Let d represents dimes and q represents quarter.
Note that a dime is 10 cent, which is same as one over ten, and a quarter is one over four
[tex]\begin{gathered} d=\frac{1}{10}=0.1 \\ q=\frac{1}{4}=0.25 \end{gathered}[/tex]Given that a collection of dimes abs quarters has a value of $1.35, then this can be represented as below:
[tex]0.1d+0.25q=1.35[/tex]Multiply through by 100 to get
[tex]\begin{gathered} 100\times0.1d+100\times0.25q=100\times1.35 \\ 10d+25q=135 \end{gathered}[/tex]To get the possible combinations of dimes and quarters, lets the try different values of that will satisfy the equation.
When q is 1,
[tex]\begin{gathered} 10d+25q=135 \\ q=1 \\ 10d+25(1)=135 \\ 10d+25=135 \\ 10d=135-25 \\ 10d=110 \\ d=\frac{110}{10}=11 \end{gathered}[/tex]Therefore, 11 dimes and 1 quarter abs is a possible combination
When q is 3
[tex]\begin{gathered} 10d+25(3)=135 \\ 10d+75=135 \\ 10d=135-75 \\ 10d=60 \\ d=\frac{60}{10} \\ d=6 \end{gathered}[/tex]Also, 6 dimes and 3 quarter abs is a possible combination
When q is 5
[tex]\begin{gathered} 10d+25(5)=135 \\ 10d+125=135 \\ 10d=135-125 \\ 10d=10 \\ d=\frac{10}{10} \\ d=1 \end{gathered}[/tex]Also, 1 dime and 5 quarter abs is a possible combination
When q is 7
[tex]\begin{gathered} 10d+25(7)=135 \\ 10d+175=135 \\ 10d=135-175 \\ 10d=-40 \\ d=\frac{-40}{10}=-4 \end{gathered}[/tex]Since negative answer was gotten for dimes, 7 quater wouldn't give any possible combination.
Hence, there are It can be found that there are there are three possible combinations, these are:
11 dimes and 1 quarter abs
6 dimes and 3 quarter abs
1 dime and 5 quarter abs
Consider the following statement:
If Paul is older than Bill and Fred is younger than Bill, then Bill's age is between Paul's and Fred's.
Write the Given statement
Paul is the oldest and Fred is the youngest of the three.
What is mean by younger?Younger is a comparative adjective that generally indicates more youthful.
Similar to old, elder simply indicates older in age. It is a comparative version of old.
Given that x is a natural number, let Bill's age equal x years.
Paul's age is thus calculated as (x + a) Years, where an is any positive integer.
Fred is also younger than Bill.
So, Fred's age is equal to x - k, where k is any positive integer.
As a result, if we arranged Fred, Bill, and Paul's ages, they would be
Bill, Fred, and Paul
x-k < x < x+a
As a result, we can conclude that Paul is the oldest and Fred is the youngest of the three.
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a group orders three large veggie pizzas each slice represents eighth of an entire Pizza the group eats 3/4 of a piece of how many slices of pizza are left
1 pizza contain 8 slices, so we can state a rule of three as:
[tex]\begin{gathered} 1\text{ Pizza ------ 8 slices} \\ \frac{3}{4}\text{ pizza ------ x} \end{gathered}[/tex]then, x is given by
[tex]x=\frac{(\frac{3}{4})(8)}{1}\text{ slices}[/tex]which gives
[tex]\begin{gathered} x=\frac{3}{4}\times8 \\ x=\frac{3\times8}{4} \\ x=3\times2 \\ x=6\text{ slices} \end{gathered}[/tex]that is, 3/4 of pizza is equivalent to 6 slices. So, there are 8 - 6 = 2 slices left of one pizza.
However, they bought 3 large pizzas and ate almost one of them. So, there are 2x8 = 16slices plus 2 slices, that is, 18 slices are left.
Write an equation of variation to represent the situation and solve for the missing information The time needed to travel a certain distance varies inversely with the rate of speed. If ittakes 8 hours to travel a certain distance at 36 miles per hour, how long will it take to travelthe same distance at 60 miles per hour?
The time needed to travel a certain distance varies inversely with the rate of speed, so:
[tex]\begin{gathered} let\colon \\ t=\text{time} \\ v=\text{rate of speed} \\ t\propto\frac{1}{v} \end{gathered}[/tex]8hours----------------------------->36mi/h
xhours----------------------------->60mi/h
[tex]\begin{gathered} \frac{8}{x}=\frac{36}{60} \\ \text{ Since the it varies inversely:} \\ \frac{8}{x}=(\frac{36}{60})^{-1} \\ \frac{8}{x}=\frac{5}{3} \\ \text{solve for x:} \\ x=\frac{3\cdot8}{5} \\ x=4.8h \end{gathered}[/tex]4.8 hours or 4 hours and 48 minutes
4 Use the sequence below to complete each task. -6, 1, 8, 15, ... a. Identify the common difference (a). b. Write an equation to represent the sequence. C. Find the 12th term (a) our Wilson (All Things Algebral. 2011 Enter your answer(s) here
we have
-6, 1, 8, 15, ...
so
a1=-6
a2=1
a3=8
a4=15
a2-a1=1-(-6)=7
a3-a2=8-1=7
a4-a3=15-8=7
so
the common difference is
d=7
Part 2
write an equation
we have that
The equation of a general aritmetic sequence is equal to
an=a1+(n-1)d
we have
d=7
a1=-6
substitute
an=-6+(n-1)7
an=-6+7n-7
an=7n-13
Part 3
Find 12th term
we have
n=12
a12=-7(12)-13
a12=71
Please answer last oneTo graph F using a graphing utility…Either A,B,C, or DLet me know which option
We have to graph the function F(x) defined as:
[tex]F(x)=\frac{x^2-11x-12}{x+6}[/tex]We can graph it as:
To see the complete graph we have to show the horizontal axis from x = -30 to x = 30 and the vertical axis from y = -80 to y = 80.
Answer: Option B
A line passes through the point (-2,-7) and has a slope of 4
Answer:
y= 4x +1
Step-by-step explanation:
The equation of a line, in slope-intercept form, is given by y= mx +c, where m is the slope and c is the y-intercept.
Given that the slope is 4, m= 4.
Substitute m= 4 into y= mx +c:
y= 4x +c
To find the value of c, substitute a pair of coordinates the line passes through.
When x= -2, y= -7,
-7= 4(-2) +c
-7= -8 +c
c= -7 +8
c= 1
Substitute the value of c into the equation:
Thus, the equation of the line is y= 4x +1.
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A) 14x + 7y > 21 B) 14x + 7y < 21 C) 14x + 7y 5 21 D) 14x + 7y 221match with graph
As all the options are the same equation
so, we need to know the type of the sign of the inequality
As shown in the graph
The line is shaded so, the sign is < or >
The shaded area which is the solution of the inequaity is below the line
So, the sign is <
So, the answer is option B) 14x + 7y < 21
What is the length of the dotted line in the diagram below? Round to the
nearest tenth.
Answer:
12.1 units
Step-by-step explanation:
what is 12 + 0.2 + 0.006 as a decimal and word form
twelve and two hundred six thousandths
The sum of 19 and twice a number
Answer:
2x+19
Step-by-step explanation:
Let x be the number
2x+19
17. 19yd. 28in.- 16yd. 31in.18. 61wk. 4da.- 18wk. 6da.21. 8tbsp. 2tsp. * 15
We need to solve the next expressions:
17. 19yd. 28in.- 16yd. 31in
We need to solve subtract each expression.
Then:
19yd. 28in.- 16yd. 31in =
19yd - 16yd and 28in-31in
3yd -3in
Then, we have the next equivalent.
1 yard = 36 in
So:
36 in - 3 in = 33 in
Therefore
19yd. 28in.- 16yd. 31in = 2 yard 33
18 61wk. 4da.- 18wk. 6da.
We need to subtract both expression:
Then
61wk - 18wk = 43kw
4da-6da = -2da
Where 1 week = 7 days
Then
7 da - 2da = 5 da
Hence, 43kw -1 wk = 42 wk.
The result is:
42 wk 5 da
21. 8tbsp. 2tsp. * 15
We need to convert 2ts into tbsp and then multiply the result by 15.
If
1 tsp ------- 0.333tbsp
Then
2tp ------ 2(0.333tbsp)= 0.66666 tbsp
Now
(8tbsp + 0.6666 ) * 15 = 130 tbsp
Percents build on one another in strange ways. It would seem that if you increased a number by 5% and thenincreased its result by 5% more, the overall increase would be 10%.7. Let's do exactly this with the easiest number to handle in percents.(a) Increase 100 by 5%(b) Increase your result form (a) by 5%.(C) What was the overall percent increase of the number 100? Why is it not 10%?
Answer:
a) 105
b) 110.25
c) Increase of 10.25%. It is not 100% because the second increase of 5% is over the first increased value, not over the initial value.
Step-by-step explanation:
Increase and multipliers:
Suppose we have a value of a, and want a increse of x%. The multiplier of a increase of x% is given by 1 + (x/100). So the increased value is (1 + (x/100))a.
(a) Increase 100 by 5%
The multiplier is 1 + (5/100) = 1 + 0.05 = 1.05
1.05*100 = 105
(b) Increase your result form (a) by 5%.
1.05*105 = 110.25
(C) What was the overall percent increase of the number 100? Why is it not 10%?
110.25/100 = 1.1025
1.1025 - 1 = 0.1025
Increase of 10.25%. It is not 100% because the second increase of 5% is over the first increased value, not over the initial value.
solve for x. then find the missing piece(s) of the parallelogram for #7
Let us find the angles of the parallelogram below
[tex]\begin{gathered} 2x+30 \\ x=40 \\ 2(40)+30 \\ 80+30 \\ 110^0 \end{gathered}[/tex][tex]\begin{gathered} 2x-10 \\ 2(40)-10 \\ 80-10 \\ 70^0 \end{gathered}[/tex]Theorem+: opposite angles of a parallelogram are the same
Hence the angles of the parallelogram are 110, 70, 110, and 70
A. The measure of the angle can not be determined B. 70 degreesC. 110 degreesD. 180 degrees
Okay, here we have this:
Considering the provided graph, we are going to find the measure of the angle "3", so we obtain the following:
Since angle 3 and 4 form a straight angle, that is to say that these two angles are supplementary, then we have:
[tex]\begin{gathered} m\angle3+m\angle4=180 \\ m\angle3+70=180 \\ m\angle3=180-70 \\ m\angle3=110\text{ degre}es \end{gathered}[/tex]Finally we obtain that the correct answer is the option C.
factoring out: 25m + 10
Answer:
5(5m + 2)
Explanation:
To factor out the expression, we first need to find the greatest common factor between 25m and 10, so the factors if these terms are:
25m: 1, 5, m, 5m, 25m
10: 1, 2, 5, 10
Then, the common factors are 1 and 5. So, the greatest common factor is 5.
Now, we need to divide each term by the greatest common factor 5 as:
25m/5 = 5m
10/5 = 2
So, the factorization of the expression is:
25m + 10 = 5(5m + 2)
The beginning mean weekly wage in a certain industry is $789.35. If the mean weekly wage grows by 5.125%, what is the new mean annual wage? (1 point)O $829.80O $1,659.60O $41,046.20$43,149.82
Given:
The initial mean weekly wage is $ 789.35.
The growth rate is 5.125 %.
Aim:
We need to find a new annual wage.
Explanation:
Consider the equation
[tex]A=PT(1+R)[/tex]Let A be the new annual wage.
Here R is the growth rate and P is the initial mean weekly wage and T is the number of weeks in a year.
The number of weeks in a year = 52 weeks.
Substitute P=789.35 , R =5.125 % =0.05125 and T =52 in the equation.
[tex]A=789.35\times52(1+0.05125)[/tex][tex]A=43149.817[/tex][tex]A=43149.82[/tex]The new mean annual wage is $ 43,149.82.
Final answer:
The new mean annual wage is $ 43,149.82.
Find the domain of the function f(x)=√100x²
The domain of the function √(100x²) will be (-∞,∞) as the definition of domain states that the set of inputs that a function will accept is known as the domain of the function in mathematics.
What is domain?The range of values that we are permitted to enter into our function is known as the domain of a function. The x values for a function like f make up this set (x). A function's range is the set of values it can take as input.
What is function?A function in mathematics from a set X to a set Y assigns exactly one element of Y to each element of X. The sets X and Y are collectively referred to as the function's domain and codomain, respectively.
Here,
The function is √(100x²).
The domain would be (-∞,∞).
The set of inputs that a function will accept is known as the domain of the function in mathematics, and the domain of the function √(100x²) will be (-∞,∞), according to the definition of domain.
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1. flight 1007 will hold 300 passengers the airline has booked 84% of the plane already. How many seats are open for the last-minute Travelers?2.Carly interviewed students to ask their favorite kind of television programs. 12 students claimed that they preferred comedies, 18 like drama 13 enjoy documentaries, and 7 voted for news programs what percentage of the students selected comedies?
The total passengers in the flight is P=300.
Determine the passenger who booked the seat in the flight.
[tex]\begin{gathered} Q=\frac{84}{100}\cdot300 \\ =252 \end{gathered}[/tex]The number of seats booked by passenger is 252.
Determine the seats available for last-minutes travelers.
[tex]\begin{gathered} S=300-252 \\ =48 \end{gathered}[/tex]So 48 seats available for the last minute travelers.
Identify the vertex of the function below.f(x) - 4= (x + 1)2-onSelect one:O a. (-4,1)O b.(1,-4)O c. (-1,-4)O d.(-1,4)
The standard equation of a vertex is given by:
[tex]f(x)=a(x-h)^2+k[/tex]where (h,k) is the vertex.
Comparing with the given equation after re-arranging:
[tex]f(x)=(x+1)^2+4[/tex]The vertex of the function is (-1, 4)
Add the rational expression as indicated be sure to express your answer in simplest form. By inspection, the least common denominator of the given factor is
Notice that the least common denominator is 9*2=18, therefore:
[tex]\begin{gathered} \frac{x-3}{9}+\frac{x+7}{2}=\frac{2(x-3)}{9\cdot2}+\frac{9(x+7)}{9\cdot2}, \\ \frac{x-3}{9}+\frac{x+7}{2}=\frac{2x-6}{18}+\frac{9x+63}{18}, \\ \frac{x-3}{9}+\frac{x+7}{2}=\frac{2x-6+9x+63}{18}, \\ \frac{x-3}{9}+\frac{x+7}{2}=\frac{11x+57}{18}\text{.} \end{gathered}[/tex]Answer:
[tex]\frac{x-3}{9}+\frac{x+7}{2}=\frac{11x+57}{18}\text{.}[/tex]The half-life of radium is 1690 years. If 70 grams are present now, how much will be present in 570 years?
Solution
Given that
Half life is 1690 years.
Let A(t) = amount remaining in t years
[tex]\begin{gathered} A(t)=A_0e^{kt} \\ \\ \text{ where }A_{0\text{ }}\text{ is the initial amount} \\ \\ k\text{ is a constant to be determined.} \\ \end{gathered}[/tex]SInce A(1690) = (1/2)A0 and A0 = 70
[tex]\begin{gathered} \Rightarrow35=70e^{1690k} \\ \\ \Rightarrow\frac{1}{2}=e^{1690k} \\ \\ \Rightarrow\ln(\frac{1}{2})=1690k \\ \\ \Rightarrow k=\frac{\ln(\frac{1}{2})}{1690} \\ \\ \Rightarrow k=-0.0004 \end{gathered}[/tex]So,
[tex]A(t)=70e^{-0.0004t}[/tex][tex]\Rightarrow A(570)=70e^{-0.0004(570)}\approx55.407\text{ g}[/tex]Therefore, the answer is 55.407 g
Point P is in the interior of
∵ m< OZQ = m[tex]\because m\angle OZP=62[/tex]Substitute the measures of the given angles in the equation above
[tex]\therefore125=62+m\angle PZQ[/tex]Subtract 62 from both sides
[tex]\begin{gathered} \therefore125-62=62-62+m\angle PZQ \\ \therefore63=m\angle PZQ \end{gathered}[/tex]The measure of angle PZQ is 63 degrees
(9 •10^9)•(2•10)^-3)
First, let's distribute the exponent -3 for 2 and ten, like this:
[tex]\begin{gathered} 9\times10^9\times(2\times10)^{-3}^{} \\ 9\times10^9\times2^{-3}\times10^{-3} \end{gathered}[/tex]Now, we can apply the next property when we have a number raised to a negative power:
[tex]a^{-b}=\frac{1}{a^b}[/tex]Then:
[tex]\begin{gathered} 9\times10^9\times2^{-3}\times\frac{1}{10^3} \\ 9\times2^{-3}\times\frac{10^9}{10^3} \end{gathered}[/tex]And when we have a division of the same number raised to different powers we can apply:
[tex]\frac{a^b}{a^c}=a^{b-c}[/tex]then:
[tex]\begin{gathered} 9\times2^{-3}\times\frac{10^9}{10^3} \\ 9\times2^{-3}\times10^{9-3} \\ 9\times2^{-3}\times10^6 \\ 9\times\frac{1}{2^3}^{}\times10^6 \end{gathered}[/tex]Now, as we know, having 10 raised to 6 means that we are multiplying ten by ten 6 times, when we do this we get:
[tex]10\times10\times10\times10\times10\times10=1000000[/tex]And with 2 raised to three we get:
[tex]2\times2\times2=8[/tex]Then we have:
[tex]\begin{gathered} 9\times\frac{1}{8^{}}^{}\times1000000 \\ \frac{9\times1000000}{8^{}}^{} \\ \frac{9000000}{8^{}}^{} \\ \frac{4500000}{4}^{}=11250000 \end{gathered}[/tex]Drag the curve to the correct location on the graph. A small museum opened in 2004 with 80 pieces of art in its collection. In 2005, the number of pieces of art in the collection was 1.2 times the initial number of pieces of art. Then, in 2006, the number of pieces of art in the collection was 1.2 times the size of the previous year's collection. The directors of the museum plan to continue acquiring pieces of art at the same rate. Complete the graph of this relationship. 1207 ces of Art f(x) 3001 240 180 120
The most appropriate choice for geometric series will be given by
The equation is [tex]y = 80 (1.2)^x[/tex] and it is a geometric function.
What is geometric series ?
The series in which the ratio between consecutive terms of the series are same is called geometric series
For example, 2, 4, 8, 16, ... is a geometric series in which ratio between two consecutive terms is 2.
Number of pieces of art in 2004 = 80
Number of pieces of art in 2005 = [tex]80 \times 1.2[/tex]
Number of pieces of art in 2006 = [tex]80\times (1.2)^2[/tex]
Number of pieces of art after x years from 2004 = [tex]80 \times (1.2)^x[/tex]
The equation is [tex]y = 80 (1.2)^x[/tex]
This is a geometric function
The graph has been attached here.
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12 + 24 =__(__+__)
Find the GCF. The first distributing number should be your GCF
A group of numbers' greatest common factor (GCF) is the biggest factor that all the numbers have in common. For instance, the numbers 12, 20, and 24 share the components 2 and 4.
Therefore, 12 and 24 have the most things in common. Figure 2: LCM = 24 and GCF = 12 for two numbers.
Find the other number if one is 12, then. What does 12 and 24's GCF stand for?
Example of an image for 12 + 24 = ( + ) Locate the GCF. You should distribute your GCF as the first number.
12 is the GCF of 12 and 24. We must factor each number individually in order to determine the highest common factor of 12 and 24 (factors of 12 = 1, 2, 3, 4, 6, 12; factors of 24 = 1, 2, 3,.
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2. Luis hizo una excursión de 20 km 75 hm 75 dam 250 m en tres etapas. En la primera recorrió 5 km 5 hm, y en la segunda 1 km 50 dam más que en la anterior. ¿Cuánto recorrió en la tercera etapa? Expresa el resultado de forma compleja
1 4/5 + (2 3/20 + 3/5) use mental math and properties to solve write your answer in simpleist form
Given data:
The given expression is 1 4/5 + (2 3/20 + 3/5).
The given expression can be written as,
[tex]\begin{gathered} 1\frac{4}{5}+(2\frac{3}{20}+\frac{3}{5}_{})=\frac{9}{5}+(\frac{43}{20}+\frac{3}{5}) \\ =\frac{9}{5}+\frac{43+12}{20} \\ =\frac{9}{5}+\frac{55}{20} \\ =\frac{36+55}{20} \\ =\frac{91}{20} \end{gathered}[/tex]Thus, the value of the given expression is 91/20.
What is the solution to the following equation?x^2+3x−7=0
Answer:
Explanation:
Given the equation:
[tex]x^2+3x-7=0[/tex]On observation, the equation cannot be factorized, so we make use of the quadratic formula.
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]Comparing with the form ax²+bx+c=0: a=1, b=3, c=-7
Substitute these values into the formula.
[tex]x=\dfrac{-3\pm\sqrt[]{3^2-4(1)(-7)}}{2\times1}[/tex]We then simplify and solve for x.
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Find the distance between the parallel lines. If necessary, round your answer to the nearest tenths.
The distance between the parallel lines is [tex]\frac{3}{5}}[/tex].
The given parallel lines are
[tex]y= $-$3x+4\\y= $-$3x+1[/tex]
We have to find the distance between the given parallel lines.
The formula is used to solve the distance between two parallel lines [tex]ax+by+c_{1}=0[/tex] and [tex]ax+by+c_{2}=0[/tex] is
[tex]d=|c_{2} $-$c_{1}|\frac{1}{\sqrt{a^{2}+b^{2}}}[/tex]
The first given line is [tex]y= $-$3x+4[/tex]
We can write that line as [tex]3x$-$y $-$4=0[/tex]
The second given line is [tex]y= $-$3x+1[/tex]
We can write that line as [tex]3x$-$y $-$1=0[/tex]
Comparing the both given parallel lines with the standard equation of line.
After comparing we get
[tex]a=3, b= $-$1, c_{1}= $-$4, c_{2}= $-$1[/tex]
Putting the value in the formula
[tex]d=|(-1) -(-4)|\frac{1}{\sqrt{(3)^{2}+(-4)^{2}}}\\d=|-1+4|\frac{1}{\sqrt{9+16}}\\d=|3|\frac{1}{\sqrt{25}}\\d=\frac{3}{5}}[/tex]
Hence, the distance between the parallel lines is [tex]\frac{3}{5}}[/tex].
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The number of algae in a tub in a labratory increases by 10% each hour. The initial population, i.e. the population at t = 0, is 500 algae.(a) Determine a function f(t), which describes the number of algae at a given time t, t in hours.(b) What is the population at t = 2 hours?(c) What is the population at t = 4 hours?
a) Let's say initial population is po and p = p(t) is the function that describes that population at time t. If it increases 10% each hour then we can write:
t = 0
p = po
t = 1
p = po + 0.1 . po
p = (1.1)¹ . po
t = 2
p = 1.1 . (1.1 . po)
p = (1.1)² . po
t = 3
p = (1.1)³ . po
and so on
So it has an exponential growth and we can write the function as follows:
p(t) = po . (1.1)^t
p(t) = 500 . (1.1)^t
Answer: p(t) = 500 . (1.1)^t
b)
We want the population for t = 2 hours, then:
p(t) = 500 . (1.1)^t
p(2) = 500 . (1.1)^2
p(2) = 500 . (1.21)
p(2) = 605
Answer: the population at t = 2 hours is 605 algae.
c)
Let's plug t = 4 in our function again:
p(t) = 500 . (1.1)^t
p(4) = 500 . (1.1)^4
p(4) = 500 . (1.1)² . (1.1)²
p(4) = 500 . (1.21) . (1.21)
p(4) = 500 . (1.21)²
p(4) = 732.05
Answer: the population at t = 4 hours is 732 algae.