Area of a triangle
The area of a triangle of base length b and height h is:
[tex]A=\frac{b\cdot h}{2}[/tex]We are given the area as A=22 square inches and the base length b=4 inches. We are required to find the height.
Solving for h:
[tex]h=\frac{2\cdot A}{b}[/tex]Substituting:
[tex]h=\frac{2\cdot22in^2}{4in}=\frac{44in^2}{4in}=11in[/tex]The height is 11 inches
Can you please help me with this question D and E
Solution
Part d
T= -2ºC, V=15 km/hr
We can convert to F and mi/h like this:
T= 1.8*(-2)+ 32= 28.4ºF
Ms= As/0.62= 15 km/hr /0.62= 24.19 mi/hr
Replacing we have:
[tex]WC=35.74+0.6215\cdot(28.4)-35.75(24.19^{0.16})+0.4275(28.4)(24.19^{0.16})[/tex]Solving we got:
[tex]WC=35.74+17.651-59.519+20.213=14.08[/tex]Rounded to one decimal we got:
14.1
Part e
T= 10ºC = 1.8(10)+ 32=50ºF
Ms = 34 km/hr
Ms= As/0.62= 34 km/hr/ 0.62= 54.84 mi/hr
Replacing we got:
[tex]WC=35.74+0.6215\cdot(50)-35.75\cdot(54.84^{0.16})+0.4275(50)(54.84^{0.16})[/tex][tex]WC=35.74+31.075-67.847+40.566=39.53[/tex]Rounded to the nearest tenth we got:
39,5
Althought the reported temperature is 50 ºFahrenheit, because of wind it feels like 39.5 Fahrenheit
Dan paid three times as much as Greg for his dinner. (Translate the given situation into an equation. Pick the variable you use according to the context.
Let d represent Dan payment and let g represent Greg payment
"Dan paid three times as much as Greg for his dinner" can be represented as
d= 3g
Number 2
Let x represent the cost of Gym A and Y represent the cost of Gym B
Since Gym B cost twice as much as Gym A
Then y=2x
Number 3
Let d represent Diana's race and let c represent Calorine's race
Diana = 3 x Calorine
d=3c
If h(x)=-3x and g(x)=2x-1 what input value would make h(x)=12?
The question gives the function h(x) as
[tex]h(x)=-3x[/tex]To find the value that will make the function 12, we equate the function to 12, such that
[tex]-3x=12[/tex]Solving to get x, we have
[tex]\begin{gathered} x=\frac{12}{-3} \\ x=4 \end{gathered}[/tex]Therefore, h(x) = 12 when x = 4.
Which function is nonlinear?Which equation represents a nonlinear function? (Let me know if you can’t read the possible answers for the equation and I will send them)
Given:
There are 4 equation representations are given.
To find:
The nonlinear equation.
Explanation:
A)
[tex]\begin{gathered} 3x-2y=7 \\ i.e)3x-2y-7=0 \end{gathered}[/tex]Which is of the linear form,
[tex]ax+by+c=0[/tex]B)
[tex]y=\frac{2}{3}x+8[/tex]Which is of the linear form,
[tex]y=mx+c[/tex]C)
Let us consider the two points.
[tex](5,0),(6,2)[/tex]Using the two-point formula,
[tex]\begin{gathered} \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \\ \frac{y-0}{2-0}=\frac{x-5}{6-5} \\ \frac{y}{2}=\frac{x-5}{1} \\ y=2x-10 \end{gathered}[/tex]Substituting the first point to verify the linear equation,
[tex]\begin{gathered} -4=2(3)-10 \\ -4=-4 \end{gathered}[/tex]Therefore, the linear equation is satisfied for all 4 points.
Thus, options A, B, and C are linear equation.
So, the nonlinear equation must be given options D.
Final answer:
The correct option is D.
Section 1- Question 10Use the quadratic formula to solve the following equation:4m2-7m +2=57+√97A87+√17B87 + √1C87+√-63D8
The Solution:
Given:
[tex]4m^2-7m+2=5[/tex]We are required to use the Quadratic Formula to solve the equation above.
[tex]\begin{gathered} 4m^2-7m+2-5=0 \\ 4m^2-7m-3=0 \end{gathered}[/tex]The Quadratic Formula is:
[tex]m=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]In this case:
[tex]\begin{gathered} a=4 \\ b=-7 \\ c=-3 \end{gathered}[/tex]Substitute:
[tex]m=\frac{-(-7)\pm\sqrt{(-7)^2-4(4)(-3)}}{2(4)}[/tex][tex]m=\frac{7\pm\sqrt{49+48}}{8}=\frac{7\pm\sqrt{97}}{8}[/tex]Therefore, the correct answer is [option A]
What is the equation of the line shiwn graphed below
we are given the horizontal line. To determine the equation of this line let's remember that the equation of any horizontal line is of the form:
[tex]y=k[/tex]Where "k" is the point where the line touches the y-axis, in this case, the equation of the line is:
[tex]y=4[/tex]An arch is in the shape of a parabola. It has a span of 96 feet and a maximum height of 8 feet.Find the equation of the parabola. ______________Determine the distance from the center at which the height is 2 feet. ___________
We know the equation of a parabola can be written as
y = a(x-b1) (x-b2) where a is a constant and b1 and b2 are the zeros
y = a ( x - 48) ( x - -48)
y = a ( x-48) (x+48)
Now when x = 0 y = 8
8 = a ( 0-48) ( 0+48)
8 = a (-2304)
-8/2304 = a
-1/288 = a
y = -1/288 ( x-48) (x+48)
This is the equation for the parabola
Now let y = 2
2 = -1/288 ( x-48) (x+48)
Multiply each side by -288
-576 = (x-48)(x+48)
FOIL
-576 = x^2- 2304
ADD 2304 to each side
1728=x^2
Take the square root of each side
24 sqrt(3) = x
24 sqrt(3) ft
41.569 ft
Which of the following is not a step in solving the equation 3/x = 8/7 (1) Divide both sides of the equation by 8(2) Use cross products to write the equation 3.7 =8x(3) Divide both sides of the equation by 3(4) Rewrite 3.7 as 21
Let's solve the equation:
3/x = 8/7
8x = 3 * 7 Cross product
8x = 21 Rewrite 3 * 7 as 21
Dividing both sides by 8:
8x/8 = 21/8
x = 2.625
As you can see, we do not follow step 3. Divide both sides of the equation by 3.
Use the graph to respond to the questions. 6a. What is the height of the ball before its thrown? b. Approximately how high in the air did the ball go? C. NWAOOOOO INW Approximately long did it take the ball to reach the max height? d. How long was the ball in the air? 34(0.3) 2 e. (-0.2.0) -2 -1 (3.0) t 3 4 What does the point (3, 0) tell us about this situation? 1 INV
Given data:
The given graph is shown.
a)
The initial height of the ball is 3 units.
b)
The maximum height of the ball is approx 12.5 units.
c)
The time taken by the ball to reach maximum height is 1.5 units.
d)
The time ball was in air is 3 units.
e)
The point (3,0) denoted the coordinate of time is 3 when height is zero.
When simplified, |9+ (4-3) – 17| has a value of ______.
We have the following expression:
[tex]|9+(4-3)-17|[/tex]First, let's solve the operation inside the parenthesis.
[tex]|9+1-17|[/tex]Second, we add and subtract accordingly.
[tex]\lvert-7\rvert[/tex]Third, we apply the absolute value property.
[tex]\lvert-7\rvert=7[/tex]In conclusion, the values is 7
In the figure, LMN is congruent to QRS.What is the value of x and y?
If they are congruent, then;
3x + 5 = 2x + 10
collect like term
3x - 2x = 10 - 5
x = 5
Similarly,
y + 6 = 2y - 12
collect like term
2y - y = 6 + 12
y = 18
In right triangle XYZ, the hypotenuse measures 14 inches and one leg measures 10 inches. Find the length of the other leg. Show your work and write your answer in simplest radical form.
As per given by the question,
There are given that a triangle XYZ, and the hypotenuse of the triangle is 14 in., and one leg is 10 in.
Now,
First draw the triangle XYZ.
Now,
From the triangle XYZ, XZ is hypotenuse of the triangle, XY is the first leg of the triangle, and YZ is second leg of the triangle.
So,
First leg z is 10, then find the value of second leg x.
Then,
To find the value of second leg, here use pythagoras theorem.
So,
From the pythagoras theorem;
[tex]XY^2+YZ^{2^{}^{}}=XZ^2[/tex]Now,
[tex]\begin{gathered} XY^2+YZ^2=XZ^2 \\ 10^2+x^2=14^2 \\ 100+x^2=196 \\ x^2=196-100 \\ x^2=96 \end{gathered}[/tex]Then,
[tex]\begin{gathered} x^2=96 \\ x=\sqrt{96}^{} \\ x=9.79 \end{gathered}[/tex]Hence, the length of the other leg is 9.79.
The credit remaining on a phone card (in dollars) is a linear function of the total calling time made with the card (in minutes). The remaining credit after 34minutes of calls is $25.92, and the remaining credit after 53 minutes of calls is $23.64. What is the remaining credit after 71 minutes of calls?
The remaining credit after 71 minutes is $21.48
To solve this, we can calculate the equation of the linear function that represents the remaining credit
x is the time in minutes, y is the remaining credit. the we have two points of the line and we can calculate the equation
34 min and $25.92 => (34,25.92)
53 min and 23.64 => (53,23.64)
the formula to calculate the slope of a line is
[tex]m=\frac{y_2-y_1}{x_2-x_1_{}}=\frac{23.64-25.92}{53-34}=-0.12[/tex]now we know the slope, we can get the line euqation
[tex]y=m(x-x_1)+y_1\Rightarrow\text{ y=-0.12(x-34)+25.92=-0.12x+30}[/tex]Now for wathever x we ask, just plug it in to the equation and will give you the reamining credit.
for x=71 min:
[tex]y=-0.12(71)+30=-8.52+30=21.48[/tex]theremaining credit is $21.48.
I need help answering this question and the solution set
Let's first try to rewrite the expression to see if we can write it in a fomr more familiar:
[tex]\begin{gathered} x-\sqrt{10-3x}=0, \\ x=\sqrt{10-3x}, \\ x^2=10-3x, \\ x^2+3x-10=0. \end{gathered}[/tex]This is the same equation, but now we recognize a quadratic expression equal to 0. We can use the quadratic formula to find the solutions of this equation:
[tex]x_{1,2}=\frac{-3\pm\sqrt{3^2-4\cdot1\cdot(-10)}}{2\cdot1}[/tex]And solve:
[tex]\begin{gathered} x_{1,2}=\frac{-3\pm\sqrt{9+40}}{2}=\frac{-3\pm7}{2} \\ x_1=\frac{-3+7}{2}=\frac{4}{2}=2 \\ x_2=\frac{-3-7}{2}=\frac{-10}{2}=-5 \end{gathered}[/tex]Thus, the solution set of the equation is x = -5, x = 2
ans (alpha particles) at Figure 127 Exercises 12.6. complete the following: Find the intercepts and domain and perform the symmetry test (a) 9x^2 - 16y^2 = 144
We have that the general equation of the hyperbola is the following:
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]where 'a' and '-a' are the x-intercepts.
In this case, we have the following equation:
[tex]9x^2-16y^2=144[/tex]then, we have to divide both sides by 144 to get on the right side 1, thus, we have the following:
[tex]\begin{gathered} (9x^2-16y^2=144)(\frac{1}{144}) \\ \Rightarrow\frac{9}{144}x^2-\frac{16}{144}y^2=1 \\ \Rightarrow\frac{x^2}{16}-\frac{y^2}{9}=1 \end{gathered}[/tex]now, notice that we have that:
[tex]a^2=16[/tex]then, the x-intercepts are:
[tex]\begin{gathered} a^2=16 \\ \Rightarrow a=\pm\sqrt[]{16}=\pm4 \\ a_1=4 \\ \text{and} \\ a_2=-4 \end{gathered}[/tex]therefore, the x-intercepts are 4 and -4.
Now that we know that the intercepts are on those points, we can see that the hyperbola has the following graph:
we can see that the hyperbola is not defined between -4 and 4, therefore, the domain of the hyperbola is:
[tex](-\infty,-4\rbrack\cup\lbrack4,\infty)[/tex]finally, to perform the symmetry test, we have to check first both axis simmetries by changing x to -x and y to -y:
[tex]\begin{gathered} \frac{x^2}{16}-\frac{(-y)^2}{9}=1 \\ \Rightarrow\frac{x^2}{16}-\frac{y^2}{9}=1 \\ then \\ \frac{(-x)^2}{16}-\frac{y^2}{9}=1 \\ \Rightarrow\frac{x^2}{16}-\frac{y^2}{9}=1 \end{gathered}[/tex]since the hyperbola got symmetry about the x-axis and the y-axis, we have that the hyperbola got symmetry about the origin
solve the given system of equations2y=x+93x-6y=-15
Answer:
No solutions
Explanation:
The given system of equations is
2y = x + 9
3x - 6y = -15
To solve the system, we first need to solve the first equation for x, so
2y = x + 9
2y - 9 = x + 9 - 9
2y - 9 = x
Then, replace x = 2y - 9 on the second equation
3x - 6y = -15
3(2y - 9) - 6y = -15
3(2y) + 3(-9) - 6y = -15
6y - 27 - 6y = -15
-27 = -15
Since -27 is not equal to -15, we get that this system of equation doesn't have solutions.
Find the value of 4² + 6².
Answer:
Step-by-step explanation:
52
Answer: The correct answer is 52
Step-by-step explanation:
4² + 6²
4*4 + 6*6
16 + 36 = 52
I A bar of gold is 10 cm long,5 cm wide, and 7 cm high. What is it's volume?
Assuming the bar of gold has a rectangular form, its volume would be the product between all three dimensions.
[tex]A=10\operatorname{cm}\cdot5\operatorname{cm}\cdot7\operatorname{cm}=350\operatorname{cm}^3[/tex]Therefore, the volume is 350 cubic centimeters.
The circumference of the hub cap of a tire is 79.63 centimeters. Find the area of this hub cap. Us3.14 for a. Use pencil and paper. If the circumference of the hub cap were smaller, explain how twould change the area of the hub cap.The area of this hub cap is about square centimeters.(Round the final answer to the nearest whole number as needed. Round all intermediate values to the nearestthousandth as needed.)
Recall that the circumference of a circle is given by the following formula:
[tex]C=2\pi r,[/tex]where r is the radius of the circle.
We are given that:
[tex]79.63cm=2\pi r,[/tex]therefore:
[tex]r=\frac{79.63}{2*\pi}cm.[/tex]Now, the area of a circle is given by the following formula:
[tex]A=\pi r^2,[/tex]Therefore, the area of the hub cap is:
[tex]A=\pi *(\frac{79.63cm}{2})^2*\frac{1}{\pi^2}\approx505cm^2.[/tex]Answer:
[tex]\begin{equation*} 505cm^2. \end{equation*}[/tex]Given that the radius of the circle and the circumference are proportionally related, if the circumference is smaller then the radius is smaller. The area is proportionally related to the radius squared, therefore, a smaller circumference implies a smaller radius which implies a smaller area.
Joy has $68,020 in a savings account that earns 13% annually. The interest is not compounded. How much will she have in 5 years?
A = P + I
A is the new value
P is the initial value
I is the interest
I = PRT
R is the rate in decimal
T is the time
Joy has $68020
P = 68020
The account earns 13% annually
R = 13% = 13/100 = 0.13
The time is 5 years
T = 5
Let us find I, then A
I = 68020(0.13)(5)
I = 44213
Now let us find A
A = 68020 + 44213
A = $112233
She has $112233 in 5 years
I would like someone to help me so I can understand what to do. Pls anyone
we have that
2 cups in a pint
2 pints in a quart
4 quarts in a gallon
so
4 cups in 2 pints
4 cups in a quart
16 cups in 4 quarts
16 cups in a gallon
and
we have
3 parts red and 5 parts yellow
3 parts +5 parts=8 parts
8 parts=1 gallon
red paint fraction is 3/8 gallon
yellow paint fraction is 5/8 gallon
Remember that
I need 16 cups for a gallon
Apply proportion
16/1=x/(3/8)
x=(3/8)16
x=6 cups18. Light Bulbs: The mean lifespan of a standard 60 watt incandescent light bulb is 875 hours with a standard deviation of 80 hours. The mean lifespan of a standard 14 watt compact fluorescent light bulb (CFL) is 10,000 hours with a standard deviation of 1,500 hours. These two bulbs put out about the same amount of light. Assume the lifespan’s of both types of bulbs are normally distributed to answer the following questions. (a) I select one incandescent light bulb and put it in my barn. It seems to last forever and I estimate that it has lasted more than 2000 hours. What is the probability of selecting a random incandescent light bulb and having it last 2000 hours or more. Did something unusual happen here? (b) I select one CFL bulb and put it in the bathroom. It doesn’t seem to last very long and I estimate that it has lasted less than 5,000 hours. What is the probability of selecting a random CFL and having it last less than 5,000 hours. Did something unusual happen here? (c) Compare the the lifespan of the middle 99% of all incandescent and CFL light bulbs. (d) Is there much of a chance that I happen to buy an incandescent light bulb that lasts longer than a randomly selected CFL?
Set
[tex]\begin{gathered} \mu_1=875,\sigma_1=80 \\ \text{and} \\ \mu_2=10000,\sigma=1500 \end{gathered}[/tex]a) The Z-score formula is
[tex]Z=\frac{x-\mu}{\sigma}[/tex]Therefore, in our case, if x=2000
[tex]\Rightarrow Z=\frac{2000-875}{80}=\frac{1125}{80}=14.0625[/tex]Using a Z-score table,
[tex]\begin{gathered} P(z\ge2000)=1-P(z<2000)\approx1-1=0 \\ \Rightarrow P(z\ge2000)=0 \end{gathered}[/tex]A value of 2000 hrs is 14 standard deviations away from the mean. The probability is practically zero.
b) Similarly, set x=5000; then,
[tex]Z=\frac{5000-10000}{1500}=-\frac{5000}{1500}=-3.333\ldots[/tex]Thus, using a z-score table
[tex]P(z\le5000)=0.0004[/tex]The probability is 0.0004=0.04%. It is quite improbable but not an impossible event.
c) According to the empirical rule 99.8% of the data lies within 3 standard deviations; thus,
[tex]\begin{gathered} Incandescent \\ \mu_1\pm3\sigma_1=\lbrack875-240,875+240\rbrack=\lbrack635,1115\rbrack \\ \text{CFL} \\ \mu_2\pm3\sigma_2=\lbrack10000-4500,10000+4500\rbrack=\lbrack5500,14500\rbrack \end{gathered}[/tex]The lifespan of 99% of all incandescent bulbs is between 635 and 1115 hrs, whereas that of all CFL bulbs is between 5500 and 14500 hrs.
d) If we randomly select a CFL, the most probably lifespan is the mean of the distribution, in other words, 10000 hrs.
The probability of an incandescent bulb lasting 10000 hrs is
[tex]\begin{gathered} Z=\frac{10000-875}{80}=114.0625 \\ \Rightarrow P(z\ge10000)=1-P(z<10000)=0 \end{gathered}[/tex]The event is practically impossible.
Consider the following graph. Explain why it is possible to find an Euler circuit for this graph. Then, using Fleury's algorithm, identify and draw an Euler circuit starting with point A and choosing the first edge as A to F. Show all work using the required method. Draw the graph properly. (If you choose not to draw on my graph, then draw a LARGE graph on your paper.)
The Euler circuit for the graph using Fleury's algorithm is attached below.
The following steps are to be used to form the circuit.
Of the three edges coming from vertex "2" should I choose? We avoid choosing the edge "2-3" because it is a bridge and we cannot return to "3".
Any of the two remaining edges is ours to choose. Let's say we choose "2-0." By eliminating this edge, we can travel to vertex "0."
We pick the one edge from vertex "0" that is present, remove it, and then move to vertex "1." The Euler tour changes to "2-0 0-1."
We take the lone edge from vertex "1," delete it, and then move to vertex "2." The Euler tour is now "2-0 0-1 1-2."
Once more, vertex 2 has just one edge, so we pick it, remove it, and continue on to vertex 3.
Hence we get the required circuit.
The term "Eulerian graph" has two different meanings in graph theory. One interpretation applies to a graph with an Eulerian circuit, whereas the other applies to a graph where each vertex has an even degree. These definitions apply to linked graphs.
To learn more about Euler circuit visit:
https://brainly.com/question/22089241
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What is the parallel slope to this equation? f(x)=-5x + 6
Remember the equation of a line with slope m and y-intercept b:
[tex]y=mx+b[/tex]Which can be expressed in terms of functions by replacing y=f(x)
[tex]f(x)=mx+b[/tex]Any line with the same slope will be parallel to that equation.
By comparing with:
[tex]f(x)=-5x+6[/tex]We can know that its slope is equal to -5.
Since the "parallel slope" is just the slope, then
Solve 9abs(3n-2) + 6 > 51 and graph on the number line
Solving the inequality,
[tex]\begin{gathered} 9\lvert3n-2\rvert+6>51 \\ \rightarrow9\lvert3n-2\rvert>45 \\ \rightarrow\lvert3n-2\rvert>5 \end{gathered}[/tex]Remember that if
[tex]\lvert u\rvert>a,a>0\Rightarrow u<-a\text{ or }u>a[/tex]This way, we get:
[tex]\begin{gathered} 3n-2<-5\rightarrow3n<-3\rightarrow n<-1 \\ 3n-2>5\rightarrow3n>7\rightarrow n>\frac{7}{3} \end{gathered}[/tex]As an interval, we get:
[tex]\: \mleft(-\infty\: ,\: -1\mright)\cup\mleft(\frac{7}{3},\: \infty\: \mright)[/tex]In the number line, we get:
whats the constant proportionality of the table. x 30 24 9 y 10 8 3
Let:
y = kx
Where:
k = constant proportionality of the table
If x = 30, y = 10:
10 = k30
Solving for k:
k = 10/30 = 1/3
Verify the answer:
If x = 24, y = 8:
y = kx = 1/3*24 =24/3 = 8
Therefore, the constant proportionality of the table is 1/3
a. Make a scatter plot of the data in the table below.b. Does it appear that a linear model or an exponential model is the better fit for the data?
a) Option C
b) A linear model
Explanation:To determine the correct plot from the options, we trace the x and y values given
Option A:
When x = 5, y = 1
when x = 6.6, y = 2
We see this is exact opposite of the values in the given table
Option B:
when x = 5, y = 5
when x = 6.6, y = 6.6
This is also different from the given values in the table
Option C:
when x = 1, y = 5
when x = 2, y = 6.6
when x = 3, y = 8.4
This is the same as the given values in the table.
Hence, the correct scatter plot is option C
We need to check if the points are linear:
[tex]\begin{gathered} u\sin g\text{ any two points on the table,} \\ \text{rate = }\frac{6.6\text{ - 5}}{2-1}\text{ = 1.6/1 = 1.6} \\ \text{rate = }\frac{8.4\text{ - 6.6}}{3-2}\text{ = 1.8/1 = 1.8} \\ \text{rate = }\frac{10.2\text{ - 8.4}}{3\text{ - 2}}\text{ = 1.8/1 = 1.8} \\ \text{rate = }\frac{12-10.2}{4-3}\text{ = 1.2/1 = 1.2} \end{gathered}[/tex]A linear model fit the data
Find the midpoint and distance on points (5,5) and (-1,3)
Given:
Two points (5, 5) and (-1, 3).
To find the midpoint and distance:
Using the midpoint formula,
[tex]\begin{gathered} m=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ =(\frac{5-1}{2},\frac{5+3}{2}) \\ =(\frac{4}{2},\frac{8}{2}) \\ =(2,4) \end{gathered}[/tex]Thus, the mid-point formula is (2, 4).
Using the distance formula,
[tex]\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2_{}} \\ =\sqrt[]{(-1-5)^2+(3-5)^2} \\ =\sqrt[]{(-6)^2+(-2)^2} \\ =\sqrt[]{36+4} \\ =\sqrt[]{40} \\ =2\sqrt[]{10}\text{ units} \end{gathered}[/tex]Thus, the distance between the two points is,
[tex]2\sqrt[]{10}units[/tex]Elisa's school is selling tickets to a choral performance. On the first day of ticket sales the school sold 8 senior citizen tickets and 5 child tickets for a total of $94. The school took in $152 on the second day by selling 4 senior citizen tickets and 10 child tickets. What is the price each of one senior citizen ticket and one child ticket?
Let's define the next variables:
x: price of one senior ticket
y: the price of one child ticket
On the first day of ticket sales, the school sold 8 senior citizen tickets and 5 child tickets for a total of $94. That is:
8x + 5y = 94 (eq. 1)
The school took in $152 on the second day by selling 4 senior citizen tickets and 10 child tickets. That is:
4x + 10y = 152 (eq. 2)
Multiplying equation 2 by 2, we get:
2(4x + 10y) = 2*152
8x + 20y = 304 (eq. 3)
Subtracting equation 1 to equation 3, we get:
8x + 20y = 304
-
8x + 5y = 94
------------------------
15y = 210
y = 210/15
y = 14
Substituting this result into equation 1:
8x + 5(14) = 94
8x + 70 = 94
8x = 94 - 70
8x = 24
x = 24/8
x = 3
Each senior citizen ticket cost $3 and each child ticket cost $14
Last week you worked 28 hours, and earned $266. What is your hourly pay rate? $ per hour Give your answer in dollars and cents (like 5.50)
1) Gathering the data
Last week
28 worked hours = $266
What is your hourly pay rate?
2) To find out that, let's set a proportion
Worked hours $
28------------------------ 266
1--------------------------- x
Let's use the fundamental property of a proportion that allows us to cross multiply the ratios of that proportion
So the hourly pay rate is approximately $11.08 per hour.