We have to calculate the height of the stack of hay bales.
We can start by calculating the volume as the number of bales times the volume of one hay:
[tex]\begin{gathered} V=n*V_0=8*(10+\frac{2}{3}) \\ V=8*10+8*\frac{2}{3} \\ V=80+\frac{16}{3} \\ V=80+\frac{15}{3}+\frac{1}{3} \\ V=80+5+\frac{1}{3} \\ V=85+\frac{1}{3} \end{gathered}[/tex]Now, we know that this volume will be the area of the base times the height.
The area of the base can be calculated as the product of the length and the width:
[tex]\begin{gathered} A_b=L*W \\ A_b=4*(1+\frac{1}{3}) \\ A_b=4+\frac{4}{3} \\ A_b=\frac{4*3+4}{3} \\ A_b=\frac{12+4}{3} \\ A_b=\frac{16}{3} \end{gathered}[/tex]We then can calculate the height as the volume divided by the base area:
[tex]\begin{gathered} h=\frac{V}{A} \\ h=\frac{85+\frac{1}{3}}{\frac{16}{3}} \\ h=\frac{85*3+1}{3}*\frac{3}{16} \\ h=\frac{256}{3}*\frac{3}{16} \\ h=\frac{256}{16} \\ h=16 \end{gathered}[/tex]Answer the height is 16 feet.
can some one clarify this question, i think ik the answer but i need some elses opinion Find m
hello
to solve this question, we simply need to add two quadrants that make up mto get m[tex]\begin{gathered} m<\text{WYV}=60^0 \\ m<\text{VYU}=85^0 \end{gathered}[/tex][tex]\begin{gathered} m<\text{WYU}=<\text{WYV}+m<\text{VYU} \\ m<\text{WYU}=60^0+85^0 \\ m<\text{WYU}=145^0 \end{gathered}[/tex]from the calculations above, the value of m
Jan and her brother mel go to different schools. Jan goes 6 kilometer east from home. Mel goes 8 kilometer north. How many kilometer apart are their schools.
Jan goes 6 km east from her home, and Mel goes 8 km north from the same home. So we need to find the distance between both schools, and we do such using the Pithagorean theorem, since we are in the presence of a right angle triangle for which we know the two legs, and need to find the measure of the hypotenuse.
I am going to represent the problem with a diagram below, so you see the right angle triangle I am talking about.
So the two legs are represented by the distances each student travels, and the segment in red is the distance between the schools which appears as the HYPOTENUSE of the right angle triangle.
Therefore, we use the Pythagoras theorem for the hypotenuse:
[tex]\begin{gathered} \text{Hypotenuse}=\sqrt[]{leg1^2+\text{leg}2^2} \\ \text{hypotenuse}=\sqrt[]{6^2+8^2} \\ \text{Hypotenuse}=\sqrt[]{36+64} \\ \text{Hypotenuse}=\sqrt[]{100}=10 \end{gathered}[/tex]Therefore, the distance between the schools is 10 km
Tony is a hiring director at a large tech company in Chicago, and he gets hundreds of resumes each week. How long does Tony MOST likely spend looking over each resume?30 seconds50 seconds3 minutes30 minutes
The time needed to look over the resumes depends on how many papers is the resume
But it is convenient to have a speed looking on each one
so, the answe will be 50 seconds
Will you ever completely remove the drug from your system? Explain your reasoning.
Answer
The drug cannot be completely eliminated from one's system.
This is because the kidney removes 25% of the drug, leaving 75% at any time; the 75% of any number will give a smaller number, but never zero.
So, the amount of the drug in the body system can become extremely low, but it can never be 0.
The mathematical proof is shown under explanation.
Explanation
We are told that the kidney filters off 25% of the drug out of the system every 4 hours.
This means that 75% of the dosage of the drug remains in the person's system every 4 hours.
If one starts with A₀ of the drug and classify every 4 hour time period as n
At n = 1,
A₁ = 0.75 (A₀)
A₂ = 0.75 (A₁) = 0.75² (A₀)
Aₙ = 0.75ⁿ (A₀)
For this question, we start wit 1000 mg
A₀ = 1000 mg
We are then asked to calculate if Aₙ, the amount of drug in the system after n time periods, can ever be 0
Aₙ = 0.75ⁿ (A₀)
0 = 0.75ⁿ (1000)
To solve for n, if there's an n for when the value of Aₙ = 0, we first divide both sides by 1000
0 = 0.75ⁿ (1000)
0 = 0.75ⁿ
We then take the natural logarithms of both sides
In 0 = In (0.75ⁿ)
In (0.75ⁿ) = In 0
n (In 0.75) = In 0
But, since In 0 does not exist, it shows that there is no value of n that can make the value of Aₙ go to 0.
Hope this Helps!!!
Use slope to determine if lines AB and CD are parallel, perpendicular, or neither 6. A(-3, 8), B(3, 2), C(7,1), D(5,-1)m(AB) m(CD) Types of Lines
In a class of 10 boys and 12 girls, a committee of 4 members is to be formed. What is the probability to form a committee consisting of 2 boys and 2 girls?0.30400.40600.50600.2060
Consider all the different possible combinations of 4 members of the committee (b,b,b,b), (b,b,b,g),...(g,g,g,g). We need to use the binomial distribution given below
[tex]P(k)=(nbinomialk)p^k(1-p)^{n-k}[/tex]In our case
[tex]k=2,n=4,p=\frac{10}{10+12}=\frac{10}{22}=\frac{5}{11}[/tex]Then,
[tex]\begin{gathered} P(2)=(\frac{4!}{2!(4-2)!})(\frac{5}{11})^2(\frac{6}{11})^2 \\ \Rightarrow P(2)=6\cdot\frac{900}{14641} \\ \Rightarrow P(2)=0. \end{gathered}[/tex]solve the equation by completing the square. Show all solutions8x^2 + 16x = 42
8x² + 16x = 42
x² + 16/8x = 42/8 dividing by 8 at both sides
x² + 2x = 5.25
x² + 2x - 5.25 = 0
If we compute (x + 1)², we get:
(x + 1)² = x² + 2*x*1 + 1² = x² + 2x + 1
Then,
x² + 2x - 5.25 + 1 - 1 = 0
(x² + 2x + 1) + (-5.25 - 1) = 0
(x + 1)² - 6.25 = 0
(x + 1)² = 6.25
x + 1 = √6.25
This has 2 solutions,
x + 1 = 2.5 or x + 1 = -2.5
x = 2.5 - 1 x = -2.5 - 1
x = 1.5 x = -3.5
According to Debt.org the average household has $7,281 in credit card debt. Estimate how much interest the average household accumulates over the course of 1 year. We are going to assume the APR is 16.99%.
In order to estimate the interest the average househould accumulates in 1 year, you use the following formula:
A = Prt
where P is the initial credit card debt ($7,281), r is the interest rate per period (16.99%) and t is the number of time periods. In this case the value of r is given by the APR, then, there is one period of 1 year.
To use the formula it is necessary to express 16.99% as 0.1699. Thus, you have:
I = 7,281 x 0.1699 x 1
I = 1,237.04
Hence, the interest accumulated is of $1,234.04
Determine whether the equation represents an exponential growth function, anexponential decay function, and give the percent growth or decay.17. y = 18(1.3)^t
A exponential growth or decay function has the next general form:
[tex]y=a(1\pm r)^t[/tex]If it is :
(1+r) , (>1) the function growth
(1-r) , (<1) the function decay
------
The given equation:
[tex]y=18(1.3)^t[/tex]As the (1+r) is equal to 1.3 (> 1) then it is a exponential growth function.In (1+r) the r is the percent of growth, then for the given equation you have:
[tex]\begin{gathered} 1+r=1.3 \\ r=1.3-1 \\ \\ r=0.3 \end{gathered}[/tex]The percent of decay is 0.3 or 30%review the rental and purchase property information to answer the question: calculate the difference in total move-in cost between the two properties. $31,497.35 $35,842.95$39,285.45$4,976.55
Let us calculate the move-in costs of both properties.
Rental Property
The monthly rent is $1,350.
The move-in costs are:
First month = $1,350
Last month = $1,350
Security deposit = 55% of one month's rent
[tex]\Rightarrow\frac{55}{100}\times1350=742.5[/tex]Therefore, the move-in cost is:
[tex]\Rightarrow1350+1350+742.5=3442.5[/tex]Purchase Property
The purchase price is $195,450.
The move-in costs are:
Down payment of 18% of purchase price:
[tex]\Rightarrow\frac{18}{100}\times195450=35181[/tex]Closing costs of 2.1% of purchase price:
[tex]\Rightarrow\frac{2.1}{100}\times195450=4104.45[/tex]Therefore, the move-in cost is:
[tex]\Rightarrow35181+4104.45=39285.45[/tex]Difference in Total Move-In Cost
This is calculated to be:
[tex]\Rightarrow39285.45-3442.5=35842.95[/tex]ANSWER
The difference in total move-in cost is $35,842.95
A small radio transmitter broadcasts in a 60 mile radius. If you drive along a straight line from a city 75 miles north of the transmitter to a second city 76 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?
Answer:
Explanation:
We start by having a diagrammatic representation as follows:
A brownie recipe asks for two and two thirds times as much sugar as chocolate chips. If four and one third cups of sugar is used, what quantity of chocolatechips would then be needed, according to the recipe?0308X5?
Let's call C to the cups of chocolate chips and S to the cups of sugar. We are told that the cups of sugar are 2 2/3 times the cups of chocolate, then we can formulate the following equation:
[tex]S=2\frac{2}{3}C[/tex]In the case 4 1/3 of sugar is added, we can replace 4 1/3 for S to get:
[tex]4\frac{1}{3}=2\frac{2}{3}C[/tex]By dividing both sides by 2 2/3 we get:
[tex]\begin{gathered} 4\frac{1}{3}\div2\frac{2}{3}=2\frac{2}{3}C\div2\frac{2}{3} \\ 4\frac{1}{3}\div2\frac{2}{3}=C \end{gathered}[/tex]We can rewrite the mixed fractions to get:
[tex]\begin{gathered} \frac{4\times3+1}{3}\div\frac{2\times3+2}{3}=C \\ \frac{12+1}{3}\div\frac{6+2}{3}=C \\ \frac{13}{3}\div\frac{8}{3}=C \end{gathered}[/tex]By changing the division symbol to a multiplication symbol and flipping the 8/3, we get:
[tex]\begin{gathered} \frac{13}{3}\times\frac{3}{8}=C \\ \frac{13}{8}=C \\ \frac{8+5}{8}=C \\ \frac{8}{8}+\frac{5}{8}=C \\ 1+\frac{5}{8}=C \\ 1\frac{5}{8}=C \\ C=1\frac{5}{8} \end{gathered}[/tex]Then, 1 5/8 cups of chocolate chips are needed
Determine (Freshman) Small Cafeteria). Interpret this answer in the context of the situation.
Step 1:
[tex]\text{Probability = }\frac{N\text{umber of required outcomes}}{N\text{umber of total possible outcome}}[/tex]Step 2:
a)
Total possible outcome = 2640
Total number of freshman = 625
[tex]\begin{gathered} P(\text{Freshman) = }\frac{625}{2640} \\ \text{= }\frac{125}{528} \\ \text{= 0.237} \end{gathered}[/tex]Step 3:
b)
Total number of senior and large cafeteria = 350
[tex]\begin{gathered} P(\text{senior and large cafeteria) = }\frac{350}{2640} \\ =\text{ }\frac{70}{528} \\ =\text{ }\frac{35}{264} \\ =\text{ 0.132} \end{gathered}[/tex]Step 4:
c)
Number of Sophomore or student center = 650 + 595 - 125 = 1120
[tex]\begin{gathered} P(\text{Sophomore or student center) = }\frac{1120}{2640} \\ =\text{ }\frac{112}{264} \\ =\text{ 0}.424 \end{gathered}[/tex]Step 5:
d)
[tex]\begin{gathered} p(\text{freshman}|\text{small cafeteria) = }\frac{n(freahman\text{ and small cafeteria)}}{n(small\text{cafeteria)}} \\ =\frac{435}{860}\text{ } \\ =\text{ }\frac{87}{172} \\ =\text{ 0.506} \end{gathered}[/tex]Find the indicated function given f(x)=2x^2+1 and g(x)=3x-5. When typing your answer if you have an exponent then use the carrot key ^ by pressing SHIFT and 6. Type your simplified answers in descending powers of x an do not include any spaces between your characters.f(g(2))=Answerf(g(x))=Answerg(f(x))=Answer (g \circ g)(x) =Answer (f \circ f)(-2) =Answer
Given the functions
[tex]\begin{gathered} f(x)=2x^2+1 \\ g(x)=3x-5 \end{gathered}[/tex]1) To find f(g(2))
[tex]\begin{gathered} f(g(x))=2(3x-5)^2+1 \\ f(g(x))=2(9x^2-15x-15x+25)+1=2(9x^2-30x+25)+1 \\ f(g(x))=18x^2-60x+50+1=18x^2-60x+51 \\ f(g(2))=18(2)^2-60(2)+51=18(4)-120+51 \\ f(g(2))=72-120+51=3 \\ f(g(2))=3 \end{gathered}[/tex]Hence, f(g(2)) = 3
2) To find f(g(x))
[tex]\begin{gathered} f(g(x))=2(3x-5)^2+1 \\ f(g(x))=2(9x^2-15x-15x+25)+1=2(9x^2-30x+25)+1 \\ f(g(x))=18x^2-60x+50+1=18x^2-60x+51 \\ f(g(x))=18x^2-60x+51 \end{gathered}[/tex]Hence, f(g(x)) = 18x²-60x+51
3) To find g(f(x))
[tex]\begin{gathered} g(f(x))=3(2x^2+1)-5 \\ g(f(x))=6x^2+3-5=6x^2-2 \\ g(f(x))=6x^2-2 \end{gathered}[/tex]Hence, g(f(x)) = 6x²-2
4) To find (gog)(x)
[tex]\begin{gathered} (g\circ g)(x)=3(3x-5)-5=9x-15-5=9x-20 \\ (g\circ g)(x)=9x-20 \end{gathered}[/tex]Which angles are adjacent to <2? Select all that apply.
help meeeeeeeeee pleaseee !!!!!
The values of the functions are:
a. (f + g)(x) = 9x + 1
b. (f - g)(x) = -7x - 17
c. (f * g)(x) = 8x² - 55x - 72
d. (f/g)(x) = (x - 8)/(8x + 9)
How to Evaluate Functions?To evaluate a function for a given value of input, substitute the value of the input, x, into the function equation, evaluate and simplify.
Given the following:
f(x) = x - 8
g(x) = 8x + 9
a. (f + g)(x) = (x - 8) + (8x + 9)
Combine like terms
(f + g)(x) = 9x + 1
b. (f - g)(x) = (x - 8) - (8x + 9)
(f - g)(x) = x - 8 - 8x - 9
Combine like terms
(f - g)(x) = -7x - 17
c. (f * g)(x) = (x - 8)(8x + 9)
Expand
(f * g)(x) = x(8x + 9) -8(8x + 9)
(f * g)(x) = 8x² - 55x - 72
d. (f/g)(x) = (x - 8)/(8x + 9)
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My name is nessalovetrillo i am prepping and studying to test out of my algebra class this is for a study guide Please see attached picture
Given:
S={(5,6),(-2,-9),(-9,6)}
To find the domain and range:
The domain is,
{-9, -2, 5}
The range is,
{-9, 6}
As cashier, you need to record all over times you worked in hours. If you worked 330 mnts of over time how many hours will you record ?
First, we need the next equivalence
1 hour = 60 min
we have 330 min in order to know the number of hours we need to divide the 330 min between 60
[tex]\frac{330}{60}=5.5[/tex]He will record 5.5 hours
An ordinary (Pair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in successionand that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of andereCompute the probability of each of the following svents.Event A: The sum is greater than 7.Event B: The sum is divisible by 3 or 6 (or both).Write your answers as fractions
1) We are going to tackle this question starting with the total outcomes of dice rolled twice in succession.
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
So we can see that there are 36 possibilities.
2) Let's examine the events.
a) P (>7)
Let's bold the combinations of outcomes whose sum is greater than 7
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
So, we can see that there are 15 favorable outcomes.
Now, we can find the Probability of rolling the dice twice and get a sum greater than 7:
[tex]P(A)=\frac{15}{36}=\frac{5}{12}[/tex]b) Now, for the other event: The sum is divisible by 3 or 6, or both:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Hence, the favorable outcomes are: 12
So now, let's find the probability of getting a sum that way:
[tex]P(B)=\frac{12}{36}=\frac{1}{3}[/tex]There are 16 appetizers available at a restaurant. From these, Pablo is to choose 12 for his party. How many groups of 12 appetizers are possible?
EXPLANATION
This is a combinatory, as there are 12 groups, the combinatory will be as follows:
16C12 = 16!/[12!*(16-12)!] = 1820
In conclusion, there will be 1820 possible groups of 12 appetizers.
solve the following system of inequalities graphically on the set of axes below?witch of the coordinates points would be in the solution set
Using a graphing calculator to find local extrema of a polynomial function
The given function is:
[tex]f(x)=3x^4-5x^3-4x^2+5x-2[/tex]By using a graphing calculator, we found that the local maximum is located at:
x=0.41, then f(0.41)=-0.88
The answer is (0.41, -0.88)
The graph of polynomial f is shown. Select all the true statements about the polynomial.aThe degree of the polynomial is even.bThe degree of the polynomial is odd.cThe leading coefficient is positive.dThe leading coefficient is negative.eThe constant term of the polynomial is positive.fThe constant term of the polynomial is negative.
Explanation:
From the graph,
we can see that the graph is symmetric about the y axis
Hence,
We can say that the Polynomial is even
Also, Because th opwning of the function is downwards,
Hence the leading coefficient is negative
Also we can see that the y-intercept is positive
That is when x=0, y=3
Hence,
The constant term of the polynomial is positive.
Therefore,
The final answers are OPTION A,OPTION D,OPTION E
8.1 km to miles and feet
Given
[tex]8.1\operatorname{km}[/tex]It should be noted that
[tex]\begin{gathered} 1\operatorname{km}=0.621371miles \\ 1\text{mile}=5280\text{feet} \end{gathered}[/tex][tex]\begin{gathered} \text{convert 8.1km to miles} \\ 1\operatorname{km}=0.621371\text{miles} \\ 8.1\operatorname{km}=8.1\times0.621371 \\ 8.1\operatorname{km}=5.0331051\text{miles} \end{gathered}[/tex][tex]\begin{gathered} 8.1\operatorname{km}=5\text{miles}+0.0331051\text{miles} \\ \text{convert 0.0331051miles to fe}et \\ 1\text{miles}=5280ft \\ 0.0331051\text{miles}=0.0331051\times5280feet \\ 0.0331051\text{miles}=174.79feet \end{gathered}[/tex]Hence, 8.1km is 5 miles and 174.79 feet
What is the value of x? A pair of intersecting lines is shown. The angle above the point of intersection is labeled left parenthesis 7 x minus 8 right parenthesis degrees. The angle directly opposite below the point of intersection is labeled left parenthesis 6 x plus 11 right parenthesis degrees. (1 point)
A –19
B 125
C 19
D 55
The value of x in the angles is 19.
How to find angles in intersecting lines?When lines intersect, angle relationships are formed such as vertically opposite angles, adjacent angles etc.
Therefore, let's find the value of x in the intersecting lines.
Hence,
7x - 8 = 6x + 11 (vertically opposite angles)
Vertically opposite angles are congruent and they share the same vertex point.
Hence,
7x - 8 = 6x + 11
subtract 6x from both sides of the equation
7x - 8 = 6x + 11
7x - 6x - 8 = 6x - 6x + 11
x - 8 = 11
add 8 to both sides of the equation
x - 8 + 8 = 11 + 8
x = 19
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From the given proportional relationship, which of the following points lie on the same line?
As per given by the question,
There are given that a graph of line.
Now,
h
What is the radius of a hemispherewith a volume of 281,250 cm??
Given:
The volume of the hemisphere = 281,250
Find-:
Radius of hemisphere
Explanation-:
The volume of the hemisphere is:
[tex]V=\frac{2}{3}\pi r^3[/tex]Given volume is 281250
[tex]\begin{gathered} V=\frac{2}{3}\pi r^3 \\ \\ 281250=\frac{2}{3}\pi r^3 \\ \\ r^3=\frac{281250\times3}{2\times\pi} \\ \\ r^3=134286.9832 \\ \\ r=51.209 \end{gathered}[/tex]So, the radius is 51.209 cm
1. Identify the vertex (locator point) of the above parabola2 po(1,2)(3,0)(3,0)(2,1)2. Identify the vertex from the quadratic function y=-5(x-6) 2+82 point
Answer:
(2,1)
Step-by-step explanation:
The vertex of a parabola is it's highest point(if it is concave down), or it's lowest point, if it's concave up.
In this question:
It's concave down, so the vertex is the highest point.
It happens when x = 2, at which y = 1.
So the vertex is the point (2,1)
Please help I need by today only questions 5 and 6 need to show work
Part a: Pot the points X, Y and Z are obtained on graph.
Part b: Distances; XY = 3, YZ = 5 and XZ = √34 units.
Part c: Measure of angles; X = 59.04 degrees and Z = 30.96 degrees.
What is termed as the Pythagorean theorem?The Pythagorean theorem states that the sum of a squares just on legs of a right triangle equals the square just on hypotenuse.For the given question, Triangle XYZ with the vertexes are given.
Part a: Pot the points.
X = (6, 6)
Y = (6, 3)
Z = (1, 3)
The points on the graph are plotted.
Part b: Distances;
XY = 6 - 3 = 3 units
YZ = 6 - 1 = 5 units
XZ , use Pythagorean theorem.
XZ² = XY² + YZ²
Put the values.
XZ² = 3² + 5²
XZ² = 9 + 25
XZ² = 34
XZ = √34 units.
Part c: Measure of angles.
In right triangle XYZ
cos X = XY/XZ
cos X = 3/ √34
X = 59.04 degrees.
cos Z = ZY/XZ
cos Z = 5/ √34
Z = 30.96 degrees.
Thus, the value of the triangle are obtained.
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What is the slope of the line passing through the points (−1, 7) and (4, −1)? −5/62−8/5−2
Given the points:
(−1, 7) and (4, −1)
The slope of the line passing through the points is given by:
[tex]slope=\frac{rise}{run}=\frac{y_2-y_1}{x_2-x_1}=\frac{-1-7}{4-(-1)}=\frac{-8}{5}[/tex]So, the answer will be Slope = -8/5