we have the function
[tex]V(x)=12-2x-5[/tex]Combine like terms and rewrite the given function
[tex]V(x)=7-2x[/tex]Evaluate for x=-2
[tex]\begin{gathered} V(-2)=7-2(-2) \\ V(-2)=7+4 \\ V(-2)=11 \end{gathered}[/tex]Evaluate for x=0
[tex]\begin{gathered} V(0)=7-2(0) \\ V(0)=7 \end{gathered}[/tex]Evaluate for x=5
[tex]\begin{gathered} V(5)=7-2(5) \\ V(5)=7-10 \\ V(5)=-3 \end{gathered}[/tex]Danny deposits $12,500 into a pension fund that invests in stocks. After a successful two years in investing on the stock market, the fund agrees to pay a simple interest rate of 12% per year. What will the balance on the account be after two years? Give your answer in dollars to the nearest dollar. Do not include commas or the dollar sign in your answer. For example if your answer is $1,234.56 enter 1235.
To obtain the final amount after 2 years of simple interest, subtitute the values in the following formula:
[tex]A=P(1+rt)[/tex]where A is the final amount of the investment, P is the principal or the starting amount, r is the rate in decimals, and t is the time in years.
From the problem, we have the following given:
[tex]\begin{gathered} P=12500 \\ r=12\%=0.12 \\ t=2 \end{gathered}[/tex]Substitute the values into the formula.
[tex]\begin{gathered} A=P(1+rt) \\ A=12500\lbrack1+(0.12)(2)\rbrack \end{gathered}[/tex]Simplify the right side of the equation.
[tex]\begin{gathered} A=12500(1+0.24) \\ =12500(1.24) \\ =15500 \end{gathered}[/tex]Therefore, after 2 years, the value of the investment will be $15500.
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
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
Express the given equation in standard form by solving for x. Simplify your answer
SOLUTION
Recall that a linear equation in one variable is in standard form if it is in the form:
[tex]ax+b=0[/tex]Hence the equation:
[tex]x+1=0[/tex]Is in tandrd form
Solving for x gives
'
[tex]x=-1[/tex]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%which inequality is shown in the graph below? A x<2 B x>2 C y>2 S y<2
In the picture, we can see that the solution to the inequality are those x-values greater than 2 (2 is not included) no matter which value the y-variable has. This corresponds to: x > 2.
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
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
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.
Solve the problem below, inputting your answer in decimal form.32 3/4 + 12 1/2
Answer:
45.25
Explanation:
To find the value of the expression given, we first convert the mixed numbers into fractions .
[tex]32\frac{3}{4}=32+\frac{3}{4}[/tex]multiplying 32 by 4/4 gives
[tex]undefined[/tex]people leaving a football match with acid be supported in Manchester United or Newcastle
How many people were asked the questions in total?
40 + 2 + 8 + 20 = 70.
Find the measure of ZGHJ and ZGI.68°H31°.K115°angle GHJ =degreesangle GIJ =degrees
We are asked to determine angles GHJ and GIJ. To do that we need to have into account that these two angles are half the measure of their respective intercepted arc. Since both intercepted arcs are the same then the angles are equal. The intercepted arc is given by:
[tex]\begin{gathered} \theta=360-68-31-115 \\ \theta=146 \end{gathered}[/tex]Therefore, the angles are:
[tex]\angle GHJ=\angle GIJ=\frac{\theta}{2}=\frac{146}{2}=73[/tex]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]James’ dealership uses a one-price, “no haggle” selling policy. The dealership averages 13% profit on new car sales. If the dealership pays $15,600 for a Rancho Turbo, find the selling price after adding the profit to the dealer’s cost.
Help me and I will give you 5 stars!!!:):):)
The selling price after adding the profit to the dealer’s cost is $17628
The dealership averages 13% profit on new car sales
If the dealership pays $15,600 for a Rancho Turbo,
The profit is 13% of the dealership
profit =(13/100) 15600
profit = 2028
Selling proce = cost price + profit
= 15600 + 2028
= 17628
Therefore, the selling price after adding the profit to the dealer’s cost is $17628
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Sonic runs down a ladder of 19 ft against a wall and the base of the ladder is 30 degrees to the ground. What is the distance from the base of the ladder to the wall?
The distance from the base of the ladder to the wall is 16.45 ft.
Given,Sonic runs down a ladder of 19 ft against a wall and the base of the ladder is 30 degrees to the ground.
we are asked to determine the distance from the base of the ladder to the wall=?
Since we have given that
Angle of elevation with the ground = 30°
Here, AC is the ladder .
Distance between the foot ladder from the wall = ?
length of the ladder = 19 ft.
So, we will use "Cosine Rule":
cos 30° = BC/AC
√3/2 = BC/19
2BC = √3×19
2BC = 32.9
BC = 32.9/2
BC = 16.45 ft
Hence the distance from the base of the ladder to the wall is 16.45 ft.
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2. What type of quadrilateral do the following points represent? A (2,1) B (4,3) C (8,3) D (6, 1)
The quadrilateral is a parallelogram (the opposite sides are parallel and equal)
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]Suppose that $4000 is placed in a savings account at an annual rate of 9%, compounded monthly. Assuming that no
withdrawals are made, how long will it take for the account to grow to $6216?
Do not round any intermediate computations, and round your answer to the nearest hundredth.
_ years
Answer:
below
Step-by-step explanation:
The equation to use
FV = PV ( 1 + i)^n FV = 6216 PV = 4000
i = decimal interest per period = .09/12
n = how many months?
6216 = 4000 ( 1 + .09/12)^n
6216/4000 = (1 + .09/12)^n
1.554 = 1.0075 ^n
log 1.554 / log(1.0075) = n = 59 months (approx 5 years )
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.
Mr. Fawcett is building a ramp for loading motorcycles onto atrailer. The trailer is 2.8 feet off the ground. To avoid makingit too difficult to push a motorcycle off the ramp, Mr. Fawcettdecides to make the angle between the ramp and the ground15°. To the nearest hundredth of a foot, find the length ofathe ramp.
Solution
- The illustration described can be sketched as follows:
- From the above diagram, we can observe that the ramp forms a right-angled triangle with the ground.
- The Opposite of the triangle is 2.8 feet, the angle made by the ramp with the ground is 5 degrees., whilethe length of the ramp is labeled as x.
- Thus, we can apply SOHCAHTOA to find the value of x as follows:
[tex]\begin{gathered} \sin\theta=\frac{Opposite}{Hypotenuse} \\ \\ \theta=15\degree,Opposite=2.8,Hypotenuse=x \\ \text{ Thus, we have:} \\ \sin15\degree=\frac{2.8}{x} \\ \\ \therefore x=\frac{2.8}{\sin15\degree} \\ \\ x=10.8183692544...\approx10.82ft \end{gathered}[/tex]Final Answer
The length of the ramp is 10.82 feet
The tin can shown below has the indicated dimensions.1.5in.3.25in.A cylinder is shown. The radius of the top circular base is labeled (1 .5) inches and the altitude is labeled (3.25) inches.Estimate the number of square inches of tin required for its construction. (Hint: Include the lid and the base in the result. Use your calculator value of . Round your answer to two decimal places.)in2
Given:
A cylinder having radius 1.5 inches and a height of 3.25 inches
Required:
Estimate the number of square inches of tin required for its construction.
Explanation:
To calculate the number of square inches of tin required for its construction we have to calculate the area of the lid and base and the curved surface then add them all.
[tex]\begin{gathered} \pi r^2+\pi r^2+2\pi rh \\ \Rightarrow\frac{22}{7}\times1.5^2+\frac{22}{7}\times1.5^2+2\times\frac{22}{7}\times1.5\times3.25 \\ =44.76769531\text{ in}^2 \\ =44.77\text{ in}^2 \end{gathered}[/tex]Final Answer:
44.77 inches square
Determine the remainder when 6x^3+ 23x^2 - 6x -8 is divided by 3x-2. What information does the remainder provide about 3x-2? Explain.
we have
6x^3+ 23x^2 - 6x -8 : (3x-2)
step 1
Verify if (3x-2) represents a factor
If (3x-2) is a factor
then
3x-2=0 ------> x=2/3
Substitute the value of x=2/3 in the given expression
6(2/3)^3+23(2/3)^2-6(2/3)-8
6(8/27)+23(4/9)-4-8
(16/9)+(92/9)-12
12-12=0
that means
(3x-2) is a zero of the given function
therefore
when divide (6x^3+ 23x^2 - 6x -8 ) by (3x-2), the remainder is zerowhat two intergers does the square root of 15 fall between
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
2 integers ===> √15
Step 02:
[tex]\sqrt[]{15}=\text{ }3.87[/tex]integer 1 = 3
integer 2 = 4
That is the solution.
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:
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
Difference of Squares gives which complex factors for the expression x2 +11?A. (x + W11)(x - 111)B. (x+in/11)(x +111)C. (x + 111)2(x - in 11)D. (x - iw/11)(x-in 11)SUBMIT
we have that
[tex](x+i\sqrt[\square]{11})\cdot(x-i\sqrt[\square]{11})=x^2-(i^2)(11)=x^2+11[/tex]answer is the first option
option A
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.
4. Enter the total area of the figure ABCD in square centimeters 8 cm А 6 cm F C 15 cm 8 cm D O 268 O 336 168 O 37
The figure ABCD has the shape of a Rhombus with diagonals AC and BD.
To determine the area of a Rhombus you have to multiply the length of both diagonals and divide the result by 2, following the formula:
[tex]A=\frac{pq}{2}[/tex]Where
p represents the horizontal diagonal
q represents the vertical diagonal
For the quadrilateral ABCD, the lengths of the diagonals are:
AC=6cm + 15cm =21cm
BD= 8cm + 8cm=16cm
[tex]\begin{gathered} A=\frac{AC\cdot BD}{2} \\ A=\frac{21\cdot16}{2} \\ A=\frac{336}{2} \\ A=168\operatorname{cm}^2 \end{gathered}[/tex]The area of the figure is 168cm²
How many solutions does the system of equations below have?4x − 8y = –17x − 14y = 4No solutionOne solutionInfinitely solutions
start clearing the x in the first equation
[tex]\begin{gathered} 4x=-1+8y \\ x=-\frac{1}{4}+2y \end{gathered}[/tex]insert this equation into the second one
[tex]\begin{gathered} 7\cdot(-\frac{1}{4}+2y)-14y=4 \\ -\frac{7}{4}+14y-14y=4 \\ -\frac{7}{4}\ne4 \end{gathered}[/tex]the system has no solution
Solve the system of two linear inequalities graphically.ſr<62-3Step 2 of 3 : Graph the solution set of the second linear inequality.AnswerKeyboThe line will be drawn once all required data is provided and will update whenever a value is updated. The regions will be added once the line is drawn.Enable Zoom/PanChoose the type of boundary line:Solid (-) Dashed (-)Enter two points on the boundary line:510-35Select the region you wish to be shaded:
Answer and Explanation:
Given the system of two linear inequalities;
[tex]\begin{gathered} x<6 \\ x\ge-3 \end{gathered}[/tex]To solve the above system of inequalities graphically, we follow the below steps;
Step 1: Graph the first inequality;
Since the first inequality has a less than sign, we'll shade the region to the left of the line.
Also, the first inequality does not have an equality sign, so the line will be a dashed line.
See below the graph of the first inequality;
Step 2: Graph the second inequality on the same grid;
Since the inequality has an equality sign, the line will be a solid line.
Also, the inequality has the greater than sign, so we'll shade the region to the right of the line
See below the image of the graph;
Step 3: The solution set of the two systems of inequalities is the region where the shading overlaps.
As can be seen in the above graph, the shaded region between the dashed line and the solid line is the solution of the system of inequalities.
The graph of the solution set of the second inequality is as shown below;
On the boundary line we can select the below points;
[tex]\begin{gathered} \lparen-3,5) \\ \lparen-3,-5) \end{gathered}[/tex]