The ratio wins to losses of each team are:
[tex]\begin{gathered} \text{Leopards:}\frac{15}{10}=\frac{3}{2} \\ \text{ Pirates:}\frac{12}{8}=\frac{3}{2} \\ \text{Knights:}\frac{14}{7}=2 \\ \text{Lions:}\frac{18}{10}=\frac{9}{5} \end{gathered}[/tex]Then, Leopards and Pirates have the same ratio
a tree that is 20 feet tall casts a shadow 30 feet long. a girl standing next to the tree has a shadow 9 feet long. how tall is the girl?
The shadow of the three is proportional to the shadow of the girl, then we can use a rule of three:
if x is the height of the girl: 20 is to 30 is the same as x is to 9
This can be written as:
[tex]\frac{20}{30}=\frac{x}{9}\Longrightarrow x=\frac{9\cdot20}{30}=\text{ 6}[/tex]Answer:
The girl is 6 feet tall
Precalculus:Consider the right triangle where a = 3mi and alpha = 45 degrees Find an approximate value (accurate up to three or more decimals) of each of the following. Give the angle in degrees.
Given:
[tex]\begin{gathered} a=3mi \\ \alpha=45\degree \end{gathered}[/tex]Required:
To find the value of beta, b and c.
Explanation:
The given triangle is right triangle.
Therefore,
[tex]\begin{gathered} \sin\alpha=\frac{a}{c} \\ \\ \sin45=\frac{3}{c} \\ \\ 0.7071=\frac{3}{c} \\ \\ c=\frac{3}{0.7071} \\ \\ c=4.2426mi \end{gathered}[/tex]Now
[tex]\begin{gathered} \tan\alpha=\frac{a}{b} \\ \\ \tan45=\frac{3}{b} \\ \\ 1=\frac{3}{b} \\ \\ b=3 \end{gathered}[/tex]The sum of the angle in triangle is 180 degree.
Therefore
[tex]\begin{gathered} 90+45+\beta=180 \\ \beta=180-90-45 \\ \beta=45\degree \end{gathered}[/tex]Final Answer:
[tex]\begin{gathered} b=3mi \\ c=4.2426mi \\ \beta=45\degree \end{gathered}[/tex]How is a number related to its opposite?
The opposite of any given number is simply that number multiplied by negative 1. For instance, the opposite of 15 is 15 x (-1) and that is -15.
Therefore, a number is always a negative of its opposite
What is the real part of 4 – 5i? 54-5i-5
Every imaginary number have the following form:
[tex]a\text{ +bi (1)}[/tex]Where a= real part , b= imaginary part
Basically on this case the real part would be the number without the i and the imaginary part the number with i
The real part is:
[tex]4[/tex]And the imaginary part:
[tex]-5[/tex]Final answer:
[tex]4[/tex]Sam is building a model of an antique car. The scale of his model to the actual1car is 1:10. His model is 15 1/2 inches long. How long is the actual car?215The length of the actual car isinches.
The car is 10 times the size of the model.
So, we have to multiply the length of the model by 10.
15 1/2 x 10 =
(15x2+1)/2 x 10=
31/2 x10 =310/2 = 155
The length of the actual car is 155 inches
Options for all the three boxes are:neither student, Becky and Laura
The parent cosine function is given to be:
[tex]f(x)=\cos (x)[/tex]The graph of this function is shown below:
Laura's Function:
Laura's function is given to be horizontally compressed by a factor of 1/3 and reflected over the x-axis.
The rule for horizontal compression by a factor of 1/a is given to be:
[tex]f(x)\Rightarrow f(\frac{1}{a}\cdot x)[/tex]and the rule for reflection over the x-axis is given to be:
[tex]f(x)\Rightarrow-f(x)[/tex]Therefore, Laura's function will be:
[tex]f(x)=-\cos (3x)[/tex]The graph of the function will be:
Becky's Function:
Becky's function is given to be:
[tex]f\mleft(x\mright)=3\cos \mleft(x-\pi\mright)[/tex]The graph is given to be:
ANSWERS:
BECKY's GRAPH is the FIRST GRAPH
The SECOND GRAPH belongs to NEITHER
LAURA's GRAPH is the THIRD GRAPH
At the city museum, child admission is $5.90 and adult admission is $9.40. On Saturday, 170 tickets were sold for a total sales of $1332.00. How many adulttickets were sold that day?Number of adult tickets:
In this case, we'll have to carry out several steps to find the solution.
Step 01:
data:
city museum
Step 02:
system of equations:
x = # child tickets
y = # adult tickets
equations:
x + y = 170 eq. 1
5.90x + 9.40y = 1332 eq.2
x + y = 170 * (- 5.90)
5.90x + 9.40y = 1332
-5.90x - 5.90y = - 1003
5.90x + 9.40y = 1332
--------------------------------
3.5y = 329
y = 329 / 3.5
y = 94
The answer is:
# adult tickets = 94
What is the perimeter of the figure2x + 3 inchesX- 6 inches
Perimeter of the figure = 6x - 6
Explanation:Perimeter = 2(length + width)
length = 2x + 3 inches
width = x- 6 inches
Perimeter = 2(2x + 3 + x- 6)
collect like terms in the bracket:
= 2(2x +x + 3 - 6)
= 2(3x - 3)
Expand the bracket:
Perimeter = 2 × 3x + 2× -3
Perimeter of the figure = 6x - 6
Mrs. Walsh assigned 16 worksheets each month, how many did she assign over 4 montns?
Data Input
Mrs. Walsh assigned 16 worksheets each month
Procedure
To calculate the number of worksheets we need to multiply the number of months by the number of worksheets per month
Total = 16*4 = 64 Worksheets
what is the lcm of 25 and 37?
SOLUTION:
Step 1:
In this question, we are meant to find the LCM of 25 and 37.
Step 2:
The least common multiple, lowest common multiple, or smallest common multiple of two integers a and b, usually denoted by lcm(a, b), is the smallest positive integer that is divisible by both a and b
The diameters of ball bearings are distributed normally. The mean diameter is 125 mm and the standard deviation is 3 mm. Find the probability that the diameter of a selected bearing is greater than 127 mm. Round your answer to four decimal places.
Explanation
Given that the mean diameter is 125 mm and the standard deviation is 3 mm. We can find the probability that the diameter of a selected bearing is greater than 127 mm below.
We will first find the z score of the given value.
[tex]z=\frac{x-\mu}{\sigma}=\frac{127-125}{3}=\frac{2}{3}=0.66667[/tex]Using the z score calculator,
[tex]P\left(x>127\right)=0.2525[/tex]Answer: 0.2525
The composition of rigid motions T(10,- 27 OT(-24,4, describes the route of a limousine in a city from its starting position. Describe the route in words. Assume that the positive y-axis points north. block(s) north, and then it drives 3 block(s) west and 4 block(s) north. First the limousine drives 2 block(s) east and (Type whole numbers.)
The problem describes two operations of translation on the car. The first operation says that the car translated "10" units on the x-coordinate and "-2" units on the y-coordinate.
When translations are performed on the "x-coordinate" if they are positive, the function is moved to the left and if they are negative the function is moved to the right. When translations are performed on the "y-coordinate" if they are positive the function is moved up and if negative, the function is moved down.
Since the first translation on the x-axis was positive, then the car moved to the left 10 units, which would be "10 blocks" on the west direction. The translation on the y-coordinate was negative, so the function moved down. The car moved "2 blocks" on the south direction.
The second translation is (-24, 4). The translation on the x-axis is negative so the car is moved 24 units to the east direction. The translation on the y-axis is positve so the car is moved 4 units on the north direction.
The final answer is:
First the limousine drives 10 blocks west and 2 blocks south and then it drives 24 blocks east and 4 blocks north.
A painting sold for $216 in 1977 and was sold again in 1985 for $467. Assume that the growth in the value V of the collectors items was exponential.
A. Find the value k of the exponential growth rate. Assume v=216
K=?
The value k of the exponential growth rate is k = 0.1012
For given situation, we use the formula for exponential growth :
V(t) = V0(1 + k)^t
where, V(t) is the value of the collectors items
V0 is the initial value of painting
k is the exponential growth rate
t is the time interval
A painting sold for $216 in 1977 and was sold again in 1985 for $467.
So, V0 = $216
t = 1985 - 1977
t = 8
V(8) = $467
We need to find k
V(8) = V0(1 + k)^8
467 = 216 * (1 + k)^8
2.162 = (1 + k)^8
taking 8th root on both the sides,
1 + k = ± (2.162)^(1/8)
1 + k = (2.162)^(1/8) or 1 + k = -(2.162)^(1/8)
k = 0.1012 or k = -2.1012
Therefore, the value k of the exponential growth rate is k = 0.1012
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Help with number one a and b is both parts of number one
Design A. The garden is 9 ft by 12 ft.
Design B. The garden is a circle of radius 5 ft
Area of garden A:
[tex]A_A=9\times12=108[/tex]Area of garden B:
[tex]A_B=\pi\left(5\right)^2=25\pi\approx78.54[/tex]The areas are:
Design A = 108 square ft
Design B ≈ 78.54 square ft
Note: The area of garden B can also be expressed in exact form as 25π square ft
Jose is riding his bicycle. He rides for 14.4 kilometers at a speed of 9 kilometers per hour. For how many hours does he ride?hours:
Given:-
Jose rides 14.4 kilometers at a speed of 9 kilometers.
To find:-
The amount of hours he ride.
The formula used to find the time is,
[tex]time=\frac{Dis\tan ce}{speed}[/tex]Subsituting the value we get,
[tex]\begin{gathered} t=\frac{14.4}{9} \\ t=1.6 \end{gathered}[/tex]So the amount of hours jose rides is 1.6 hours.
A survey wanted to determine if there was a relationship between the number of joggers who used a local park for exercise and the temperature outside. The data in the table display their findings.Use graphing technology to create a scatter plot of the data.What is the slope of the line of best fit and what does it mean in this context? Is it realistic?
Step 1
In order to find the slope of the line of the best fit, we must graph the points given.
Step 2
From the image above we can conclude that the equation of the line is;
[tex]y=0.410045x-2.8757[/tex]The slope is, therefore;
[tex]0.410045[/tex]The Slope (or Gradient) we call m, represents the change in y-value per unit change in x-value. In the case of this survey, the slope represents the increase in the number of joggers per degree rise in temperature.
From the calculated slope, there are about 0.4 more joggers for every degree rise in temperature. Using whole numbers to represent this, we can multiply the slope by 5:
[tex]0.4\times5=2[/tex]Therefore, there are 2 more joggers for every 5°F increase in temperature.
What is the congruence correspondence, if any, that will prove the given triangles congruent?A. SASB. AASC. noneD. ASA
None (option C)
Explanation:Congruent triangles have same shape and size
Rules that can be used to prove triangle congruency:
SAS - side angel side
AAS = Angle angle Side
SSS = side-side-side
ASA = Angle-side-angle
From the triangles given, we see they are not of the same size even though they look alike.
Also the corresponding angles of both triangles do not look the same.
As a result, we do not have conguruence correspondence that will prove the triangles are congruent.
None
The following dataset represents the dollar amounts of donations collected at the entrance to a free museum during onehour:5, 10,5,5, 15, 1, 10, 10,5,600,5Is the mean a reasonably good measure of central tendency for this dataset? What if the outlier were removed fromconsideration?
Given the data set :
5, 10, 5, 5, 15, 1, 10, 10,5, 600, 5
As seen most of th data is between 1 and 15 except one data is 600
so, 600 represents an outlier for the data
so, the answer is option 4
i have no idea how to do this please help
Answer:
D. -1
Explanation:
• There are 5 x(s)
,• There are 6 1-boxes.
The equation modeled is:
[tex]5x+6=1[/tex]We solve for x.
[tex]\begin{gathered} 5x=1-6 \\ 5x=-5 \\ x=-\frac{5}{5} \\ x=-1 \end{gathered}[/tex]The value of x that makes the equation true is -1.
Consider the diagram and angle measures shown below.m∠1 = (3x +25)m∠2 = (7x+5)m∠3 =(-2x + 70)4322 || 7What is the value of m∠3 ?
First, we need to find the value of x
m<4 = m<1 = (3x + 25)° ( corresponding angle)
Let the angle between m<2 and m<3 be m<5
m< 5 = m<4 = (3x + 25)° ( vertical angle)
m<2 + m<5 + m<3 = 180° (angles on a straight line
(7x+5)° + (3x+25)° + (-2x + 70)° = 180°
7x + 5 + 3x + 25 -2x + 70 = 180
Rearrange
7x + 3x -2x + 5 + 25 + 70 = 180
8x + 100 = 180
subtract 100 from both-side of the equation
8x = 180 - 100
8x = 80
Divide both-side of the equation by 8
x = 10
m<3 = -2x + 70
substitute x = 10 in the above
m<3 = -2(10) + 70 = -20 + 70 = 50°
Part 311Use the relationships in circle O to find the missing measures in circle S.1 pointIf WR = 12 units and MN = 8 units, determine MW.Type your answerwMRNePrestipus
From the properties of secant line and the tangent to the circle
If a secant segment and tangent segment are drawn to a circle from the same external point, the length of the tangent segment is the geometric mean between the length of the secant segment and the length of the external part of the secant segment.
[tex]\frac{Whole\text{ Secant}}{\tan gent\text{ Line}}=\frac{\tan gent}{External\text{ Secant Part}}[/tex]
In the given figure we hvae :
Whole secant length (RM),
Tangnt line MN = 8 units
External secant part (WM)
Since RM = WR + WM
RM=12 + WM
Susbtitute the value:
[tex]\begin{gathered} \frac{Whole\text{ Secant}}{\tan gent\text{ Line}}=\frac{\tan gent}{External\text{ Secant Part}} \\ \frac{RM}{MN}=\frac{MN}{MW} \\ \frac{12+MW}{8}=\frac{8}{MW} \\ \text{Apply crossmultiplication:} \\ MW(12+MW)=8\times8 \\ 12MW+(MW)^2=64 \\ \text{ Let MW = x} \\ 12x+x^2=64 \\ x^2+12x-64=0 \\ \text{ Factorize:} \\ x^2+16x-4x-64=0 \\ x(x+16)-4(x+16)=0 \\ (x-4)(x+16)=0 \\ \text{ So, x = 4, -16} \\ \text{ Since measurement cannot be negative thus: x = 4 unit} \\ x\text{ = }MW=4 \end{gathered}[/tex]Answer : MW= 4 units
which system of equations is better to start up to solve using the subsition method or the elemination method?
For the set of equations, the answer will be:
The elimination method is better because both equations are in general form.
if 5 pounds of apples cost 5.50,then what is the cost of 1 pound of apples
The cost of 1 pound of apple is 1.10
Here, we want to calculate the cost of 1 pound of apples given the cost of 5 pounds
From the question;
5 pounds = 5.5
1 pound = x
Thus, we have that;
[tex]\begin{gathered} 5\text{ }\times\text{ x = 1 }\times\text{ 5.5} \\ \\ 5x\text{ = 5.5} \\ \\ x\text{ = }\frac{5.5}{5} \\ \\ x\text{ = 1.1} \end{gathered}[/tex]find the correct area
we have that the area of the semicircle is 100,530964915 m^2
[tex]\begin{gathered} A\text{ = }\frac{\pi r^2}{2} \\ =\text{ }\frac{\pi(8m)^2}{2}=\text{ }32\pi m^2 \\ =\text{ }100,530964915m^2 \end{gathered}[/tex]Now we need to find the area of the triangle, we now that the area of the non rectangle triangle is 96 m^2
[tex]\begin{gathered} A\text{ = }\frac{b\cdot h}{2}\text{ = }\frac{16m\cdot\text{ 12m}}{2} \\ =\text{ }\frac{192m^2}{2}=96m^2 \end{gathered}[/tex]So the area of the figure is 196,530964915, the answer is 196.5 m^2.
The number of needles in each packet are 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 7,and 8. Find the range of the given data.
Given that
The number of needles in the packet is 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 7, 8
And we have to find the range of this data.
Explanation -
To find the range first we will arrange the numbers in the order as
1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 6, 7, 8
Now we will subtract the lowest number from the highest to get the range.
So
nge = 8 - 1 = 7
Final answer -
Hence the range is 7, So the final answer is 7.The figure shows a person estimating the height of a tree by looking at thetop of the tree with a mirror. Assuming that both the person and the tree formright angles with the ground, which of the following proportions can be usedto estimate the height of the tree?6 ft3 ft7 fthft
SOLUTIONS
Using the rule of SOHCAHTOA:
The height of the tree compared to the height of the boy is tangent:
[tex]tan\theta=\frac{opp}{adj}[/tex]For the small triangle:
[tex]tan\theta=\frac{6}{3}.........(1)[/tex]For the Big triangle:
[tex]tan\theta=\frac{h}{7}..........(11)[/tex]Equating (1) and (11
[tex]\begin{gathered} \frac{6}{3}=\frac{h}{7} \\ \frac{6}{h}=\frac{3}{7} \end{gathered}[/tex]Hence the correct answer = Option A
a crate is 24cm by 18cm by 15cm. it has to be packed with boxes that are 8cm by 6cm by 5cm. what is the largest number of boxes that can fit in the crate.
please help its level 2
show answer and working out.
The maximum no. of small boxes that can fit is 27 boxes.
What is a cuboid?A cuboid is a three-dimensional closed figure which has volume along with surface area.
The volume of a cuboid is the product of its length, width, and height.
The total surface area of a cuboid is 2( lw + wh + wl) and the lateral surface area is 2(l + w)×h.
Given, A crate is 24cm by 18cm by 15cm and it has to be packed with boxes that are 8cm by 6cm by 5cm.
Now, the maximum no of small boxes that can fit is the volume of the larger box divided by the volume of the sampler box which is,
= (24×18×15)/(8×6×5).
= 3×3×3.
= 27 boxes.
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1990, the cost of tuition at a large Midwestern university was $101 per credit hour. In 2005, tuition had risen to $236 per credit hour.
Solution:
According to the problem, the linear equation would be:
[tex]C(X)=101+9(x-1990)[/tex]now, if x=2013, the tuition will be:
[tex]C(2013)=101+9(2013-1990)=101+9(23)[/tex]this is equivalent to
[tex]C(2013)=101+9(23)=308[/tex]The function h is defined below.xh (x) =2+5x-- 142x- 81Find all values of × that are NOT in the domain of h.If there is more than one value, separate them with commas.
We want to find the value of x that is not in the domain of the function;
[tex]h(x)=\frac{x^2+5x-14}{x^2-81}[/tex]The value of x that will make h undefined is the value that makes the denominator zero.
Therefore;
[tex]\begin{gathered} x^2-81=0 \\ x^2=81 \\ x=\pm9_{} \end{gathered}[/tex]Therefore, the values not in the domain is +9 and -9
In the expansion of (2a+4b)^8, which of the following are possible variable terms? Explain your reasoning?
In the expansion of (2a+4b)⁸ the possible variable terms are a⁸, a⁵b³, a³b⁵, a⁷b, and b⁸.
What is an Expression?
In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.
We need to find the possible variable terms in the expansion of (2a+4b)^8
The given expression can be written as:
(2a + 4b)⁸ = (2a + 4b)² x (2a + 4b)² x (2a + 4b)² x (2a + 4b)²
= (4a² + 8ab + 16b²) x (4a² + 8ab + 16b²) x (4a² + 8ab + 16b²) x (4a² + 8ab + 16b²)
Now, further simplify the above expression.
(16a⁴ + 32a³b + 64a²b² + 32a³b + 64 a²b² + 128b⁴ + 64a²b² + 128ab³ + 256b⁴) x (16a⁴ + 32a³b + 64a²b² + 32a³b + 64 a²b² + 128b⁴ + 64a²b² + 128ab³ + 256b⁴)
Hence, In the expansion of (2a+4b)⁸ the possible variable terms are a⁸, a⁵b³, a³b⁵, a⁷b, and b⁸.
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