Given the word problem, we can deduce the following information:
Fixed Cost = $60,000
Production Cost = $10 for each unit produced
To solve the following based on the given information, we let x be the number of units. So,
a)
The get the cost function, we note first that the cost would be:
Cost = Fixed Cost + Production Cost per unit produced
Hence, the cost function is:
C(x)=60,000+10x
b)
The revenue is the gross sales, so the revenue function would be:
R(x) = 15x
c)
We also note that the formula to get the profit is:
Profit= Revenue - Cost
So the profit function would be:
P(x)= 15x-(60,000+10x) = 15x-60,000-10x
Simplify
P(x)= 5x-60,000
d)
To compute the profit or loss, we plug in x=10,000 and x = 14,000 into the profit function:
When x =10,000:
[tex]\begin{gathered} P(x)=5x-60000 \\ P(10000)=5(10000)-60000 \\ \text{Simplify} \\ P(10000)=-10,000 \end{gathered}[/tex]It means that when x=10,000, there would be a $10,000 loss.
When x=14,000:
[tex]\begin{gathered} P(x)=5x-60000 \\ P(14000)=5(14000)-60000 \\ \text{Calculate} \\ P(14000)=10000 \end{gathered}[/tex]It means that when the production is 14,000 units/month, there would be a $10,000 profit.
O GRAPHS AND FUNCTIONSVertical line testFor each graph below, state whether it represents a function.Graph 1Graph 2Graph 34Function?Yes ΟΝΟGraph 4Yes O NoGraph 5Yes ΟΝΟGraph 6YesFunction?NOYesNOYesNOх?
For a relation to be a function, use the Vertical Line Test:
Draw a vertical line anywhere on the graph, and if it never hits the graph more than once, it is a function. If your vertical line hits twice or more, it's not a function.
So, look at the graphs:
Graph 1: Not a function
Graph 2: Not a function
Graph 3: Not a function
Graph 4: IS a function
Graph 5: IS a function
Graph 6: Not a function
One linear equation is defined by the points (2,4) and (1,1), while the other is defined by the points (2,-2) and (-1,-5). Which point represents the solutions to this system of equations?
Explanation:
one linear equation points: (2,4) and (1,1)
the 2nd equation points: (2,-2) and (-1,-5)
we find the equation of line of both
equation of line: y = mx + c
m = slope, c = intercept
[tex]m\text{ = }\frac{y_2-y_1}{x_2-x_1}[/tex]for the 1st one: (2,4) and (1,1)
m = (1-4)/(1-2) = -3/-1
m = 3
To get c, we use the slope and any point. Using point (2,4):
4 = 3(2) + c
4 = 6 + c
c = 4-6
c = -2
The equation of the line: y = 3x - 2
For the second one:
Bree is climbing a mountain. She knows that the faster she climbs, the less time the climb will take.
We know that the faster she climbs, the less time the climb will take.
As you can observe, the given relationship is between her speed and the height of the climb. We can deduct that this situation can be modeled by the speed in function of the time.
However, let's analyze each answer choice.
Choice A can't be right because the height of the mountain is not variable, that is, the mountain won't change its height.
Choice B is relating time and speed which, as we said before, the situation involves a relationship between time and speed.
Therefore, the right answer is B.Mark solved the equation in the box, using the steps shown.8-√x = 108-√x = 10-√x = 2X = (-2)²X= 4Is the solution x = 4 correct? State yes or no, and justify your answer.O YESO NOO I DO NOT KNOW√√x = -2
Given:
[tex]8-\sqrt{x}=10[/tex]Required:
To check whether the solution x=4 is correct or wrong.
Explanation:
Now put x=4 in to th original equation, we get
[tex]\begin{gathered} 8-\sqrt{4}=10 \\ 8-2=10 \\ 6\ne10 \end{gathered}[/tex]Therefore x=4 is not a solution.
Final Answer:
NO.
x=4 is not a solution for the given equation.
Someone can give me the answer please ?Question : “determine the equation of the line that represents moonbeam drive . Write the equation in slope intercept form”
Let:
[tex]\begin{gathered} (x1,y1)=(0,16) \\ (x2,y2)=(20,0) \end{gathered}[/tex]The slope m of the line is given by:
[tex]m=\frac{y2-y1}{x2-x1}=\frac{0-16}{20-0}=-\frac{4}{5}[/tex]Using the point-slope equation:
[tex]\begin{gathered} y-y1=m(x-x1) \\ y-16=-\frac{4}{5}(x-0) \end{gathered}[/tex]Since we need to write the equation in the slope-intercept form, let's solve for y:
[tex]\begin{gathered} y-16=-\frac{4}{5}x \\ y=-\frac{4}{5}x+16 \end{gathered}[/tex]you start at (3,1) you move left 2 units where do u end
Solution
For this case we have the original point given (3,1)
and we move 2 units to the left so then we can do this:
(3-2 =1, 1)
And the final point would be:
(1,1)
A data set whose original x values ranged from 41 through 78 was used to generate a regression equation of ŷ=5.3x – 21.9. Use the regression equation to predict the value of y when x=70.392.9Meaningless result402.1349.1
We have the equation, and we will replace the x-value to find the regression value, as follows,
[tex]\begin{gathered} y=5.3x-21.9 \\ y=5.3(70)-21.9 \\ y=349.1 \end{gathered}[/tex]The answer is d.349.1
what is an approximate solution of the system of equations? y=-0.5x+4 y=1+2x1.25,3.42.25,3.43.4,2.253.4,1.25
Combining the two equations gives us
[tex]-0.5x+4=2x+1[/tex]We add 0.5x to both sides and get
[tex]4=2.5x+1[/tex]and subtracting 1 from both sides gives
[tex]2.5x=3[/tex]And finally, dividing both sides by 2.5 gives
[tex]x=1.2[/tex]Putting this value of x into y = 2x + 1 gives
[tex]y=2(1.2)+1[/tex][tex]y=3.4[/tex]Hence our solution is
[tex]\textcolor{#FF7968}{(1.2,3.4)}[/tex]How do I factor this equation? [tex] {x}^{2} + 10x + 25[/tex]
Factorize:
[tex]x^2+10x+25[/tex][tex]\begin{gathered} \text{Step 1: Multiply the coefficient of x}^2\text{ and the constant term} \\ \text{The coefficient of x}^2\text{ in this case is 1 and the constant term is 25} \\ \text{ Therefore, 1 }\times25\text{ = 25} \end{gathered}[/tex][tex]\begin{gathered} \text{ Step 2: Find all the possible two-factor pairs of 25, obtained in step1 } \\ \text{The possible factors pairs are } \\ 1\text{ and 25} \\ 5\text{ and 5} \end{gathered}[/tex][tex]\begin{gathered} \text{Step 3: Find the two-factor pairs of 25 that will sum up to give} \\ \text{ the coefficient of x which is 10} \\ \text{The desirable two-factor pairs is 5 and 5} \\ \text{ Because, 5+5 =10} \end{gathered}[/tex][tex]\begin{gathered} \text{ Step 4: Substitute 10x for addition of the desirable pairs into the expression } \\ x^2+5x+5x+25 \\ \text{Factorize by grouping them} \\ (x^2+5x)+(5x+25) \\ x(x+5)+5(x+5) \\ =(x+5)(x+5) \end{gathered}[/tex]Therefore, the factors of the
[tex]x^2+10x+25\text{ = (}x+5)(x+5)[/tex]Evaluate f(x)={Vx - 4 if x2 4{ 2×-6if x<4For X = -5and x = 9
Answer:
For x = -5: -16
For x = 9: √5
Explanation:
The function is given by
f(x) = √(x - 4) when x ≥ 4 and
f(x) = 2x - 6 when x < 4
Since -5 is lower than 4, we need to use f(x) = 2x - 6 to evaluate f(x) for x = -5. So, replacing x by -5, we get:
f(x) = 2x - 6
f(-5) = 2(-5) - 6
f(-5) = -10 - 6
f(-5) = -16
On the other hand, 9 is greater than 4, so we will use the function f(x) = √(x - 4). Then, replacing x = 9, we get
f(x) = √(x - 4)
f(9) = √(9 - 4)
f(9) = √5
Therefore, the answers are
For x = -5: -16
For x = 9: √5
Aubrey left her house and drove to the store .she stopped and went inside .from there she drove in the same direction until she got to the bank .then she drove home .the graph below shows the number of blocks away from home or Aubrey is X minutes after she left her house , until she got back home .
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
Graph
x: time elapsed (min)
y: distance (blocks)
After 11 minutes:
distance = ?
Step 02:
We must analyze the graph to find the solution.
Graph
11 minutes ===> 4 blocks
The answer is:
The distance after 11 minutes is 4 blocks from her house
Which statement is TRUE for the given equations?A] Both equations have 2 solutions.Bx2 = 324 has 2 solutions and x =-512 has onesolutionx2 = 324 has 1 solution and x'= -512 has twosolutions.x2 = 324 has 2 solutions and x' =-512 has nosolutions
You didn't repond on the value of the other equation but the second equation which is
[tex]\begin{gathered} x^2=324 \\ \text{square root both side} \\ \sqrt{x^2}=\sqrt{324} \\ x=18\text{ or -18} \\ \text{Therefore, the equation x}^2=324\text{ has 2 solution.} \end{gathered}[/tex]what is -6(w-4)+8w=2(w+9)
Answer:
Explanation:
The equation
[tex]-6(w-4)+8w=2(w+9)[/tex]can be solved for w by first expanding both sides
Now,
[tex]-6(w-4)+8w=2w+24[/tex]and
[tex]2(w+9)=2w+18[/tex]therefore, our equation becomes
[tex]2w+24=2w+18[/tex]Subtracting 2w from both sides gives
[tex]24=18[/tex]We see a contradiction here. 24 is not equal to 18; therefore, the equation given has no solutions.
29.3212. IF VS = 12 m, find the length of UT.13. If JH = 21 in, fnd the length of KIG.RH5912785U1114. If FG = 27 yd, find the length of FED.15. If WS = 4.5 mm, find the length of TS.80P12811L7.62Gina Wilson (All Things Algebra), 2015
hello
12.
from a careful observation, one would notice that line VS is the diameter of the circle and ut is the lenght of an arc. however, we have to find the angle UT
[tex]\begin{gathered} vs=180^0 \\ vs=vr+sr \\ vr=127^0 \\ 180=127+sr \\ sr=180-127 \\ sr=53^0 \end{gathered}[/tex]note: vs is equal to 180 degrees because angle on a straight line is equal to 180 degree
[tex]\begin{gathered} ur=180^0 \\ ur=uv+vr \\ 180=uv+127 \\ uv=180-127 \\ uv=53^0 \end{gathered}[/tex]now we can solve for angle UT easily
[tex]\begin{gathered} vs=vu+ut+ts \\ 180=53+ut+90 \\ 180=143^{}+ut \\ ut=180-143 \\ ut=37^0 \end{gathered}[/tex]now we have the value of angle msince we know the value of angle UT, we can use this information to solve for length of the arc of UT
[tex]\begin{gathered} L_{ut}=\frac{\theta}{360}\times2\pi r \\ r=\frac{vs}{2}=\frac{12}{2}=6 \\ L_{ut}=\frac{37}{360}\times2\times3.142\times6 \\ L_{ut}=3.875m \end{gathered}[/tex]from the calculations above, the length of the arc UT is equal to 3.87m
I don't understand please helpFirst box options: Increase of decreased
The initial investment in 1985 was $31400, while the final amount in 1990 was $38900. Then the value of this investment increased at a rate of (38900 - 31400)/(1990 - 1985) = 1500 dollars per year.
Caitlin likes to finish your homework at early as you can the probability that she has a math homework tonight is 0.63 and the probability that she have science homework tonight 0.22 If the probability that Caitlin had both math and science homework tonight is 0.13 then what is the probability that she has neither math homework nor science homework tonight?a. 0.12b. 0.15c. 0.28d. 0.41
QUESTION: Caitlin likes to finish her homework as early as she can. The probability that she has a math homework tonight is 0.63 and the probability that she has science homework tonight is 0.22. If the probability that Caitlin had both math and science homework tonight is 0.13 then what is the probability that she has neither math homework nor science homework tonight?
GIVEN:
P(Math) = 0.63
P(Science) = 0.22
P(Math & Science) = 0.13
ASKED: P(Neither Math nor Science)
FORMULA: P(Neither Math nor Science) = 1 - P(Math or Science)
SOLUTION:
In order to find what is asked, we needed to find first the P(Math or Science).
P(Math or Science) = P(Math) + P(Science) - P(Math and Science)
P(Math or Science) = 0.63 +0.22 -0.13
P(Math or Science) = 0.72
Next, we should substitute the value we got to the formula given above.
P(Neither Math nor Science) = 1 - P(Math or Science)
P(Neither Math nor Science) = 1 - 0.72
P(Neither Math nor Science) = 0.28
ANSWER: The probability that Caitlin has neither Math homework nor Science homework is 0.28. In the choices, it is the letter C.
What is the 99% confidence interval for a sample of 52 seat belts that have a mean length of 85.6 inches long and a population standard deviation of 3.8 inches?
We need o find the 99% confidence interval for a sample with:
[tex]\begin{gathered} n=52 \\ \\ \overline{x}=85.6\text{ in} \\ \\ \sigma=3.8\text{ in} \end{gathered}[/tex]A 99% confidence interval has a z-value z = 2.576.
And the confidence interval is given by:
[tex]\overline{x}\pm z\times\frac{\sigma}{\sqrt{n}}[/tex]Thus, we obtain:
[tex]\begin{gathered} 85.6\text{ in}\pm2.576\times\frac{3.8\text{ in}}{\sqrt{52}} \\ \\ \cong85.6\text{ }\imaginaryI\text{n}\pm1.4\text{ in} \end{gathered}[/tex]Approximating to the nearest tenth, the 99% confidence interval is:
Answer
[tex]\lbrack84.2\text{ in},87.0\text{ in}\rbrack[/tex]Simplify n^6 * n^5 / n^4 * n^3 /n^2 * nThere is a picture too.
Given the following expression:
[tex]n^6\cdot n^5\frac{\cdot}{\cdot}n^4\cdot n^3\frac{\cdot}{\cdot}n^2\cdot n[/tex]to simplify, we have to operate from left to right, then, we get the following:
[tex]\begin{gathered} n^6\cdot n^5\frac{\cdot}{\cdot}n^4\cdot n^3\frac{\cdot}{\cdot}n^2\cdot n=n^{11}\frac{\cdot}{\cdot}n^4\cdot n^3\frac{\cdot}{\cdot}n^2\cdot n^{} \\ =n^7\cdot n^3\frac{\cdot}{\cdot}n^2\cdot n=n^{10}\frac{\cdot}{\cdot}n^2\cdot n=n^8\cdot n=n^9 \end{gathered}[/tex]therefore, the simplified expression is n^9
In the figure below, points J, K, and L are the midpoints of the sides of XYZ. Suppose JK = 38, XZ = 96, and YX = 56. Find the following lengths KL, YZ and LZ
We have the following:
We must calculate each of the sides with the attached image, since they are the midpoints, the value of the inside triangle (JKL) is half.
For KL
[tex]\begin{gathered} KL=\frac{YX}{2} \\ KL=\frac{56}{2} \\ KL=28 \end{gathered}[/tex]For YZ
[tex]\begin{gathered} \frac{YZ}{2}=JK \\ YZ=38\cdot2 \\ YZ=76 \end{gathered}[/tex]For LZ
[tex]\begin{gathered} LZ=\frac{Y\Z}{2} \\ LZ=\frac{76}{2} \\ LZ=38 \end{gathered}[/tex]The answer is KL is 28, YZ is 76 and LZ is 38
1. A particle is traveling along the x-axis and its position from the origin can be modeled by x(t)=t^3 - 2t^2+ 4twhere x is centimeters and t is seconds.a. On the interval 0< t <4, find when the particle is farthest to the right.b. On the same interval, what is the maximum speed?
Given the relation between x and t :
where: x is the position from the origin and t is the seconds
[tex]x(t)=t^3-2t^2+4t[/tex]a. On the interval 0< t <4, find when the particle is farthest to the right.
There are two methods to find the answer : by graph or by derivatives
So, we will use derivatives to find the answer
[tex]x^{\prime}(t)=3t^2-4t+4[/tex]so, at the farthest point the speed = 0
The solution of the equation will be imaginary solutions
Which mean distance will always increase
So, the farthest point will be at t = 4
So, the farthest =
[tex]x(4)=4^3-2\cdot4^2+4\cdot4=64-32+16=48\operatorname{cm}[/tex]b. On the same interval, what is the maximum speed?
The speed is given by x'(t)
So, at the same interval , the speed will be maximum at t = 4
so,
[tex]\begin{gathered} x^{\prime}(t)=3t^2-4t+4 \\ \\ x^{\prime}(4)=3\cdot4^2-4\cdot4+4=3\cdot16-16+4 \\ x^{\prime}(4)=36\text{ cm/sec} \end{gathered}[/tex]Also, see the following figure which represents the graph solutions:
The position equation is the blue curve
The speed equation is the violate curve
The line x = 4 is the time at t = 4 seconds
The intersect with the curves give the answers
The graph of a linear function is given below.f(x)86-6-8What the ero of the function?А-3
Given:
The objective is to find the zero of the function.
Explanation:
The zero of a function represents the point where the graph intersects at the x-axis.
By observing the given figure, the straight line intersects the x-axis at the point,
[tex](0,-2)[/tex]Thus, the zero of the function is -2.
Hence, option (D) is the correct answer.
progress bar may be uneven because questions can be worth more or less (including zero) dependingRewrite (123 + 456) + 789 using the Associative Law of Addition.(456 + 123) + 789789 + (123 + 456)123 + (456 + 789)(123 + 789) + 456
ANSWER:
[tex]123+(456+789)[/tex]STEP-BY-STEP EXPLANATION:
We have the following expression:
[tex]\mleft(123+456\mright)+789[/tex]Associative laws state that when you add or multiply any three real numbers, the group (or association) of the numbers does not affect the result. Therefore:
[tex]123+(456+789)[/tex]An expression of the fifth degree is written with a leading coefficient of seven and a constant of six. Which expression is correctly written for these conditionsGroup of answer choices65+4+7 76−64+5 67−5+5 75+2+6
Given an expression of the fifth degree is written with a leading coefficient of seven and a constant of six
To Determine: Correct expression for the given polynomial
Solution:
Note that:
(i) The fifth degree means that the highest power of the polynomials is 5
i.e.
[tex]x^5[/tex](ii) The leading coefficient is the coefficient of the highest power of the polynomial, which is the cofficient of the fifth degree. i.e.
[tex]7x^5[/tex](iii) The constant is the coefficient of the variable with the power of 0. so we have
[tex]\begin{gathered} 6x^0 \\ x^0=1 \\ 6x^0=6\times1=6 \end{gathered}[/tex]So, it should be known that the given polynomials must have the 7x⁵ as the first term and must have 6 as the last term
From the options provided, it can be concluded that the correct expression for the given condition is 7x⁵+2x+6
Solve for y4(2Y - 2) = 7(y + 5)
Answer:
y = 43
Explanation:
To solve for y, we will apply the distributive property as:
[tex]\begin{gathered} 4(2y-2)=7(y+5) \\ (4\cdot2y)-(4\cdot2)=(7\cdot y)+(7\cdot5) \\ 8y-8=7y+35 \end{gathered}[/tex]So, subtracting 7y from both sides:
[tex]\begin{gathered} 8y-8-7y=7y+35-7y \\ y-8=35 \end{gathered}[/tex]Adding 8 to both sides:
[tex]\begin{gathered} y-8+8=35+8 \\ y=43 \end{gathered}[/tex]Therefore, the answer is y = 43
Mrs. Jameson had coupons for the following amounts off of her food bill: $0.49, $0.38. $0.33, $0.71, $1.04, $0.55, $0.25, $0.83, $0.78, and $0.61. Her food bill was $39.95before the grocery clerk applied the coupons. How much was her food bill with the coupons?
hello
the first step in solving this question is to add the sum of the coupons she presented to the grocery clerk
[tex]\text{0}.49+0.38+0.33+0.71+1.04+0.55+0.25+0.83+0.78+0.61=5.97[/tex]the sum discount on her coupons was $5.97
she bought food worth $39.95
let's subtract the discount from the coupon from the worth of the food
[tex]39.95-5.97=33.98[/tex]her food bill with coupons would cost her $33.98
the answer to this question is option A
Which expression is equivalent to (2x^5 + 7x³) - (5x² - 4x³)?a)-3x³ +11b) -3x³ +3c) 2x5 + 11x³5x²d) 2x5-3x³-5x2²
Given:
[tex](2x^5+7x^3)-(5x^2-4x^3)[/tex]Required:
We need to find the equivalent expression of the given.
Explanation:
Distribute the minus sign.
[tex](2x^5+7x^3)-(5x^2-4x^3)=2x^5+7x^3-5x^2+4x^3[/tex]Add like terms.
[tex](2x^5+7x^3)-(5x^2-4x^3)=2x^5+7x^3+4x^3-5x^2[/tex][tex](2x^5+7x^3)-(5x^2-4x^3)=2x^5+11x^3-5x^2[/tex]Final answer:
[tex]2x^5+11x^3-5x^2[/tex]Solve the following system of a circle and a parabola: x^2 + y^2 = 9 and 5x^2 - 6y= 18.(1,897,0)(2.4, 1.8)(-2.4, 1.8)(-1.897,0)(0,-3)
Solve the following system of a circle and a parabola: x^2 + y^2 = 9 and 5x^2 - 6y= 18.
(1,897,0)
(2.4, 1.8)
(-2.4, 1.8)
(-1.897,0)
(0,-3)
we have
x^2 + y^2 = 9 ------> equation A
5x^2 - 6y= 18 -----> equation B
Solve the system by graphing
Remember that the solution of tghe system are the intersection point both graphs
so
using a graphing tool
see the attached figure
please wait a minute
the solutions are
(-2.4, 1.8), (2.4, 1.8) and (0,-3)whats the answer.[tex]1[/tex]
Input data
4224 feet
Procedure
Definition: A mile (symbol: mi or m) is a unit of length in the imperial and US customary systems of measurement. It is currently defined as 5,280 feet, 1,760 yards, or exactly 1,609.344 meters.
ratio = 1 mile/5280 feet
Result: 4224 foot = 0.8 mile
How do you figure out slope and what is the formula?
Given the following two points
[tex](x_1,y_1),(x_2,y_2)[/tex]The formula to find a slope of a line is given by
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Where m represents the slope.
f(x) = 12x + 2 - 11x ^ 2 Then the equation of the tangent line to the graph of f(x) at the (0, - 9) is given by y = pi*nx + b for
Given:
[tex]f(x)=12x+2-11e^x[/tex]We will find the equation of the line tangent to f(x) at the point (0, -9)
the slope of the tangent line = the first derivative f'(x)
the first derivative will be as follows:
[tex]f^{\prime}(x)=12-11e^x[/tex]substitute x = 0 to find the slope of the line tangent at (0,-9)
[tex]m=f^{\prime}(0)=12-11e^0=12-11=1[/tex]So, the equation of the line will be: y = x - 9
so, the answer will be:
m = 1
b = -9