Answer:
C. 1655
Step-by-step explanation:
+10, +4, -21, +23, -8
Adding all those terms together we get 8
1647 + 8 = 1655
f(x) = -2x² + 2x
Find f(-8)
Answer:
f(-8)= -112
Step-by-step explanation:
-2(-8)^2 + 2(-8)
-2(-64) + 16
-128 + 16
= -112
hope this helped
have a good day :)
Divide the polynomial by the binomial. (Simplify your answer completely.)
(q² + 5q + 20) / (q + 8)
Answer:
It's already in its simplest form.
Step-by-step explanation:
Factorising QuadraticsI'm writing a book/document on this topic but it's not finished. I suspect there is an error in this question because it's practically impossible to factorise it into integers.
The quadratic polynomial in the numerator has imaginary roots because 5 squared is less than 20 times 4.
ax² + bx + c;
if ( b² < 4ac ) { 'solution is imaginary' }
The quadratic equation will explain the above.
two ships leave a port at the same time. the first ship sails on a bearing of 55° at 12 knots (natural miles per hour) and the second on a bearing of 145° at 22 knots. how far apart are they after 1.5 hours (round to the nearest nautical mile)
Given that one of the ships travels at 12 nautical miles per hour, then after 1.5 hours, it will travel 12*1.5 = 18 miles
The other ship will travel 22*1.5 = 33 miles
The angle of 145° is measured with respect to the positive x-axis. Then, respect the negative x-axis, its measure is 180° - 145° = 35°
We have to use trigonometric functions to find x1, y1, x2, and y2, as follows:
sin(55°) = y1/18
sin(55°) *18 = y1
14.7 = y1
cos(55°) = x1/18
cos(55°)*18 = x1
10.3 = x1
sin(35°) = y2/33
sin(35°)*33 = y2
18.9 = y2
cos(35°) = x2/33
cos(35°)*33 = x
-27 = x2 (the minus sign comes from the graph)
The distance between two points (x1, y1) and (x2, y2) is computed as follows:
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]Substituting with the values found,
[tex]\begin{gathered} d=\sqrt[]{(-27-10.3)^2+(18.9-14.7)^2} \\ d=\sqrt[]{(-37.3)^2+4.2^2} \\ d=\sqrt[]{1391.29+17.64} \\ d\approx38\text{ miles} \end{gathered}[/tex]A collection of dimes and quarters has a value of $1.35. List all possible combinations of dimes and quarters. Remember to write a let statement
3 combinations:
3 quarters and 6 dimes
5 quarters and 1 dime
1 quarter and 11 dimes
1) Remember that a dime corresponds to $0.10 and a quarter to $0.25. And the value we want to find is $1.35
2) As we can see the last digit on $1.35 is 5 then we can infer that we're going to need an odd number of quarters ($0.25). Also, notice that we need whole numbers for the quantities of each coin. In other words, multiples of 0.10 and 0.25 whose sum yields to $1.35. So let's do it step by step:
So, we can write out the following list of combinations:
q (quarter) 3 q = 3 x 0.25 = $ 0.75
d (dimes) 6 d = 6 x 0.10 = $ 0.60
0.60 + 0.75 = 1.35
2.2) Another possible combination:
q (quarter) 5 q = 5 x 0.25 = $ 1.25
d (dimes) 1 d = 1 x 0.10 = $ 0.10
0.10+1.25= 1.35
2.3)
q (quarter) 1 q = 1 x 0.25 = $ 0.25
d (dimes) 11 d = 11x 0.10 = $ 1.10
0.25+1.10 = 1.35
3) Hence, considering that we need to combine dimes and quarters and their sum must be lesser than $1.35 We have three combinations with whole numbers of dimes and quarters:
3 quarters and 6 dimes
5 quarters and 1 dime
1 quarter and 11 dimes
y=x-6y=x+2 how to solve this problem
No solutions
Explanation
[tex]\begin{gathered} y=x+6\text{ Eq(1)} \\ y=x+2\text{ Eq(2)} \end{gathered}[/tex]
Step 1
replace Eq(1) in Eq (2)
[tex]\begin{gathered} y=x+2\text{ Eq(2)} \\ x+6=x+2 \\ \end{gathered}[/tex]Step 2
subtract x in both sides
[tex]\begin{gathered} x+6=x+2 \\ x+6-x=x+2-x \\ 6=2 \end{gathered}[/tex]6=2 is false, it means there is not solution for both equations,
Gary is saving money to buy a ticket to a New York Jets game that costs $225. Healready has saved $18. What is the least amount of money Gary must save each week, sothat at the end of 9 weeks he has enough money to buy the ticket? (Only an algebraic- solution will be accepted.)
lillyvong13, this is the solution:
Cost of the ticket to a New York Jets game = $ 225
Savings up to now = $ 18
Difference = 225-18 = 207
Number of weeks = 9
Let x to represent the amount of money Gary must save each week for buying the ticket, as follows:
x = 207/9
x = 23
Gary must save $ 23 at the end of 9 weeks to have enough money to buy the ticket
The equation for a straight line (deterministic model) is y = Bo + B,X.If the line passes through the point ( - 10,3), then x = - 10, y = 3 must satisfy the equation; that is, 3 = Bo + B1(-10).Similarly, if the line passes through the point (11,4), then x = 11, y = 4 must satisfy the equation; that is, 4 = Bo+B1(11).Use these two equations to solve for Bo and By; then find the equation of the line that passes through the points (-10,3) and (11,4)...Find Bo and B,B1 =Bo(Simplify your answers. Type integers or simplified fractions.)
To find the equation of the line we just need to find the beta constants. In order to do this we have (they provided us with) the following system of equations:
[tex]\begin{cases}3=\beta_0+\beta_1(-10) \\ 4=\beta_0+\beta_1(11)\end{cases}[/tex]Let us subtract the second equation to the first one:
This implies that
[tex]\beta_1=\frac{-1}{-21}=\frac{1}{21}[/tex]Now, let us replace this value we just got into the second equation to find beta_0:
[tex]\begin{gathered} 4=\beta_0+\frac{1}{21}\cdot11, \\ 4=\beta_0+\frac{11}{21}, \\ \beta_0=4-\frac{11}{21}=\frac{4\cdot21}{21}-\frac{11}{21}=\frac{84-11}{21}=\frac{73}{21} \end{gathered}[/tex]At last,
[tex]\beta_1=\frac{1}{21},\beta_0=\frac{73}{21}[/tex]Then, the equation of the line is just
[tex]y=\frac{73}{21}+\frac{1}{21}x[/tex]Find the sum of the first 39 terms of the following series, to the nearest integer.2,7, 12,...
The sequence 2,7,12,... given is an arithmetic progression. This is because it has a common difference.
Given:
first term, a = 2
common difference, d = second term - first term = 7 - 2 = 5
d = 5
n = 38
The sum of an arithmetic progression is given by;
[tex]\begin{gathered} S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack \\ S_{38}=\frac{38}{2}\lbrack2(2)+(38-1)5\rbrack \\ S_{38}=19\lbrack4+37(5)\rbrack \\ S_{38}=19\lbrack4+185\rbrack \\ S_{38}=19(189) \\ S_{38}=3591 \end{gathered}[/tex]Therefore, the sum of the first 39 terms of the series is 3,591
7) The water park is a popular field trip destination. This year the senior class at High School A and thesenior class at High School B both planned trips there. The senior class at High School A rented andfilled 1 van and 14 buses with 309 students. High School B rented and filled 4 vans and 14 buseswith 354 students. Each van and each bus carried the same number of students. Find the number ofstudents in each van and in each bus.C) Van: 19 Bus: 29 D) Van: 15, Bus: 21
Given
The water park is a popular field trip destination. This year the senior class at High School A and the senior class at High School B both planned trips there. The senior class at High School A rented and filled 1 van and 14 buses with 309 students. High School B rented and filled 4 vans and 14 buses with 354 students. Each van and each bus carried the same number of students.
Answer
Let students in Van be x
And students in bus be y
A/Q
x + 14y = 309 (1)
4x + 14y = 354 (2)
Subtracting (1) and (2)
3x = 45
x = 15
Put in eq (1)
15 + 14 y = 309
14y = 309 - 15
14 y = 294
y = 21
St
A businesswoman buys a new computer for $4000. For each year that she uses it the value depreciates by $400. The equation y=-400x+4000 gives the value y of the computer after x years. What does the x-intercept mean in this situation ? Find the x-intercept. After how many years will the value of the computer be $2000 ?
We have the following:
The intersection with the x-axis is when the value of y is 0, that is, the computer already has a value of 0 and has no commercial value.
we found it like this
[tex]\begin{gathered} 0=-400x+4000 \\ 400x=4000 \\ x=\frac{4000}{400} \\ x=10 \end{gathered}[/tex]which means that in 10 years, the computer does not represent a commercial value
to calculate the number of years that have passed when the computer has a value of 2000, y = 2000, therefore we replace
[tex]\begin{gathered} 2000=-400x+4000 \\ 400x=4000-2000 \\ x=\frac{2000}{400} \\ x=5 \end{gathered}[/tex]This means that in a total of 5 years the value of the computer will be $ 2000
Find extreme values of function f(x) = x³ - (3/2)x² on interval [- 1,2]
we have the function
[tex]f(x)=x^3-\frac{3}{2}x^2[/tex]Find out the first derivative of the given function
[tex]f^{\prime}(x)=3x^2-3x[/tex]Equate to zero the first derivative
[tex]\begin{gathered} 3x^2-3x=0 \\ 3x(x-1)=0 \end{gathered}[/tex]The values of x are
x=0 and x=1
we have the intervals
(-infinite,0) (0,1) (1,infinite)
Interval (-infinite,0) -----> f'(x) is positive
interval (0,1) ---------> f'(x) is negative
interval (1,infinite) -----> f'(x) is positive
that means
x=0 is a local maximum
x=1 is a local minimum
Find out the y-coordinates of the extreme values
For x=0 -----> substitute in the function f(x) ---------> f(x)=0
For x=1 ------> substitute in the function f(x) ------> f(x)=-0.5
therefore
The extreme values are
local maximum at (0,0)local minimum at (1,-0.5)-2 decimal 3/% because if you divide the principal in half its 1% of 3
Last year, Susan had $20,000 to invest. She invested some of it in an account that paid 10% simple interest per year, and she invested the rest in an account that paid 7% simple interest per year. After one year, she received a total of $1790 in interest. How much did she invest in each account?
Last year, Susan had $20,000 to invest. She invested some of it in an account that paid 10% simple interest per year, and she invested the rest in an account that paid 7% simple interest per year. After one year, she received a total of $1790 in interest. How much did she invest in each account?
Let
x ------> amount invested in an account that paid 10% simple interest per year
20,000-x ------> amount invested in an account that paid 7% simple interest per year
so
The formula of simple interest is equal to
I=P(rt)
In this problem we have that
10%=0.10
7%=0.07
x*(0.1)+(20,000-x)*(0.07)=1,790
solve for x
0.10x+1,400-0.07x=1,790
0.03x=1,790-1,400
0.03x=390
x=$13,000
therefore
amount invested in an account that paid 10% simple interest per year was $13,000and amount invested in an account that paid 7% simple interest per year was $7,000Brenda is preparing food for a picnic. She must make at least 6 sandwiches and at least 3 dozen cookies. It takes her 4 minutes to make a sandwich
For 6 sandwiches and 3 dozen cookies, the total time taken by Brenda would be 33 minutes.
What is a mathematical function, equation and expression?function : In mathematics, a function from a set X to a set Y assigns to each element of X exactly one element of Y. The set X is called the domain of the function and the set Y is called the codomain of the function.expression : A mathematical expression is made up of terms (constants and variables) separated by mathematical operators.equation : A mathematical equation is used to equate two expressions.equation modelling : Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis and results of the given problemGiven is Brenda who must make at least 6 sandwiches and at least 3 dozen cookies. It takes her 4 minutes to make a sandwich, and 45 minutes to make a dozen cookies.
If she makes [x] sandwiches and [y] dozen cookies, then the total time she would take to make them can be written as -
t = 4x + 3y
For 6 sandwiches and 3 dozen cookies, we can write -
t(6, 3) = 4 x 6 + 3 x 3 = 24 + 9 = 33 minutes
t(6, 3) = 33 minutes
Therefore, for 6 sandwiches and 3 dozen cookies, the total time taken by Brenda would be 33 minutes.
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{Complete question is given below -
She must make at least 6 sandwiches and at least 3 dozen cookies. It takes her 4 minutes to make a sandwich, x, and 45 minutes to make a dozen cookies, y.}
27. A race consists of 7 women and 10 men. What is the probability that the top three finishers were(a) all men (b) all women (c) 2 men and 1 woman (d) 1 man and 2 women
Given 7 women and 10 men;
a) the top 3 are all men:
[tex]\begin{gathered} ways\text{ to choose 3 men out of 10 men is:} \\ 10C_3=\frac{10!}{(10-3)!3!} \\ \Rightarrow\frac{10!}{7!3!}=\frac{10\times9\times8\times7!}{7!\times3\times2\times1} \\ \Rightarrow\frac{10\times9\times8}{3\times2\times1}=120 \\ \text{ways to choose 3 men from 17 people(10men +7women) is:} \\ 17C_3=\frac{17!}{(17-3)!3!} \\ \Rightarrow\frac{17!}{14!\times3!}=\frac{17\times16\times15\times14!}{14!\times3\times2\times1} \\ \Rightarrow\frac{17\times16\times15}{3\times2\times1}=680 \end{gathered}[/tex]Therefore, the probability that the top 3 are all men is:
[tex]P_{all\text{ men}}=\frac{120}{680}=0.1765[/tex]b) the top 3 are all women:
[tex]\begin{gathered} \text{ways to choose 3 women from 7 women is:} \\ 7C_3=35 \\ \text{ways to choose 3 women from 17 people is:} \\ 17C_3=680 \end{gathered}[/tex]Therefore, the probability that the top 3 are all women is:
[tex]P_{\text{all women}}=\frac{35}{680}=0.0515[/tex]c) 2 men and 1 woman;
[tex]\begin{gathered} ways\text{ to choose 2 men out of 10 men is:} \\ 10C_2=45 \\ \text{ways to choose 1 woman from 7 women is:} \\ 7C_1=7 \\ \text{Thus, ways to choose 2 men and 1 woman }=45\times7=315 \end{gathered}[/tex]Therefore, the probability that the top 3 finishers are 2 men and 1 woman is:
[tex]P=\frac{315}{680}=0.4632[/tex]d) 1 man and 2 women;
[tex]\begin{gathered} \text{ways to choose 1 man from 10 men is;} \\ 10C_1=10 \\ \text{ways to choose 2 women from 7 women is:} \\ 7C_2=21 \\ \text{Thus, ways to choose 1 man and 2 women is 10}\times21=210 \end{gathered}[/tex]Therefore, the probability that the top 3 finishers are 1 man and 2 women is:
[tex]P=\frac{210}{680}=0.3088[/tex]i need help with this too
a The value of (2.3 × 10⁴) × (1.5 × 10^-2) is 3.45 × 10^2
b. The value of (3.6 × 10^-5) ÷ (1.8 × 10^2) is 2 × 10^-3. This illustrates the concept of standard form.
What is standard form?The standard form is simply used in Mathematics to illustrate the numbers that are either too large or too small.
It's important to note that the multiplication of exponents is an addition and the division of the power is subtraction.
Therefore, (2.3 × 10⁴) × (1.5 × 10^-2) will be:
= (2.3 × 1.5) × (10^(4-2)
= 3.45 × 10^2
Also, (3.6 × 10^-5) ÷ (1.8 × 10^2) will be:
= (3.6 ÷ 1.8) × 10^(-5 + 2)
= 2 × 10^-3.
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Find the equation of the line with slope = 5 and passing through (-7,-29). Write your equation in the form
y = mz+b.
Answer:
[tex]y =5x+6[/tex]
Step-by-step explanation:
In the equation, [tex]y=mx+b[/tex], the "m" represents the slope, and the b represents the y-intercept.
We know the slope is 5, so we already know part of the equation: [tex]y=5x+b[/tex]
To solve for the "b" part or y-intercept, we can simply plug in a known point on the line, which was given to be (-7, -29)
This gives us the following equation:
[tex]-29 = 5(-7) + b\\\text{simplify}\\-29 = -35 + b\\\text{add 35 to both sides}\\6=b[/tex]
Answer: y = 5x + 6
Step-by-step explanation:
1. Do point-slope formula {(y - y1) = m(x - x1})
(y - 29) = 5(x - 7)
Distribute
(y - 29) = 5x - 35
Subtract 29 on both sides
Cancel out y - 29 - 29
y = 5x + 6
Metric area unit conversion with decimal valuesTammy has a rectangular poster. The poster is 1.2 meters long and 0.9 meters wide. What is the area of the poster in square centimeters? Do not round youranswer.cmcm²cm³
We are asked to find the final area in square centimeters.
We must first convert the measurements of the poster into centimeters:
1.2m = 120cm
0.9m = 90cm
Area = length * width
= 120*90
= 10800 square centimeters.
i have test tomorrow i will love you to help me with thisClassify the following as a monomial, binomial, trinomial, or polynomial andstate the degree.a) 7^5b) 5^7 + 7^2c) 7^3 + 11^2 − 13 + 112d) + 2 + 3^2e) 11^3 − 7^3
Answer:
a) degree 5; monomial
b) degree 7; polynomial
c) degree 3; polynomial
d) degree 2; polynomial
e) degree 3 polynomial
Explanation.
A polynomial is a mathematical expression containing coefficients and variables.
An example of a polynomial is 42x^5 + 23 x^3 - 2x.
A monomial is a polynomial that contains only one term. For example, x^4, 34x^5, 2x, etc.
Having that in mind, we go through the expressions and place them in one of either category.
a) 7^5
This contains only one term, meaning it is monomial; its highest exponent is 5, meaning it is of degree 5.
b) 5^7 + 7^2
This contains two terms, meaning it is a polynomial; its highest exponent is 7, meaning its degree is 7.
c) 7^3 + 11^2 − 13 + 112
This expression contains four terms - it is a polynomial; its highest exponent is 3, meaning its degree is 3.
d) + 2 + 3^2
This contains three terms, meaning it is a polynomial; its highest exponent is 2, meaning its degree is 2.
e) 11^3 − 7^3
This contains 2 terms, meaning it is a polynomial; its highest exponent is 3, meaning it is of degree 2.
can you explain what the 8th question is asking then answer it please
Answer:
Options A and C
Explanation:
We want to find out which arithmetic sequence(s) contain the term 34.
For an arithmetic sequence to contain the term, 34, the corresponding n-value must be an integer.
Option A
Set tn = 34
[tex]\begin{gathered} t_n=6+(n-1)4 \\ 34=6+(n-1)4 \end{gathered}[/tex]Solve for n:
[tex]\begin{gathered} 34-6=4n-4 \\ 28=4(n-1) \\ n-1=\frac{28}{4}=7 \\ n-1=7 \\ n=7+1 \\ n=8 \end{gathered}[/tex]The 8th term of this sequence is 34.
Option B
[tex]\begin{gathered} t_n=3n-1 \\ 34=3n-1 \\ 34+1=3n \\ 35=3n \\ n=\frac{35}{3}=11\frac{2}{3} \end{gathered}[/tex]A sequence cannot have a decimal nth term, therefore, the sequence does not contain 34.
Option C
T1 = 12, d=5.5
[tex]\begin{gathered} 12+5.5(n-1)=34 \\ 5.5(n-1)=34-12 \\ 5.5(n-1)=22 \\ n-1=\frac{22}{5.5} \\ n=4+1 \\ n=5 \end{gathered}[/tex]The 5th term of this sequence is 34, therefore, it contains the term 34.
Option D
3,7,11,...
[tex]\begin{gathered} t_1=3 \\ d=7-3=4 \end{gathered}[/tex]Using the nth term of an arithmetic sequence formula:
[tex]\begin{gathered} t_n=t_1+(n-1)d \\ 34=3+4(n-1) \\ 34-3=4(n-1) \\ 31=4(n-1) \\ n-1=\frac{31}{4} \\ n-1=7\frac{3}{4} \\ n=8\frac{3}{4} \end{gathered}[/tex]A sequence cannot have a decimal nth term, therefore, the sequence does not contain 34.
The sequences in Options A and C contain the term 34.
What is the independent and dependent variable for k(d) =2d^2 - d +32
Given:
[tex]k(d)=2d^2-d+32[/tex]Required:
To find the independent and dependent variable.
Explanation:
In the given equation,
The dependent variable = k(d)
The independent variable = d
Final Answer:
The dependent variable = k(d)
The independent variable = d
Determine if the correlation between the two given variables is likely to be positive or negative, or if they are not likely to display a linear relationship.A child’s age and the number of hours spent napping-positive-negative-no correlation
We know that a a childresn spend more hours napping when they are youngers therefore we have a negative correlation
Leila runs at 12 mph find the number of miles she can travel if she runs for 2 hours
20 POINTS ANSWER I WILL MARK BRAINLIEST!!!!!!!!!!
Answer: 24 mph
Step-by-step explanation:
If she can run 12 mph/ 12 miles per hour
She runs for 2 hours
12 · 2 = 24
to find the height of a tree, a group of students devised the following method. A girl walks toward the tree along it's shadow until the shadow of the top of her head coincide with the shadow of the top of the tree. if the girl is 150 cm tall, her distance to the foot of the tree is 13 meters, and the length of her shadow is 3 m, how tall is the tree?
Answer: 8m
Explanation:
Given:
To find the height(h) of the tree, we can use ratio since they are similar triangles.
Triangle 1
Triangle 2
So,
[tex]\begin{gathered} \frac{1.5}{3}\text{ = }\frac{h}{16} \\ \text{Simplify and rearrange} \\ h=\text{ }\frac{1.5}{3}(16) \\ \text{Calculate} \\ h=\text{ 8 m} \end{gathered}[/tex]Therefore, the height of the tree is 8m.
#Classify each number as a natural number, a whole number, or an integer.
If a number belongs to more than one set, list all possible sets it belongs to.
a) 1
b) -20
c) -3
(a) Whole number and integer
(b) Negative integer
(c) Negative integer
What are whole numbers an integers?Positive, negative, and zero numbers all fall under the category of integers. The word "integer" is a Latin word that signifies "whole" or "intact." Therefore, fractions and decimals are not considered to be integers. A number line is a graphic representation of positive and negative integers. The use of integers on a number line facilitates mathematical procedures. The right-side number is always greater than the left-side number. Due to the fact that they are greater than 0, positive numbers are positioned to the right of zero. All natural numbers and 0 are included in a group of numbers known as whole numbers. They belong to the category of real numbers, which excludes fractions, decimals, and negative numbers. Numbers used for counting are also regarded as whole numbers.
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I need help please quickly I need help
10(2 + 3) - 8 · 3.
20+50-8 times 3
70 -24
46 is answer
Answer:
10(5)-24 ----) 50-24 -----) 26
Step-by-step explanation:
answe is 26
determine the domain and range of the piecewise function graphed below
The domain is all the possible input values, and the range is all the possible output values.
So according to this function (Given in the question).
The domain is [-3, 5] and the range is [-5, 4]
That is all to this question.
Assume that Jim Bruce and Valerie are 3 of the 17 members of the class, and that of the class members will be chosen randomly to deliver their reports during the next class meeting. What is the probability that Jim Bruce and Valerie are selected in that order?
The probability that Jim, Bruce and Valerie are selected in that order is P = 1/680
Given,
Number of students in a class = 17
Jim, Bruce and Valerie are 3 of the 17.
Three of students from the class are randomly chosen to deliver their reports during the next class meet.
We have to find the probability that Jim, Bruce and Valerie are selected in that order;
Here,
The total number of possible selection;
S = ⁿCr
Where, n = 17 and r = 3
Then,
S = ¹⁷C₃
S = 17! / 14! x 3!
S = 680
Therefore,
The probability that Jim, Bruce and Valerie are selected in that order is P = 1/680
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Which of the following is a solution to the equation c + ( 4 -3c) - 2 = 0?A. -1B. 0C. 1D. 2
Given data:
The given equation is c+(4-3c)-2=0
The given equation can be written as,
c+4-3c-2=0
-2c+2=0
-2c=-2
c=1
Thus, the value of c is 1, so option C) is correct.
L1 : y=−4x+3L2 : y=4x−1
Answer:
Assuming you're trying to find where the lines intersect (their solution), the point where they intersect is (1/2 , 1).
Step-by-step explanation:
When you are trying to find the point where two lines intersect, you have to find the x and y values of that point. To do that, just set the lines equal to each other.
First, since y must equal y:
y = y
-4x + 3 = 4x - 1
Now solve for x:
-8x + 3 = -4
-8x = -4
8x = 4
x = 1/2
We just found the x value of the shared point. Now we need to find the y value of that point. Again, since the two lines share this point, plugging in the x value will result in the same y value for both lines.
So just plug in x to any of the equations: (I think y=4x-1 is easier)
y(1/2) = 4(1/2) - 1 or y = 4(1/2) - 1 (it doesn't matter how you write it)
y(1/2) = 2 - 1 or y = 2 - 1
y(1/2) = 1
So the point is:
(1/2 , 1)
To check you can plug in the y value you find, 1 in this case, and solve for x. If you get the same x value as before, everything is correct.
So:
(1) = -4x + 3
-2 = -4x
1/2 = x or x = 1/2
Great! Everything is correct.
This may seem like a very long process, but it is very easy. Just find the x and y values that the lines share by setting the lines equal to each other.
You bought your first car for $3,000 and make payments of $150 each month. Letting B represent the balance (what still needs to be paid), which function rule represents how much money is still owed (the balance) after m months?
The function rule represents how much money is still owed is B = 3000 - 150m
What is a function?A function is used to show the relationship between the data given. This can be illustrated with the variables.
Here, the person bought the first car for $3,000 and make payments of $150 each month.
Let B represent the balance.
The function to illustrate this will be:
B = 3000 - 150(m)
B = 3000 - 150m
where B = balance
m = number of months
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