We need to find the derivative of the function
[tex]f\mleft(x\mright)=6x^{4}-7x^{3}+2x+\sqrt{2}[/tex]The derivative of a polynomial equals the sum of the derivatives of each of its terms.
And the derivative of each term axⁿ, where a is the constant multiplying the nth power of x, is given by:
[tex](ax^n)^{\prime}=n\cdot a\cdot x^{n-1}[/tex]Step 1
Find the derivatives of each term:
[tex]\begin{gathered} (6x^4)^{\prime}=4\cdot6\cdot x^{4-1}=24x^{3} \\ \\ (-7x^3)^{\prime}=3\cdot(-7)\cdot x^{3-1}=-21x^{2} \\ \\ (2x)^{\prime}=1\cdot2\cdot x^{1-1}=2x^0=2\cdot1=2 \\ \\ (\sqrt[]{2})^{\prime}=0,\text{ (since this term doesn't depend on x, its derivative is 0)} \end{gathered}[/tex]Step 2
Add the previous results to find the derivative of f(x):
[tex]f^{\prime}(x)=24x^{3}-21x^{2}+2[/tex]Answer
Therefore, the derivative of the given function is
[tex]24x^3-21x^2+2[/tex]A sofa and a love seat together costs $600. The sofa costs $75 less than double the love seat. How much do they each cost The equation
To solve this problem we need to create an equation, where the unkown variable, x, represents the cost for the love seat. We know that the sofa costs $75 less than the love seat, therefore we have:
[tex]y=x-75[/tex]The cost for both pieces of furniture together is equal to $600. So if we add them we have:
[tex]x+y=600[/tex]We can swap the expression for y on the second equation.
[tex]\begin{gathered} x+(x-75)=600 \\ x+x-75=600 \\ 2x-75=600 \\ 2x=675 \\ x=\frac{675}{2}=337.5 \end{gathered}[/tex]Now we know that the love seat costs $337.5. We will use the first equation to find the cost of the sofa.
[tex]y=337.5-75=262.5[/tex]The sofa costs $262.5.
If AABC is similar to ARST, find the value of x.
Given that
[tex]\begin{gathered} \Delta ABC\text{ is similar to }\Delta RST \\ \text{Therefore, the ratio of the corresponding sides is equal.} \\ \text{That is,} \\ \frac{AB}{RS}=\frac{BC}{ST}=\frac{AC}{RT} \end{gathered}[/tex]Given that AB = 12, BC =18, AC =24 and RS =16, RT=x
We now use the ratio of the corresponding sides to find side RT( the value of x).
Hence,
[tex]\begin{gathered} \frac{AB}{RS}=\frac{AC}{RT} \\ \frac{12}{16}=\frac{24}{x} \\ x=\frac{24\times16}{12} \\ x=32 \end{gathered}[/tex]Therefore, the value of x (RT) is 32
Simplify the expression.
9. (x^-3)^-5x^6
Answer: x to the power of 21
Step-by-step explanation:
Hi , can you help me to solve this problem please.
The polynomials are classified as shown in the image below
identify the percentage of change and increase or decrease 75 people to 25 people increase or decrease find the percent of change round to the nearest tenth of a percent
dentify the percentage of change and increase or decrease 75 people to 25 people increase or decrease find the percent of change round to the nearest tenth of a percent
we have that
75 people represent the 100 % so
Applying proportion, find out how much percentage represent the difference (75-25=50)
so
100/75=x/50
solve for x
x=(100/75)*50
x=66.7 %
therefore
Its a decrease and the percentage of change is 66.7%Recall that we can compare the vertical distance between any two points on the same vertical line to measure verticalchange. In the same way, the horizontal distance between any two points on the same horizontal line will measurehorizontal change.Suppose the linear function y = ax + b undergoes a horizontal change of 5 units. This is equivalent to what verticalchange?A) a vertical change of 5 + b unitsB)a vertical change of 5a + b unitsC)a vertical change of 5 unitsD)a vertical change of 5/a unitsE)a vertical change of 5a units
Given the linear function:
y = ax + b
And it undergoes a horizontal shift of 5 units
Let the original line be f(x) and the new line be g(x)
g(x) = f(x - 5)
The vertical change will be the horizontal change times a, using the definition of slope.
Thus, since the horizontal change here is 5 units, the vertical change is 5a units
ANSWER:
E) a vertical change of 5a units
I need help with this work question 10Find the area of each regularpolygon. Leave your answer insimplest form.
Given:
Number of sides in octagon = 8
Length of apothem = 14.1
Side length = 11.7
Required: Area
Explanation:
The area of a regular polygon is one-half the product of its apothem and its perimeter.
Here, the area of the regular octagon is
[tex]\begin{gathered} A=\frac{1}{2}ap \\ =\frac{1}{2}\times14.1\times8\times11.7 \\ =659.88 \end{gathered}[/tex]Final Answer: Area of the regular octagon is 659.88 square units.
six fifths, eight ninths, 0.5, forty percent?
Answer:
I'm assuming this is a greatest to least, but in case it was not, I put least to greatest, too.
Step-by-step explanation:
Greatest to least:
6/5, 8/9, 0.5, 40%
Least to greatest:
40%, 0.5, 8/9, 6/5
Hope this helps!
We consider the sets D = {m, n, p, q} E = {3,6,8} and the relation from D to E.R = {(m, 3), (m, 8), (n, 6), (n, 8) (p, 3), (q, 3), (q, 6)a) List the pairs of D × Eb)R is it a proper subset of D × E? Why ?c)Represent the relation R using a Cartesian network
D= {m, n, p, q}
E= {3,6,8}
a) D x E = { (m, 3), (m, 6), (m, 8), (n, 3), (n,6), (n,8), (p, 3), (p, 6), (p, 8), (q, 3), (q, 6),
(q, 8) }
b) We need to know what a proper subset is.
Proper subset
A proper subset of a set A is a subset of A that is not equal to A. In other words, if B is a proper subset of A, then all elements of B are in A but A contains at least one element that is not in B.
From the above definition, we can say R is a proper subset of D x E because there are element in D x E that is NOT in R.
Hi I need help with this math problem, i’m in high school calculus 1
Step 1:
When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is between 0 and 1, we get a vertical compression.
Step 2
Parent function y = f(x)
In general, a vertical stretch is given by the equation
y=bf(x). If b>1, the graph stretches with respect to the y-axis, or vertically. If b<1, the graph shrinks with respect to the y-axis.
The function becomes y = 1.4f(x) when trainsform vertically
The function is shifted 3 units to the left and it becomes y = 1.4f(x + 3)
Final answer
y = 1.4f(x + 3)
the area of a trapezium is 1680 sq cm. One of the parallel sides is 64 cm and the perpendicular distance between the parallel sides is 28 cm. find the length of the other parallel side.
Answer:
The missing side length is 56
Step-by-step explanation:
1680 = 1/2 · (64 + x) · 28
1680 · 2 = 28 · (64 + x)
3360 = 1792 + 28x
28x = 3360 - 1792
28x = 1568
x = 1568 ÷ 28
x = 56
Hope this helps.
The permeter of then figure below is 110cm.Find the length of the missing side.
Perimeter of a plane shape is the sum of all lenth of side of outer boundary.
Perimeter = 110cm
perimeter = 8.6 + 34.6 + 8.6 + 17.3 + 11.6 + 11.6 + 11.6 + x
110 = 103.9 + x
x = 110 - 103.9
x = 6.1cm
Which expressions are equivalent to the one below? Check all that apply.log3 3+ log3 27A. log3 81B. log3 (3^4)C. 4D. log 10
The given expression is
[tex]log_33+log_327[/tex]We will use the rule
[tex]log_ba+log_bc=log_b(ac)[/tex][tex]\begin{gathered} log_33+log_327=log_3(3\times27) \\ \\ log_3(3\times27)=log_3(81) \end{gathered}[/tex]Since 81 = 3 x 3 x 3 x 3, then
[tex]\begin{gathered} 81=3^4 \\ log_3(81)=log_3(3^4) \end{gathered}[/tex]We will use the rule
[tex]log_b(a^n)=nlog_b(a)[/tex][tex]undefined[/tex]Hello. I am trying to help my 9th grade daughter with text corrections. It has been over 20 yrs since I had Algebra 1 and Im a bit rusty. She gets easily frustrated especially in math so Im trying to do some of the leg work before going over how to do it with her. I appreciate your help in advance.
The half-life of a radioactive substance is given 3 hours.
Given the initial amount of substance is 800 grams. After 3 hours, the substance becomes half that is 400 grams. Then again after 3 more hours, the substance becomes half again that is 200 grams. Again after three hours, the substance becomes half that is 100 grams.
Thus, the amount of radioactive material after 9 hours is 100 grams.
What is the value of x in the solution to the system of equations below?2x+3y=112x+y=1
Answer:
x=-2
Explanation:
Given the system of equations:
[tex]\begin{gathered} 2x+y=1 \\ 2x+3y=11 \end{gathered}[/tex]We use the method of elimination by subtracting.
This gives us:
[tex]\begin{gathered} -2y=-10 \\ y=\frac{-10}{-2} \\ y=5 \end{gathered}[/tex]We then substitute y=5 into any of the equations to solve for x.
[tex]\begin{gathered} 2x+y=1 \\ 2x+5=1 \\ 2x=1-5 \\ 2x=-4 \\ x=-\frac{4}{2} \\ x=-2 \end{gathered}[/tex]Therefore, the value of x in the solution to the system of equations is -2.
2x<=-3y+9. graph solution set for this inequality
We have to graph the solution set for the inequality:
[tex]2x\le-3y+9[/tex]The first step is to graph the function that divides the solution region from the other region. This line correspond to the equality within this inequality:
[tex]2x=-3y+9[/tex]If we rearrange it we can find two points to graph it:
[tex]\begin{gathered} 2x=-3y+9 \\ 2x+3y=9 \end{gathered}[/tex]When x=0, then y is:
[tex]\begin{gathered} 2\cdot0+3y=9 \\ y=\frac{9}{3} \\ y=3 \end{gathered}[/tex]Then, the y-intercept is at y=3.
When y=0, then x is:
[tex]\begin{gathered} 2x+3\cdot0=9 \\ x=\frac{9}{2} \end{gathered}[/tex]Now we now that the x-intercept is at x=9/2.
We have two points from the line, so we can graph it as:
Now, we know the line that limits the solution region.
As the inequality includes the equal sign, we know that this limit is included in the solution region.
The only thing left is to find is if the solution region is above this line or if it is below.
One easy way to test it is to select a point from one of the regions and replace (x,y) in the inequality: if the inequality stands true, then this point is in the solution region and we then now on which side the solution region is.
In this case, we can test with point (0,0) to make it easier:
[tex]\begin{gathered} (x,y)=(0,0)\Rightarrow2\cdot0\le-3\cdot0+9 \\ 0\le-0+9 \\ 0\le9\to\text{True} \end{gathered}[/tex]As the inequality is true for this point, we know that the solution region includes (0,0).
Then, we know that the solution region is below the line.
We then can graph it as:
Find the area of the shaded region assume all angles are right angles
The given figure is of a rectangle which is enclosed in the large rectangle.
Area of rectangle = Length x Width
Dimension of large rectangle, 10 and 20.
Area of larger rectangle = 10 x 20
Area of larger rectangle = 200
Dimension of the small rectangle, 14 and 6.
Area of small rectangle = 14 x 6
Area of small rectangle = 84
Area of shaded region = Area of large rectangle - Area of small rectangle
Area of shaded region = 200 - 84
Area of shaded region = 116
Area of shaded region is 116 unit²
Sally's wallet contains:5 quarters3 dimes• 8 nickels• 4 penniesA coin is drawn from the purse and replaced 240 times. How many times can you predict that a nickle or apenny will be drawn?
The total number of coins in the wallet, is:
[tex]5+3+8+4=20[/tex]Since there are 8 nickels and 4 pennies, there are 12 coins which are either nickels or pennies. Then, the probability of picking a nicle or a penny, is:
[tex]\frac{12}{20}=\frac{3}{5}[/tex]Multiply 3/5 by 240 to find the expected amount of times that a nicke or penny will be drawn:
[tex]\frac{3}{5}\times240=144[/tex]For the polynomial function ƒ(x) = .5x3 + .25x2 + .125x + .0625, find the zeros. Then determine the multiplicity at each zero and state whether the graph displays the behavior of a touch or a cross at each intercept.x = .5, touchx = −.5, touchx = .5, crossx = −.5, cross
Given:
The polynomial is
[tex]f(x)=.5x^3+.25x^2+.125x+0.0625[/tex]Required:
Find the zeros. Then determine the multiplicity at each zero and state whether the graph displays the behavior of a touch or a cross at each intercept.
Explanation:
The zeros of polynomial are
[tex]\begin{gathered} x\approx0.5 \\ x=\pm0.5i \end{gathered}[/tex]Now,
So, graph is crossing at -0.5
Answer:
Hence, fourth option is correct.
For the data set 1,7,7,7,8, the mean is 6. What is the mean absolutedeviation?O A. The mean absolute deviation is 10.O B. The mean absolute deviation is 6.O c. The mean absolute deviation is 2.O D. The mean absolute deviation is 5.
The mean absolute deviation is given by:
[tex]\frac{\sum ^{}_{}\lvert x_i-\bar{x}\rvert}{n}[/tex]where xi represent each data, x bar the mean and n the number of data we have. Then:
[tex]\begin{gathered} \frac{\lvert1-6\rvert+\lvert7-6\rvert+\lvert7-6\rvert+\lvert7-6\rvert+\lvert8-6\rvert}{5} \\ =\frac{\lvert-5\rvert+\lvert1\rvert+\lvert1\rvert+\lvert1\rvert+\lvert2\rvert}{5} \\ =\frac{5+1+1+1+2}{5} \\ =\frac{10}{5} \\ =2 \end{gathered}[/tex]Therefore the mean absolute value is 2 and the answer is C.
list the first 5 multiples of the denominator and each fraction in order of least to greatest
The fraction given is 2/6.
The first five multiples of the denominator are as follows;
[tex]\begin{gathered} \frac{2}{6}, \\ 6,12,18,24,30 \end{gathered}[/tex]The other fraction is 7/10.
The first five multiples of the denominator are as follows;
[tex]\begin{gathered} \frac{7}{10}, \\ 10,20,30,40,50 \end{gathered}[/tex]Basically, you simply multiply the denominator by any series of numbers, in this case from 1 to 5. Therefore you'll have
6 x 1 = 6, 6 x 2 = 12, and so on. The same principle applies to the other denominator, that is 10.
The Nut Shack sells hazelnuts for $6.80 per pound and peanuts nuts for $4.80 per pound. How much of each type should be used to make a 44 pound mixture that sells for $5.94 per pound?
18.92 pounds of peanut and 25.08 pounds of nut shack should be used to make the mixture
Explanation:the cost per pound for the nut shack = $6.80
let the amount of pounds of nut shack used in the mixture = n
the cost per pound for the peanuts = $4.80
let the amount of pounds for the peanuts used in the mixture = p
We want to obtain 44 pounds of mixture which sells for $5.94 per pound
sum of pounds mixture = 44
amount of pounds of nut shack used in the mixture + amount of pounds for the peanuts used in the mixture = 44
[tex]n+p=44\text{ }....\mleft(1\mright)[/tex]cost per pound for the nut shack (amount used) + cost per pound for the peanuts (amount used) = cost per pound of the mixture (amount of mixture)
6.80(n) + 4.80(p) = 5.94(44)
[tex]6.8n+4.8p=261.36\text{ }\ldots\mleft(2\mright)[/tex]using substitution method:
from equation 1, we can make n the subject of formula
n = 44 - p
substitute for n in equation (2):
[tex]\begin{gathered} 6.8(44\text{ - p) + 4.8p = 261.36} \\ 299.2\text{ - 6.8p + 4.8p = 261.3}6 \\ 299.2\text{ - 2p = 261.3}6 \end{gathered}[/tex][tex]\begin{gathered} collect\text{ like terms:} \\ 299.2\text{ - 261.36 - 2p = 0} \\ \text{add 2p to both sides:} \\ 37.84\text{ = 2p} \\ \text{divide both sides by 2:} \\ \frac{37.84}{2}\text{ = p} \\ p\text{ = 18.9}2 \end{gathered}[/tex]substitute for p in equation 1:
[tex]\begin{gathered} n\text{ + 18.92 = 44} \\ n\text{ = 44 - 18.9}2 \\ n\text{ = 25.0}8 \end{gathered}[/tex]18.92 pounds of peanut and 25.08 pounds of nut shack should be used to make the mixture
One of the roofers claims that the roof area of each pillar is the same as the area of a square with edges of 21.5 feet.The roofer is correct or incorrect?
SOLUTION
We have been given the height of each lateral triangular face of the roof h as 13.4 ft and the length of the square base of the pyramid as 21.5 feet
We want to know if the area of the square base is the same as the area of each triangular lateral face
Area of the square base is
[tex]21.5\times21.5=462.25\text{ ft}^2[/tex]Area of the four triangular lateral face becomes
[tex]\begin{gathered} 4(\frac{1}{2}\times b\times h) \\ =4\times\frac{1}{2}\times21.5\times13.4 \\ =2\times21.5\times13.4 \\ =576.2\text{ ft}^2 \end{gathered}[/tex]From our calculations, the area of the square base is 462.25 square-feet,
While the area of the four lateral face triangle of the roof is 576.2 square-feet
Hence the roofer is incorrect
Consider the function f(x) = 5 - 4x ^ 2, - 5 <= x <= 1 .
Given: A function-
[tex]f(x)=5-4x^2,\text{ }-5\leq x\leq1[/tex]Required: To determine the absolute maxima and minima of the function.
Explanation: The given function is-
[tex]f(x)=5-4x^2[/tex]Differentiating the function,
[tex]f^{\prime}(x)=-8x[/tex]Setting f'(x)=0 gives-
[tex]\begin{gathered} -8x=0 \\ \Rightarrow x=0 \end{gathered}[/tex]So we have to check the function at the boundary points of the interval [-5,1] and x=0 as follows-
Hence, the absolute maximum is 5 at x=o, and the minimum is -95 at x=-5.
Final Answer: The absolute maximum value is 5, and this occurs at x=0.
The absolute minimum value is -95, and this occurs at x=-5.
What percent of the data is greater than the median?A box-and-whisker plot. The number line goes from 0 to 20. The whiskers range from 2 to 19 and the box ranges from 6.5 to 18. A line divides the box at 17.a.20%c.50%b.25%d.80%
1) Consider that the 1st Quartile corresponds to 25%, the Median is equivalent to 50% of the data, the Third Quartile to 75% of the data as the sketch below:
Notice that the Median is that bar inside the box, also known as the 2nd Quartile.
2) So the percentage of the data greater than the median is:
[tex]75\%-50\%=25\%[/tex]Your team has carefully researched and selected two possible painting companies. Pro Painters charge $200 per hour plus $6000 in material fees. Illusion Ltd charges $150 per hour plus $8000 in material fees.Create a graph of the cost for both companies using the grid below. Circle the point of intersection. Be sure your lines are properly identified.
Given:
• Pro Painters:
Charge per hour = $200
Material fees = $6000
• Illusion Ltd:
Charge per hour = $150
Material fees = $8000
Let's create a graph of the cost for both companies.
Represent each situation using the slope-intercept form:
y = mx + b
In this case, y represents the total charge, m is the charge per hour, x represents the number of hours, and b represents the material fees.
We have the following:
• Equation for Pro Painters:
y = 200x + 6000
• Equation for Illsion Ltd:
y = 150x + 8000
To graph let's create two points on each equation.
We have:
• Pro painters:
y = 200x + 6000
When x = 1: y = 200(10) + 6000 = 8000
When x = 3: y = 200(30) + 6000 = 12000
We have the points:
(x, y) ==> (10, 8000), (30, 12000)
Plot the points and connect them using a straight line.
• Illusion Ltd:
y = 150x + 8000
When x = 2: y = 150(20) + 8000 = 11000
When x = 4: y = 150(40) + 8000 = 14000
We have the points:
(x, y) ==> (20, 11000), (40, 14000)
Plot the points and connect them using a straight line.
We have the graph below:
The green line represents the cost for Pro Painters
The blue line represents the cost for Illusion Ltd.
From the graph, the point of intersection is (40, 14000).
This means at 40 hours, the cost for both companies will be the same ($14,000)
ANSWER:
• Equation for Pro painters: , y = 200x + 6000
,• Equation for Illusion Ltd: , y = 150x + 8000
,• Point of intersection: (40, 14000)
Can you explain.Use the intermediate value theorem for polynomials to show that the polynomial function has a real zero between the numbers given.f(x) = -6x^4+5x^2+4;-2 and -1
SOLUTION:
We are to show that the given polynomial function has a real zero between the numbers given.
[tex]f(x)=-6x^4+5x^2\text{ + 4}[/tex]At x = -2, we substitute -2 for x in the given function;
[tex]\begin{gathered} f(-2)=-6(-2)^4+5(-2)^2\text{ + 4} \\ f(-2)\text{ = -6(16) + 5(4) + 4} \\ f(-2_{})\text{ = -96 + 20 + 4} \\ f(-2)\text{ = -72} \end{gathered}[/tex]At x = -1, we substitute -1 for x in the given function;
[tex]\begin{gathered} f(-1)=-6(-1)^4+5(-1)^2\text{ + 4} \\ f(-1)\text{ = -6(1) + 5(1) + 4} \\ f(-1)\text{ = -6 + 5 + 4} \\ f(-1)\text{ = 3} \end{gathered}[/tex]CONCLUSION:
Since the function f went from -72 to +3 over the interval of -2 to -1, that means it must have passed through zero.
Refer to the figure below to answer the following questions: (a) When placed in Quadrant ), name the coordinates of point T that forms parallelogram QTRS. (b) When placed in Quadrant II, name the coordinates of point T that forms parallelogram QRST. (c) When placed in Quadrant IV, name the coordinates of point T. that forms parallelogram QRTS. Given Points Q(-1,3), R(3.0), and S(-2,-1) Q T. S
A parallelogram is a quadilateral that has two pairs of parallel sides. The opposite sides of a parallelogram are equal.
Given the points:
Q(-1,3), R(3,0), and S(-2,-1)
a) When placed in quadrant I, let's find the point T that forms a parallellogram.
Here the distance QS and RT must be equal.
Use the distance formula:
[tex]d=\sqrt[]{(x2-x1)^2+(y2-y1)^2}[/tex]The point of T that forms a parallellogram when placed in quadrant I is:
T(4, 4)
From point R
b) When placed in Quadrant II, let's find the point T that forms a parallellogram.
We have:
T(-6, 2)
From point Q, make a movement 5 units left and 1 unit down
The point of T that forms a parallellogram when placed in quadrant II is:
T(-6, 2)
c) When placed in quadrant IV, let's find the point T that forms a parallelogram.
We have:
T(2, -4)
From point R, make a movement of down 4 units and left 1 unit.
The point of T, that forms a parallelogram when placed in quadrant IV is:
T(2, -4)
ANSWER:
a) (4, 4)
b) (-6, 2)
c) (2, -4)
Solve for h: A = (1/2)*b*h*O h = 2*A*bO h = A *(b/2)O h = (2*A)/b0 h = (2+b)/A
Calculate the density of the cube.240 grams4 cm3 cm5 cm
Answer:
4 g / cm^2
Explanation:
The density is defined is
[tex]p=\frac{M}{V}[/tex]where m is the mass of the object and V is its volume.
Now in our case, we see that the cube weighs M = 240 g and has a volume of
[tex]V=3\operatorname{cm}\times5\operatorname{cm}\times4\operatorname{cm}=60\operatorname{cm}^3[/tex]With the value of M and V in hand, we now calculate the density
[tex]p=\frac{240g}{60\operatorname{cm}^3}[/tex][tex]p=\frac{40g}{\operatorname{cm}^3}[/tex]which is our answer!