : 12
Equate the two ratio together
3/4 = 9/12
Introduce cross multiplication
We have,
3 x 12 = 4 x 9
36 = 36
Therefore, the ratio which katie enlarged the photo results to a proportion
For Jacob
Jacob enlarged the photo to 11 x 12
Equating the two ratios
3/4 = 11/12
3 x 12 = 4 x 11
36 = 44
This does not give us a proportion
Therefore, Katie is correct while Jacob is wrong
its composition of fractions in pre-calculus.I know how to do these types of questions, im just not sure how u would set it up if there are 2 x's in one of the equations.
Answer:
(f o g)(x) = x
Explanation:
Given f(x) and g(x) defined below:
[tex]\begin{gathered} f(x)=\frac{1-x}{x} \\ g(x)=\frac{1}{1+x} \end{gathered}[/tex]The composition (f o g)(x) is obtained below:
[tex]\begin{gathered} (f\circ g)(x)=f\lbrack g(x)\rbrack \\ f(x)=\frac{1-x}{x}\implies f\lbrack g(x)\rbrack=\frac{1-g(x)}{g(x)} \end{gathered}[/tex]Substitute g(x) into the expression and simplify:
[tex]\begin{gathered} f\lbrack g(x)\rbrack=\frac{1-g(x)}{g(x)}=\lbrack1-g(x)\rbrack\div g(x) \\ =(1-\frac{1}{1+x})\div(\frac{1}{1+x}) \\ \text{ Take the LCM in the first bracket} \\ =\frac{1(1+x)-1}{1+x}\div\frac{1}{1+x}\text{ } \\ \text{Open the bracket} \\ =\frac{1+x-1}{1+x}\div\frac{1}{1+x}\text{ } \\ =\frac{x}{1+x}\times\frac{1+x}{1}\text{ } \\ =x \end{gathered}[/tex]Therefore, the composition (f o g)(x) is x.
find the intercepts and graph the equation by plotting points. 13^2 + 4y = 52
ANSWER
[tex]y-intercept:(0,-\frac{117}{4})[/tex]Graph:
EXPLANATION
Given:
[tex]13^2+4y=52[/tex]Desired Results:
Intercepts and graph the equation
Solve for y
[tex][/tex]How do I understand Standard Form of a Line? I don't know how to do it.
There are several forms in which one can write the equation of a line. Have in mind that TWO variables should be included in the equation. These two variables are: x and y.
If you type the equation in a form that looks like:
A x + B y = C
where the A, B, and C are actual numbers (like for example: 3 x - 2 y = 5)
This is the standard form of a line. to recognize it notice that bith variables x an y appear in separate terms on the LEFT of the equal sign., and a pure number (no variables) appears on the right of the equal sign.
Another form of writing the equation of a line is in the so called "solpe-intercept" form. This form looks like:
y = m x + b
Notice that in this case the variable ÿ" appears isolated on the left , and on the right of the equal sign you get a term with the variable x, and another constant (pure number) term (b). Like for example in the case of:
y = 3 x
I’ve already done this problem, but I’m being told it’s wrong and I need to simplify but I don’t know how to do it with this question.
Sarkis OganesyanCombine Like Terms (Basic, Decimals)May 20, 11:02:29 AMA triangle has side lengths of (1.1p + 9.59) centimeters, (4.5p - 5.2r)centimeters, and (5.3r + 5.4q) centimeters. Which expression represents theperimeter, in centimeters, of the triangle?14.89 + 5.6p + 0.2rO 0.1r + 5.6p + 14.99Submit Answer-0.7pr + 10.7qr + 10.6pq9.7qr + 10.9pr
The sides of the triagle have lengths:
1.1 p + 9.5 q
4.5 p - 5.2 r
5.3 r + 5.4 q
Or:
1.1 p + 0 r + 9.5 q
4.5 p - 5.2 r + 0 q
0 p + 5.3 r + 5.4 q
If we want to calculate the perimeter of the triangle, we just need to sum the three lenghts:
(1.1 + 4.5) p + (-5.2 + 5.3) r + (9.5 + 5.4) q
= 5.6 p + 0.1 r + 14.9 q
Given the following confidence interval for a population mean compute the margin of error E
Given that the Confidence Interval for a population mean:
[tex]11.81<\mu<13.21[/tex]In this case, you can set up these two equations:
[tex]\bar{x}+E=13.21\text{ \lparen Equation 1\rparen}[/tex][tex]\bar{x}-E=11.81\text{ \lparen Equation 2\rparen}[/tex]Because by definition:
[tex]\bar{x}-E<\mu<\bar{x}+E[/tex]Where "ME" is the margin of error and this is the mean:
[tex]\bar{x}[/tex]In this case, in order to find the "ME", you need to follow these steps:
1. Add Equation 1 and Equation 2:
[tex]\begin{gathered} \bar{x}+E=13.21 \\ \bar{x}-E=11.81 \\ -------- \\ 2\bar{x}=25.02 \end{gathered}[/tex]2. Solve for the mean:
[tex]\begin{gathered} \bar{x}=\frac{25.02}{2} \\ \\ \bar{x}=12.51 \end{gathered}[/tex]3. Substitute the mean into Equation 1 and solve for "ME":
[tex]12.51+E=13.21[/tex][tex]\begin{gathered} E=13.21-12.51 \\ E=0.7 \end{gathered}[/tex]Hence, the answer is:
[tex]E=0.7[/tex]х3,2y=x?(x, y)00(0,0)2.4(2, 4)For which value of x is the row in the table of values incorrect?3The function is the quadratic function y = -x?4366를18(3,6)(5,18 )5
Since the given equation is
[tex]y=\frac{3}{4}x^2[/tex]If x = 0, then
[tex]y=\frac{3}{4}(0)^2=0[/tex]Then x = 0 is correct because it gives the same value of y in the table
If x = 2
[tex]\begin{gathered} y=\frac{3}{4}(2)^2 \\ y=\frac{3}{4}(4) \\ y=3 \end{gathered}[/tex]Since the value of y in the table is 4
Then x = 2 is incorrect
Suppose you want to have $ 749,791 for retirement in 13 years. Your account earns 9.5 % interest monthly. How much interest will you earn?$_________ (Round to the nearest DOLLAR)
ANSWER
$530,663
EXPLANATION
The amount the account will have in t years is given by,
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Where n = 12, t = 13 years, r = 0.095 and A = 749,791. We have to find P,
[tex]P=\frac{A}{(1+\frac{r}{n})^{nt}}[/tex]Replace with the values and solve,
[tex]P=\frac{749,791}{(1+\frac{0.095}{12})^{12\cdot13}}\approx219,128[/tex]The interest earned is the difference between the initial deposit P and the final amount A,
[tex]i=A-P=749,791-219,128=530,663[/tex]Hence, the interest earned would be $530,663.
cos(alpha + beta) = cos^2 alpha - sin^2 beta
The trigonometric identity cos(α + β)cos(α - β) = cos²(α) - sin²(β) is verified in this answer.
Verifying the trigonometric identityThe identity is defined as follows:
cos(α + β)cos(α - β) = cos²(α) - sin²(β)
The cosine of the sum and the cosine of the subtraction identities are given as follows:
cos(α + β) = cos(α)cos(β) - sin(α)sin(β).cos(α - β) = cos(α)cos(β) + sin(α)sin(β).Hence, the multiplication of these measures is given as follows:
cos(α + β)cos(α - β) = (cos(α)cos(β) - sin(α)sin(β))(cos(α)cos(β) + sin(α)sin(β))
Applying the subtraction of perfect squares, it is found that:
(cos(α)cos(β) - sin(α)sin(β))(cos(α)cos(β) + sin(α)sin(β)) = cos²(α)cos²(β) - sin²(α)sin²(β)
Then another identity is applied, as follows:
sin²(β) + cos²(β) = 1 -> cos²(β) = 1 - sin²(β).sin²(α) + cos²(α) = 1 -> sin²(α) = 1 - cos²(a).Then the expression is:
cos²(α)cos²(β) - sin²(α)sin²(β) = cos²(α)(1 - sin²(β)) - (1 - cos²(a))sin²(β)
Applying the distributive property, the simplified expression is:
cos²(α) - sin²(β)
Which proves the identity.
Missing informationThe complete identity is:
cos(α + β)cos(α - β) = cos²(α) - sin²(β)
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Linear Programming WorksheetGraph each feasible region. maximize or minimize each objective
Given:
x+2y = 8
x=2, y=0
Substitute x=2 then find value of x as,
2+2y=8
2y=6
y=3
(x,y) = (0,3)
Now, substitute y=0 then find value of y as,
x+2(0)=8
x=8
(x,y) = (8,0)
It is given that P = x+3y
(x,y) = (0,3) then P= 0+3x3
P=9
The maximum valu P=9 and vertiex (0,3)
(x,y) = (8,0) then P=8+0= 8
The mininmum val
1 mile= 1,760 yards.1 kilometer= 1,000 metersIf Jose walked 2 miles this morning, about how many kilometers did he walk?
1 mile= 1.609 km
Then,
2*1.609=3.218 km
He walked 3.218 kilometers
On a number line, let point P represent the largest integer value that is less than V380.Let point Q represent the largest integer value less than 54.What is the distance between P and Q?A. 10B. 11C. 12D. 13
We have to find P and Q first.
P is the largest integer that is less than the square root of 380.
P is 19.
Q is the largest number that is less than the square of 54.
Q is 7.
Then the distance between P and Q is |19-7|=12.
Answer: C. 12
Simplify the following equations in ax^2+bx+c=0 or ay^2+c=0 2x+y=6 4x^2+5y+y+1=0
Given the equation;
[tex]4x^2+5y^2+y+1=0[/tex]We shall begin by Subtracting 5y^2 + y from both sides;
[tex]\begin{gathered} 4x^2+5y^2+y+1-5y^2-y=0-5y^2-y \\ 4x^2+1=-5y^2-y \\ \text{Factor out -1 from the right hand side;} \\ 4x^2+1=-1(5y^2+y) \end{gathered}[/tex]Next step we subtract 1 from both sides;
[tex]\begin{gathered} 4x^2+1-1=-1(5y^2+y)-1 \\ 4x^2=-(5y^2+y)-1 \\ \end{gathered}[/tex]Next step we take the square root of both sides;
[tex]\begin{gathered} \sqrt[]{4x^2}=\pm\sqrt[]{-(5y^2+y)-1} \\ 2x=\pm\sqrt[]{-(5y^2+y)-1} \end{gathered}[/tex]We can now open the parenthesis on the right hand side;
[tex]\begin{gathered} 2x=\pm\sqrt[]{-5y^2-y-1} \\ \text{Divide both sides by 2;} \\ x=\frac{\pm\sqrt[]{-5y^2-y-1}}{2} \end{gathered}[/tex][tex]undefined[/tex]use the generic rectangle 3x-8)² and -7x⁴(3x-2) what's the product and sum?
In this case the answer is very simple .
Step 01:
Data:
eq1. (3x - 8)²
eq2. -7x⁴(3x-2)
Step 02:
Sum.
eq.1 + eq.2
(3x - 8)² + (-7x⁴(3x-2))
(9x² - 2*3x*8 - 64) + (-21x⁴ - 14x⁴)
9x² -
A manufacturer pays its assembly line workers $11.06 per hour. In addition, workers receive a piece of work rate of $0.34 per unit produced. Write a linear equation for the hourly wages W in terms of the number of units x produced per hour. Linear equation: W = _______ What is the hourly wage for Mike, who produces 17 units in one hour? Mike’s wage = _________
Let's assume the following variables.
x = number of units produced
It is stated in the problem that for every unit produced, there is an additional wage of $0.34. Hence, on top of $11.06 per hour wage, there will be an additional of $0.34x per hour. In equation, we have wage per hour:
[tex]W=11.06+0.34x[/tex]If Mike was able to produce 17 units, our x here is 17. Let's plug this value to the formula.
[tex]W=11.06+0.34(17)[/tex]Then, solve.
[tex]\begin{gathered} W=11.06+5.78 \\ W=16.84 \end{gathered}[/tex]Therefore, Mike's hourly wage is $16.84.
Please help me my answer is correct or no
Answer:
the answer is c actully
Step-by-step explanation:
iv'e took that test b4 so you welcome
Graph v (standard position) and find its magnitude. Show all work.
EXPLANATION
[tex]\mathrm{Computing\: the\: Euclidean\: Length\: of\: a\: vector}\colon\quad \mleft|\mleft(x_1\: ,\: \: \ldots\: ,\: \: x_n\mright)\mright|=\sqrt{\sum_{i=1}^n\left|x_i\right|^2}[/tex][tex]=\sqrt{2^2+5^2}[/tex][tex]=\sqrt{4+5^2}[/tex][tex]=\sqrt{4+25}[/tex][tex]=\sqrt{29}[/tex]Now, we need to graph the vector as shown as follows:
INT ALGEBRAL: 1. Write an equation that passes through (0,5) and is parallel to 3x+5y=6
Thank you for your help, and please do show work! I will be looking to give the Brainliest answer to someone!
The equation of the parallel line is y = -3/5x + 5
How to determine the line equation?The equation is given as
3x + 5y = 6
The point is also given as
Point = (0, 5)
The equation of a line can be represented as
y = mx + c
Where
Slope = m and c represents the y-intercept
So, we have
3x + 5y = 6
This gives
5y = -3x + 6
Divide
y = -3/5x + 6/5
By comparing the equations y = mx + c and y = -3/5x + 6/5, we have the following
m = -3/5
This means that the slope of y = -3/5x + 6/5 is -3/5
So, we have
m = -3/5
The slopes of parallel lines are equal
This means that the slope of the other line is -3/5
The equation of the parallel line is then calculated as
y = m(x - x₁) +y₁
Where
m = -3/5
(x₁, y₁) = (0, 5)
So, we have
y = -3/5(x + 0) + 5
Open the brackets and evaluate
y = -3/5x + 5
Hence, the parallel line has an equation of y = -3/5x + 5
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