Answer:
$1200 would be raised if 60 cars were washed.
Step-by-step explanation:
This question can be solved using a rule of three.
The total after 9 cars was of $180. How much would be raised with 60 cars washed?
9 cars - $180
60 cars - $x
Applying cross multiplication
9x = 180*60
9x = 10800
x = 10800/9
x = 1200
$1200 would be raised if 60 cars were washed.
Hello pls help meeee
The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
Given that,
The function is (-x+3)/ (3x-2)
We have to find f(1) and f'(x).
Take the function expression
f(x)= (-x+3)/ (3x-2)
Taking x as 1 value
f(1)= (-1+3)/(3(1)-2)
f(1)=2/1
f(1)=1
Now, to get f'(x)
With regard to x, we must differentiate.
f(x) is in u/v
We know
u/v=(vu'-uv')/ v² (formula)
f'(x)= ((3x-2)(-1)- (-x+3)(3))/ (3x-2)²
f'(x)= ((-3x+2)-(-3x+9))/ 9x²- 12x+4
f'(x)=(-3x+2+3x-9)/ 9x²- 12x+4
f'(x)=2-9/ (9x²- 12x+4)
f'(x)=-7/ (9x²- 12x+4)
Therefore, The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
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Find the value of 3127°B5xАC с3x + 6D127°c
From the diagram,
The angles, 127 degrees at arc AB and arc CD shows that the Chords AB and CD are equal
[tex]\begin{gathered} \text{chord AB = chord CD} \\ \text{that is} \\ 5x\text{ = 3x + 6} \\ \text{collect like terms} \\ 5x\text{ - 3x = 6} \\ 2x\text{ = 6} \\ \text{divide both sides by 2} \\ x\text{ = }\frac{6}{2} \\ x\text{ = 3} \end{gathered}[/tex]Therefore, the value of x = 3
Lynn needs a large truck toMove some furniture. She found that the cost C of renting a truck is $20 per day plus $1 per mile mWrite an equation for the cost of one day in terms of the miles drivenGraph the equation for values up to and including 100 milesEstimate the cost of driving 20 miles in one day
We can set the following equation with the information given:
[tex]C=1\cdot m+20[/tex]Notice that it is a linear equation and its graph is as follows:
which corresponds to option A.
The estimated cost of driving 20 miles in one day is:
[tex]\begin{gathered} C=1\cdot20+20 \\ C=40 \end{gathered}[/tex]If C=$25, solving the first equation on the board for m we get:
[tex]\begin{gathered} 25=m+20 \\ m=25-20 \\ m=5 \end{gathered}[/tex]Therefore, the truck was driven 5 miles if the cost for one day is $25.
Find the missing digit and check calculations.went to the store and bought 18 two-litter bottles of off-brand soda for his Super Bowl party last weekend. His receipt is smuggled, But he can see that the cost of these bottles before tax was 15.#4. how much did each bottle cost?
We have a smuggled total that is 15.#4 after buying 18 bottles.
We can estimate the cost of each bottle as:
[tex]p=\frac{15+0.1\cdot x+0.04}{18}[/tex]where x is the tenths of a dollar, and it will have an integer value between 0 and 9.
If we try the different possible values for x, we get:
[tex]\begin{gathered} x=0\Rightarrow15.04/18=0.835555555555555 \\ x=1\Rightarrow15.14/18=0.841111111111111 \\ x=2\Rightarrow15.24/18=0.846666666666667 \\ x=3\Rightarrow15.34/18=0.852222222222222 \\ x=4\Rightarrow15.44/18=0.857777777777778 \\ x=5\Rightarrow15.54/18=0.863333333333333 \\ x=6\Rightarrow15.64/18=0.868888888888889 \\ x=7\Rightarrow15.74/18=0.874444444444444 \\ x=8\Rightarrow15.84/18=0.88 \\ x=9\Rightarrow15.94/18=0.885555555555556 \end{gathered}[/tex]We can see that the only value that give a unit price in cents is for x = 8, which corresponds to a total price of $15.84 and a unit price of $0.88.
Answer: we can estimate that the unit price is $0.88.
In an Oreo factory, the mean mass of a cookie is given as40 grams with a standard deviation of 2 grams. Whatpercentage of the cookies are between 34 grams and 42grams?
The Solution:
Given:
[tex]\begin{gathered} \bar{x}=mean=40g \\ \sigma=\text{ standard deviation}=2g \end{gathered}[/tex]We are required to find the percentage that is between 34 grams and 42 grams.
By z-score statistic,
The lower limit is:
[tex]Z_1=\frac{x-\mu}{\sigma}=\frac{34-40}{2}=\frac{-6}{2}=-3[/tex]The upper limit is:
[tex]Z_2=\frac{x-\mu}{\sigma}=\frac{42-40}{2}=\frac{2}{2}=1[/tex]The probability is:
[tex]P(Z_1Converting to percent, we multiply by 100:[tex]0.840\times100=84\text{\%}[/tex]Therefore, the correct answer is 84%
Find the solution of the system by graphing.-x-4y=4y = 1/4x-3Part A: Graph the system on the coordinate plane.
the linear equations are
[tex]\begin{gathered} -x-4y=4 \\ y=\frac{1}{4}x-3 \end{gathered}[/tex]Their graph is:
The solution of the system is the point in which the lines cross in the plane.
We can see that this point is (4,-2).
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Question 12, "2.7.19 >
The amount of 20% alcohol solution is [
(Type an integer or a decimal.)
points
O Points: 0 of 1
How many ounces of a 20% alcohol solution must be mixed with 14 ounces of a 25% alcohol solution to make a 21%
alcohol solution?
ounces.
Save
You need to add 10 ounces of 20% alcohol solution.
x = ounces of 20% alcohol solution to add to the 15 oz of 25% solution
15 oz of a 25% solution = 3.75 oz of pure alcohol
The total amount can be described 15+x ounces.
We solve the problem in terms of the amount of pure alcohol
2x + 3.75 = .23(15+x)
2x + 3.75 = 3.45 + .23x
Subtracting .2x from both sides
3.75 = .03x + 3.45
Subtracting 3.45 from both sides
3 = .03x
Multiply by 100
30 = 3x
Divide by 3
10 = x
x = 10
So, you have to add 10 oz of 20% alcohol solution to 15 oz of 25% alcohol.
At the end we would have 25 oz that we believe would be 23% alcohol. If that is true, then we would have:
23 * 25 = 5.75 oz of pure alcohol in the solution.
We have shown (above) that we have 3.75 oz of pure alcohol in the 15 oz of 25% solution.
How many oz of pure alcohol is there in 10 oz of 20% alcohol?
2*10 = 2 oz
3.75 + 2 = 5.75 oz, which is exactly what we needed
You need to add 10 ounces of 20% alcohol solution.
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Which of the functions below could have created this graph?OA. F(x)=x²+2x-2OB. F(x)---4C. F(x)=3x²+2x²OD. F(x)=-²
The functions [tex]F(x)=-\frac{1}{2}x^{4}-x^{3}+x+2[/tex] could have created this graph of downward parabola.
Option D is correct because the leading coefficient of given function is negative. That's why it is making the downward parabola.
As we can see from the graph that the the parabola is downward.
And we know that if the leading coefficient is less than zero, thus the graph will be downward parabola.
Lets check all the option:
For option A:
[tex]F(x)=x^{3}+x^{2}+x+1[/tex], this is a cubic polynomial and the graph will be a cubic curve.
For option B:
[tex]F(x)=x^{2}+5[/tex], this is a quadratic polynomial and leading coefficient is positive, so the graph will be upward parabola.
For option C:
[tex]F(x)=x^{4}-2x[/tex], for x = 0, F(x) =0, so it will not intersect the y-axis. Also the graph will be upward parabola.
Hence, the option D, [tex]F(x)=-\frac{1}{2}x^{4}-x^{3}+x+2[/tex], could have created this graph of downward parabola.
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what ar the restricted value of ratio expressed in fraction.
The given rational expression is:
[tex]\frac{x^2-9}{x^2-4x}[/tex]To find the restr
Nathaniel ate 75% of his chocolate bars. If he startedwith 40 chocolate bars, how many did he eat?
We have that the 40 chocolate bars that Nathaniel had is the 100%. Then, to find out how many chocolate bars he ate, we have to multiply 40 by 75%:
[tex]40\cdot0.75=30[/tex]therefore, Nathaniel ate 30 chocolate bars,
3. SOLVE THE unear equation X - 9 = 35x
x - 9 = 35x
Subtract x from both sides:
x - 9 -x = 35x - x
-9 = 34x
Divide both sides by 34:
-9/34 = 34x/34
x = -9/34
5. Julia's test scores on the first four science tests were: 85, 77, 63, 90. Therewill be five tests. She needs an average of at least 80 in order to get a B on herreport card.Part B: What is the minimum score that Julia must earn on her final test in orderto get a B average?
Let the minimum score be x
(85 + 77 + 63 + 90 + x)/5 = 80
315 + x = 400
x = 400 - 315
= 85
Julia must get a minimum score of 85 to get a B average
the angle t is an acute angle and sin t and cos t are given. use identities to find tan t, csc t, sec t, and cot t. where necessary, rationalizing denominators.
Recall the following identities:
[tex]\begin{gathered} \tan (t)=\frac{\sin (t)}{\cos (t)} \\ \csc (t)=\frac{1}{\sin (t)} \\ \sec (t)=\frac{1}{\cos (t)} \\ \cot (t)=\frac{\cos (t)}{\sin (t)} \end{gathered}[/tex]Since sin(t)=12/13 and cos(t)=5/13, then:
[tex]\begin{gathered} \tan (t)=\frac{(\frac{12}{13})}{(\frac{5}{13})} \\ =\frac{12}{5} \end{gathered}[/tex][tex]\begin{gathered} \csc (t)=\frac{1}{(\frac{12}{13})} \\ =\frac{13}{12} \end{gathered}[/tex][tex]\begin{gathered} \sec (t)=\frac{1}{(\frac{5}{13})} \\ =\frac{13}{5} \end{gathered}[/tex][tex]\begin{gathered} \cot (t)=\frac{(\frac{5}{13})}{(\frac{12}{13})} \\ =\frac{5}{12} \end{gathered}[/tex]Find a linear equation satisfying the condition, if possible. Passes through (−1,5) and (0,10)
The linear equation has the coordinates (−1,5) and (0,10) ,
[tex]y-5=\frac{10-5}{0+1}(x+1)[/tex][tex]y-5=\frac{5}{1}(x+1)[/tex][tex]y-5=5x+5[/tex][tex]y=5x+10[/tex]Hence , the linear equation is y=5x+10.
QRS and SRT are complementary. if m QRS (8x+10)° and m SRT=(8x)°Determine m QRSm QRS=
We are given two complementary angles (QRS and SRT).
Two angles are called complementry if they sum up to 90 degrees.
Therefore, by defination
mQRS + mSRT = 90
(8x+10) + (8x) = 90
8x + 8x + 10 = 90
16x = 90-10
16x = 80
x = 80/16
x = 5
Now, put the values in both the euqations and we will get the values of both the angles.
mQRS = 8x + 10 = 8(5) + 10 = 40 + 10 = 50 degrees
mSRT = 8x = 8(5) = 40 degrees
Write your answer as a whole number and remainder. 38 : 5 = R
Answer
7 remainder 3
Explanation
38 : 5 can be written as 38/5
And we know that that will give
(38/5) = 7 remainder 3
Hope this Helps!!!
In 1994, the moose population in a park was measured to be 4930. By 1999, the population was measured again to be 6180. If the population continues to change linearly: A.) Find a formula for the moose population, P, in terms of t, the years since 1990. P(t) B.) What does your model predict the moose population to be in 2006?
We define the following variables for our problem:
P = population of mooses
t = number of years since 1990
m = growth ratio
In terms of the variables that we defined above and the fact that the population of moose in terms of the year is linear, we have the following equation:
[tex]P(t)=m\cdot t+P_0[/tex]Now, we use the data of the problem:
1) In 1994 the moose population was 4930, so we have:
[tex]\begin{gathered} t=1994-1990=4 \\ P(4)=4930 \end{gathered}[/tex]2) In 1999 the moose population was 6180, so we have:
[tex]\begin{gathered} t=1999-1990=9 \\ P(9)=6180 \end{gathered}[/tex]Now, using the data above and the equation for P(t) we construct the following system of equations:
[tex]\begin{gathered} P(4)=m\cdot4+P_0=4930 \\ P(9)=m\cdot9+P_0=6180 \end{gathered}[/tex]We solve the system of equations.
First, we solve the equations for P0:
[tex]\begin{gathered} P_0=4930-4m \\ P_0=6180-9m \end{gathered}[/tex]Now, because the right-hand-side of both equations is equal to P0, we equal them and then we solve for the variable m:
[tex]\begin{gathered} 4930-4m=6180-9m \\ 9m-4m=6180-4930 \\ 5m=1250 \\ m=250 \end{gathered}[/tex]Finally, we replace the value of m in one of the equations of P0 and solve for it:
[tex]\begin{gathered} P_0=4930-4\cdot m \\ P_0=4930-4\cdot250 \\ P_0=3930 \end{gathered}[/tex]A) The formula for the moose population, P, in terms of t, the years since 1990 is:
[tex]P(t)=250t+3930[/tex]B) We want to know the value of the moose population in 2006.
First, we compute the value of t:
[tex]t=2006-1990=16[/tex]Now, we replace the value of t in the equation of P(t) above:
[tex]P(6)=250\cdot16+3930=7930[/tex]Answer: 7930
5. The quotient of a and b is negative. Decide if each statement about a and b is true or false.
(4 pts)
True False
a. The quotient b + a is positive.
b. The product ab is negative.
c. Either a or b must be negative.
d. The quotient -a + b is negative.
Deciding if each statement about a and b is true or false.
a)False
b)True
c)True
d)False
Given:
The quotient of a and b is negative.
a.
If quotient is negative means the dividend or divisor any one is negative so when the quotient b+a is always negative so given statement is false.
b.
The product ab is always negative because if the quotient is negative means either a or b is negative so ab is negative.
So given statement is true.
c.
Either a or b must be negative is true because if no element a or b is not negative we cannot produce a negative quotient hence either a or b must be negative.
d.
The quotient -a+b is negative is false because if the quotient is negative the values of a and b is one positive and one negative the positive number that is b greater than -a in that case the statement is false.
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what are the coordinates of the point on the directed line segment from (2,-1) to (9,6) that partitions the segment into a ratio of 5 to 2
Ok, so:
Let x and y be coordinate of the point C that partitions the segment.
And Let A = ( 2, -1 ) and B = ( 9 , 6 ).
So, given that C partitions the segment into a ratio of 5 to 2, we have:
Total parts of the segment: 5+2 = 7.
So, the point C is 5/7 of way from A to B.
Let me draw the situation:
Now, we know that the right distance is 7 and the upper distance is 7.
Now we multiply 5/7 per both distances.
5/7 * 7 = 5
5/7 * 7= 5
Now, we take the initial point A ( 2, -1 ), and sum 5 to each coordinate.
Then, the point C = ( 7 , 4 )
Formulas, walkthrough, something. I can't figure it out.
Answer:
[-16,16]U[20,+oo)
Step-by-step explanation:
1) x-16 ≥ 0
x ≥ 16
2) x-20 ≥ 0
x ≥ 20
3) x+16 ≥ 0
x ≥ -16
- 16. 16. 20
1) - - +. +
2) - - - +
3) - +. +. +
- +. - +
Solution : -16≤x≤ 16 ∨ x≥20
An orange has about I cup of juice. How many oranges are needed to make 2 cups of juice?
An orange has about 1/4 cup of juice. To find how many oranges are needed to make 2 1/2 cups of juice, we can use the next proportion
[tex]\frac{\frac{1}{4}\text{cup}}{2\frac{1}{2}\text{ cup}}=\frac{1\text{ orange}}{x\text{ oranges}}[/tex]Solving for x:
[tex]undefined[/tex]When the length of each edge of a cube is increased by 1 cm, the volume is increased by 19 cm3.A cube is shown.The length is labeled e.The width is labeled e.The height is labeled e.What is the length (in centimeters) of each edge of the original cube? cm
Solution
solution given;
let length be x
its
volume be x ³
we have when length is increased by 1cm volume increased by 19 cm³ so
(x+1)³=x³+19cm³
x³+1³+3x²+3x=x³+19
3x²+3x-18=0
3(x²+x-6)=0
x²+3x-2x-6=0
x(x+3)-2(x+3)=0
(x+3)(x-2)=0
either
x=-3 rejected
or
x=2cm
Hence the correct answer for the length of each cube = 2cm
1) Find the probability of rolling at least one 6 2)Find the probability of rolling exactly one 6
ANSWER:
For item 14: 1/36 (there is only one possible way of having both dice as 3)
For item 15: 1/18 (because there are 2 ways that one is 5 and one is 6 so it is 2/36 or 1/18)
For Item 16: 5/9 (because there 20 ways we can have a 5 or 6. 20/36)
For item 17: 11/36 (because there are 11 ways of having at least one six)
For item 18: 25/36 (because there are 25 ways having numbers other than 6)
For item 19: 5/18 (because there are 10 ways having exactly one six 10/36)
Review the following table verify that the calculations are correct. If there are errors note the day where the air exist and what the correct calculation should be.
Given
Number of patients = 6,663
- For Sunday
Patients = 1,187
Percent is:
[tex]\frac{1187}{6663}\times100\%=0.178\times100\%=17.8\%[/tex]- For Monday
Patients = 755
Percent:
[tex]\frac{755}{6663}\times100\%=11.3\%[/tex]Monday is correct.
- For Tuesday
Patients = 1,085
Percent:
[tex]\frac{1085}{6663}\times100\%=16.3\%[/tex]Tuesday is correct.
- For Wednesday
Patients = 1,031
Percent:
[tex]\frac{1031}{6663}\times100\%=15.5\%[/tex]Wednesday is correct.
- For Thursday
Patients = 1,024
Percent:
[tex]\frac{1024}{6663}\times100\%=15.4\%[/tex]- For Friday
Patients = 808
Percent:
[tex]\frac{808}{6663}\times100\%=12.1\%[/tex]Friday is correct.
- For Saturday
Patients = 773
Percent:
[tex]\frac{773}{6663}\times100\%=11.6\%[/tex]Saturday is correct.
Answer:
Errors
Sunday ---> 17.8%
Thursday ---> 15.4%
For the real-valued functions f(x) = 2x+10 and g(x) = x-1, find the composition f•g and specify its domain using interval notation.(f•g)(x)=Domain of f•g: (The 2x+10 is square rooted)
Okay, here we have this:
We need to meke the composition (f•g)(x), so in the function f we will replace x with the function g:
[tex]\begin{gathered} \mleft(f•g\mright)\mleft(x\mright)=\sqrt[]{2(x-1)+10} \\ =\sqrt[]{2x-2+10} \\ =\sqrt[]{2x+8} \end{gathered}[/tex]Now let's find the domain of (f•g)(x):
2X+8≥0
2X≥-8
X≥-4
Finally we obtain the following domain: [-4,∞)
Write the fraction as a decimal: 2/9
nearest tenth = 0.2
Explanation:2/9 is in its simplest form.
Hence, we use a calculator to find the fraction in decimal
2/9 = 0.2222 (a repeating decimal)
0.2222 to the nearest tenth = 0.2
Answer:
Step-by-step explanation:
2/9 as a decimal is 0.2222
4(b-1)= -4+4b what is the solution
Starting with the equation:
[tex]4(b-1)=-4+4b[/tex]Use the distributive property to expand the parenthesis in the left hand side of the equation:
[tex]4b-4=-4+4b[/tex]Use the commutative property to rewrite the right hand side of the equation, swapping the terms:
[tex]4b-4=4b-4[/tex]Since both sides of the equation are the same, any value of b is a solution to the equation.
Therefore, all numbers are a solution for the equation 4(b-1)=-4+4b.
What is nine increased by four and then doubled?
Answer:
Step-by-step explanation:
9+4=
13
13*2=
26
Answer:
Step-by-step explanation:
(9 multiplied by 4) multiplied by 2
= 72
Lori buys a $1500 certificate of deposit (CD) that earns 6% interest that compoundsmonthly. How much will the CD be worth after 10 years using simple interest?
Jimmy can jump 40 dogs in 5 hours, How many dogs can Jimmy Jump per hour?
To be able to determine how many dogs can Jimmy Jump per hour, let's divide how many dogs can Jimmy Jump by the number of hours he took to complete that certain number of jumps. Thus, we generate this equation,
[tex]\text{Jump rate of Jimmy= }\frac{No\text{. of dogs Jimmy jumped}}{No.\text{ of hours Jimmy took to jumped the no. dogs.}}[/tex]Given:
Jimmy jumped = 40 dogs
No. of hours Jimmy took to jump 40 dogs = 5 hours
We get,
[tex]\text{ Jump rate of Jimmy = }\frac{40\text{ dogs}}{5\text{ hours}}[/tex][tex]\text{ Jump per hour = 8 dogs per hour}[/tex]Therefore, Jimmy can jump 8 dogs per hour.