Rabbit's run: distance (meters) time (minutes) way 800 1 900 5 1107.5 20 1524 32.5
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Answer:
Notice that:
[tex]\begin{gathered} \frac{800}{1}=800, \\ \frac{900}{5}=180, \\ \frac{1107.5}{20}=\frac{443}{8}, \\ \frac{1524}{32.5}=\frac{3048}{65}. \end{gathered}[/tex]Since all reduced fractions are different, the distance traveled by the rabbit and the time are not proportional.
Related Questions
A translation is a type of transformation in which a figure is flipped,TrueFalse
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Rick's average score on his first three tests in math is 80. What must he score on his next test to raise his average to 84?
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SOLUTION
Now, we don't know the scores for his first three tests. But we are told that the average score for the first three tests was 80.
So, let the scores of the first three tests be a, b, and c. That means
[tex]\frac{a+b+c}{3}=80[/tex]Also, let's assume the total score for his first three tests was x, This means that
[tex]\begin{gathered} a+b+c=x \\ or \\ x=a+b+c \end{gathered}[/tex]Comparing with the first equation it means that
[tex]\begin{gathered} \frac{a+b+c}{3}=80 \\ \frac{x}{3}=80 \\ x=3\times80 \\ x=240 \end{gathered}[/tex]Now we are asked "What must he score on his next test to raise his average to 84?"
So this means the total tests becomes 4. Hence
[tex]\begin{gathered} \frac{a+b+c+d}{4}=84 \\ \frac{x+d}{4}=84 \\ \frac{240+d}{4}=84 \\ 240+d=84\times4 \\ 240+d=336 \\ d=336-240 \\ d=96 \end{gathered}[/tex]So he must score 96 to raise his average score to 84.
Hence, the answer is 96
Find the quantities indicated in the picture (Type an integer or decimal rounded to the nearest TENTH as needed.)
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Remember that 3, 4 and 5 is a Pythagorean triple, since:
[tex]3^2+4^2=5^2[/tex]Since one side of the given right triangle has a length of 3 and the hypotenuse has a length of 5, then, the remaining leg b must have a length of 4.
Therefore:
[tex]b=4[/tex]The angles A and B can be found using trigonometric identities.
Remember that the sine of an angle equals the quotient of the lengths of the side opposite to it and the hypotenuse of the right triangle.
The side opposite to A has a length of 3 and the length of the side opposite to B is 4. Then:
[tex]\begin{gathered} \sin (A)=\frac{3}{5} \\ \sin (B)=\frac{4}{5} \end{gathered}[/tex]Use the inverse sine function to find A and B:
[tex]\begin{gathered} \Rightarrow A=\sin ^{-1}(\frac{3}{5})=36.86989765\ldotsº \\ \Rightarrow B=\sin ^{-1}(\frac{4}{5})=53.13010235\ldotsº \end{gathered}[/tex]Then, to the nearest tenth:
[tex]\begin{gathered} A=36.9º \\ B=53.1º \end{gathered}[/tex]Therefore, the answers are:
[tex]undefined[/tex]the width of a rectangle is 8 inches less than its length, and the area is 9 square inches. what are the length and width of the rectangle?
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The given situation can be written in an algebraic way:
Say x the width of the rectangle and y its height.
- The width of a rectangle is 8 inches less than its length:
x = y - 8
- The area of the rectangle is 9 square inches:
xy = 9
In order to find the values of y and x, you first replace the expression
x = y - 8 into the expression xy = 9, just as follow:
[tex]\begin{gathered} xy=9 \\ (y-8)y=9 \end{gathered}[/tex]you apply distribution property, and order the equation in such a way that you obtain the general form of a quadratic equation:
[tex]\begin{gathered} (y-8)y=9 \\ y^2-8y=9 \\ y^2-8y-9=0 \end{gathered}[/tex]Next, you use the quadratic formula to solve the previous equation for y:
[tex]y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]here you have a = 1, b = -8 and c = 9. By replacing these values you obtain:
[tex]\begin{gathered} y=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(1)(-9)}}{2(1)}=\frac{8\pm\sqrt[]{64+36}}{2} \\ y=\frac{8\pm\sqrt[]{100}}{2}=\frac{8\pm10}{2}=\frac{8}{2}\pm\frac{10}{2}=4\pm5 \end{gathered}[/tex]Hence, you have two solutions for y:
y1 = 4 + 5 = 9
y2 = 4 - 5 = -1
You select only the positive solution, because negative lengths do not exist in real life. Hence, you have y = 9.
Finally, you replace the value of y into the expression x = y - 8 to obtain x:
[tex]\begin{gathered} x=y-8 \\ x=9-8 \\ x=1 \end{gathered}[/tex]Hence, the width and length of the given recgtangle are:
width = 1 in
length = 9 in
Michelle can wash dry and fold 5 loads of laundry in 3 1/2 hours. what is the average amount of time it takes Michelle to do one load of laundry
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3. You have a bad cough and have to attend your little sister's choir concert. You take cough drops that contain 100 mg of menthol in each drop. Every minute, the amount of menthol in your body is cut in half. Write a funetion that models the amount of menthol in your body over time. Use x for minutes and y for the amount of menthol, in mg, remaining in your body It is safe to take a new cough drop after the level of menthol in your body is less than 5 mg, How long will it be before you can take another cough drop?
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We have the next information
100 mg of menthol
every minute the amount of menthol in your body is cut in half
we have the next variables
x= minutes
y= amount of menthol in mg remaining in your body
so the equation that can we model is
[tex]y=100(0.5)^x[/tex]then we have that It is safe to take a new cough drop after the level of menthol in your body is less than 5 mg
y= 5mg
[tex]5=100(0.5)^x[/tex]in order to know the time we need to solve the equation above
[tex]\begin{gathered} (0.5)^x=\frac{5}{100} \\ (0.5)^x=0.05 \end{gathered}[/tex]then we isolate the x
[tex]x=4.32[/tex]after 5 minutes you can take another cough drop
You are clinic manager. You must schedule the equivalent of 1 1/2 nurses for each doctor on a shift, The friday day shift has 6 doctors scheduled How many nurses you will you need to schedule?
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Data Given:
Nurses = 1 1/2 of each doctor
This can be interpreted as
[tex]\begin{gathered} 1\frac{1}{2}\text{ = }\frac{3}{2} \\ \\ 1\text{ Doctor requires }\frac{3}{2}\text{ times nurses} \end{gathered}[/tex]If there are 6 doctors in the day shift, then there will be
[tex]\frac{3}{2}\text{ x 6 nurs}es[/tex]=>
[tex]\begin{gathered} \frac{3\text{ x 6}}{2} \\ \\ =\text{ 3 x 3 } \\ \\ =\text{ 9 nurses} \end{gathered}[/tex]This means that I will have to schedule 9 nurses for the day shift on Friday
Petrolyn motor oil is a combination of natural oil and synthetic oil. It contains 5 liters of natural oil for every 4 liters of synthetic oil. In order to make 531 litersof Petrolyn oll, how many liters of synthetic oil are needed?
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The ratio 4 : 5 means that in every 9 liters of oil, we will have 4L of synthetic oil and 5L of natural oil.
Divide the 531 by 9 to get how many times we have to amplify the ratio:
[tex]\frac{531}{9}=59[/tex]Multiply the ratio by 59:
[tex]4\colon5\rightarrow(4)(59)\colon(5)(59)\rightarrow236\colon295[/tex]Meaning that for the 531L of oil, 236L would be synthetic and 295L natural.
Answer: 236 Liters.
Find the percent increase in volume when 1 foot is added to each dimension of the prism. Round your answer to the nearest tenth of a percent.7 ft10 ft86 ft
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Solution
Step 1
The volume of a triangular prism = Cross-sectional area x Length
Step 2
[tex]\begin{gathered} Cross\text{ sectional area = area of the triangle} \\ Base\text{ = 6ft} \\ Height\text{ = 7ft} \\ Cross\text{ sectional area = }\frac{1}{2}\times\text{ 7 }\times\text{ 6 = 21 ft}^2 \\ Volume\text{ = 21 }\times\text{ 10 = 210 ft}^3 \end{gathered}[/tex]Step 3:
When 1 foot is added to each dimension of the prism.
The new dimensions becomes Base = 7, Height = 8 and length = 11
[tex]\begin{gathered} \text{Cross-sectional area = }\frac{1}{2}\text{ }\times\text{ 7 }\times\text{ 8 = 28 ft}^2 \\ Length\text{ = 11 ft} \\ Volume\text{ = 28 }\times\text{ 11 = 308 ft}^3 \end{gathered}[/tex]Step 4
Find the percent increase in volume
[tex]\begin{gathered} \text{Percent increase in volume = }\frac{308\text{ - 210}}{210}\text{ }\times\text{ 100\%} \\ \text{= }\frac{98}{210}\text{ }\times100 \\ \text{= 46.7} \end{gathered}[/tex]Final answer
46.7
please help me work through this, thank you very much!
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Given
[tex]plane-height=650m[/tex]To Determine: The angle function
Solution
The information can be represented as shown below
From the diagram below
[tex]\begin{gathered} tan\theta=\frac{650}{x} \\ \theta(x)=tan^{-1}(\frac{650}{x}) \end{gathered}[/tex]In an elementary school, 20% of the teachers teach advanced writing skills. If there are 25writing teachers, how many teachers are there in the school?
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Answer:
125 teachers
Explanation:
We were given that:
20% of teachers teach advanced writing skills = 20/100 = 0.2
Number of writing teachers = 25
The total number of teachers = x
We will obtain the number of teachers in the school as shown below:
[tex]\begin{gathered} \frac{No.of.writing.teachers}{Total.number.of.teachers}\times100\text{\%}=20\text{\%} \\ \frac{25}{x}\times100\text{\%}=20\text{\%} \\ \frac{25\times100\text{\%}}{x}=20\text{\%} \\ \text{Cross multiply, we have:} \\ x\cdot20\text{\% }=25\times100\text{\%} \\ \text{Divide both sides by 20\%, we have:} \\ \frac{x\cdot20\text{\%}}{20\text{\%}}=\frac{25\times100\text{\%}}{20\text{\%}} \\ x=\frac{2500}{20} \\ x=125 \\ \\ \therefore x=125 \end{gathered}[/tex]Hence, the total number of teachers in the school is 125
Construct a polar equation for the conic section with the focus at the origin and the following eccentricity and directrix.Conic Eccentricity Directrix1ellipsex= -75e =
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In order to find the polar equation of the ellipse, first let's find the rectangular equation.
Since the directrix is a vertical line, the ellipse is horizontal, and the model equation is:
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]Where the center is located at (h, k), the directrix is x = -a/e and the eccentricity is e = c/a.
So, if the eccentricity is e = 1/5 and the directrix is x = -7, we have:
[tex]\begin{gathered} \frac{c}{a}=\frac{1}{5}\rightarrow a=5c\\ \\ -\frac{a}{e}=-7\\ \\ \frac{a}{\frac{c}{a}}=7\\ \\ \frac{a^2}{c}=7\\ \\ \frac{25c^2}{c}=7\\ \\ 25c=7\\ \\ c=\frac{7}{25}\\ \\ a=5\cdot\frac{7}{25}=\frac{7}{5} \end{gathered}[/tex]Now, let's calculate the value of b with the formula below:
[tex]\begin{gathered} c^2=a^2-b^2\\ \\ \frac{49}{625}=\frac{49}{25}-b^2\\ \\ b^2=\frac{25\cdot49}{625}-\frac{49}{625}\\ \\ b^2=\frac{24\cdot49}{625}\\ \\ b^2=\frac{1176}{625} \end{gathered}[/tex]Assuming h = 0 and k = 0, the rectangular equation is:
[tex]\frac{x^2}{\frac{49}{25}}+\frac{y^2}{\frac{1176}{625}}=1[/tex]Now, to convert to polar form, we can do the following steps:
[tex]\begin{gathered} \frac{25x^2}{49}+\frac{625y^2}{1176}=1\\ \\ 600x^2+625y^2=1176\\ \\ 600(r\cos\theta)^2+625(r\sin\theta)^2=1176\\ \\ 600r^2\cos^2\theta+625r^2\sin^2\theta=1176\\ \\ r^2(600\cos^2\theta+625\sin^2\theta)=1176\\ \\ r^2=\frac{1176}{600\cos^2\theta+625\sin^2\theta}\\ \\ r=\sqrt{\frac{1176}{600\cos^2\theta+625\sin^2\theta}}\\ \\ r=\sqrt{\frac{1176}{600+25\sin^2\theta}} \end{gathered}[/tex]Another way of writing this equation in polar form is:
[tex]r=\frac{ep}{1+\sin^2\theta}[/tex]Where p is the distance between the focus and the directrix.
Since the foci are located at (±c, 0) = (±7/25, 0) and the directrix is x = -7, the distance is:
[tex]p=7-\frac{7}{25}=\frac{175}{25}-\frac{7}{25}=\frac{168}{25}[/tex]So the equation is:
[tex]\begin{gathered} r=\frac{\frac{1}{5}\cdot\frac{168}{25}}{1+\sin^2\theta}\\ \\ r=\frac{\frac{168}{125}}{1+\sin^2\theta}\\ \\ r=\frac{1.344}{1+\sin^2\theta} \end{gathered}[/tex]Hello Just Want to make sure my answer is correct
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So,
Let's remember that:
The three point postulate states that:
Through any three noncollinear points, there exists exactly one plane.
The Plane-Point Postulate states that:
A plane contains at least three noncollinear points.
As you can notice, the diagram illustrates that:
Given that a plane exists, then, there are three collinear points.
That's the three point postulate.
Find the area of a regularpolygon with 5 sides that has aside length of 6 inches and anapothem of 9 inches. Area = ?
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SOLUTION
Write out the formula
[tex]\text{area of regular polygon=}\frac{A\text{ }\times P}{2}[/tex]where A= apothem and P= perimeter of the regular polygon
[tex]\begin{gathered} A=9in \\ P=6(5)=30in \\ \text{perimeter of the regular polygon is sum of all the lenght} \\ \text{the number of sides }\times the\text{ lenght of a side } \end{gathered}[/tex]The area of the regular polygon is
[tex]\frac{9\times30}{2}=9\times15=135in^2[/tex]given AD is congruent to AC and AB is congruent to AE, which could be used to prove?
Answers
Answer
Option B is correct.
SAS | 2 sides and the angle between them in one triangle are congruent to the 2 sides and the angle between them in the other triangle, then the triangles are congruent.
Explanation
We have been told that the two triangles have two sets of sides that are congruent to each other.
And we can see that the angle between those congruent sides for the two triangles is exactly the same for the two triangles.
So, it is easy to see that thes two triangles have 2 sides that are congruent and the angle between these two respective sides are also congruent.
Hope this Helps!!!
Which of the following logarithmic expressions have been evaluated correctly?
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Given:
Logarithmic expressions in options.
Required:
Select correct calculated option.
Explanation:
1). ln 1 = 0
2).
[tex]log_29=3.1699250014[/tex]3)
[tex]log\frac{1}{100}=-2_[/tex]4).
[tex]log_3(-1)=NaN[/tex]5).
[tex]log_5\text{ }\frac{1}{125}=-3[/tex]Answer:
Hence, option A and E are correct.
Mrs walters had a bag full of candy she wanted to share with 18 students. If she had 335 pieces of candy how many pieces will each student get
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eln(x-3) = 9what are the steps to solve? I am so confused, what is ln even??
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Domain and range from the graph of a piecewise function
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The domain of the given function is:
[tex]\text{Domain}=\lbrack-3,-2\rbrack\cup\lbrack-1,5)[/tex]Because the domain corresponds to the set of all possible inputs, x-axis.
The range of this function is:
[tex]\text{Range}=\lbrack-5,3\rbrack[/tex]Because the range corresponds to the set of all possible outputs, y-axis.
What are inequality? When do we use inequalities?What type of inequalities are there? Which symbols are used for each type?Are the following expressions variable inequalities? Why?a. 13z=27b. x<0c 3x+5x>11d. y+5≤11e. 7-1>- 32
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Inequalities are expressions that refer to non-equivalent quantities. Inequalities can express less than, more than, less than or equal to, more than or equal to.
The type of inequalities and symbols are:
[tex]<,>,\leq,\ge[/tex]So, there are four types of inequalities, for example:
[tex]\begin{gathered} x<2 \\ x>2 \\ x\leq2 \\ x\ge2 \end{gathered}[/tex]Each inequality is different from the other, this means that the symbol used represents a type of inequality.
At last, among the choices, the inequalities are
[tex]\begin{gathered} x<0 \\ 3x+5x>11 \\ y+5\leq11 \\ 7-1>-32 \end{gathered}[/tex]However, variable inequalities mean that the inequalities must have a variable in it. So, they are:
[tex]\begin{gathered} x<0 \\ 3x+5x>11 \\ y+5\leq11 \end{gathered}[/tex]Therefore, the variable inequalities are b, c, and d.
A bucket can hold 26 litres of water when it is 8/9 full. How many litres can it hold when it is full?
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Answer:
[tex]29.25\text{ liters}[/tex]Explanation:
Here, we want to know the amount of water the bucket can hold when full
Let us have the volume as x liters
Mathematically:
[tex]\begin{gathered} \frac{8}{9}\times x\text{ = 26} \\ \\ 8x\text{ = 9 }\times\text{ 26} \\ x=\text{ }\frac{9\times26}{8} \\ \\ x\text{ = 29.25 liters} \end{gathered}[/tex]A rock has a mass of 14 g and a volume of 2 cm3. What is the density of the rock? *
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We will determine the density of the rock as follows:
[tex]\rho=\frac{14g}{2cm^3}\Rightarrow\rho=7g/cm^3[/tex]So, the density of the rock is 7 g/cm^3.
Data Set A has a Choose... interquartile range than Data Set B. This means that the values in Data Set A tend to be Choose... the median.
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The median of the given data set will be 35.
What do we mean by media?In statistics and probability theory, the median is the number that separates the upper and lower half of a population, a probability distribution, or a sample of data. For a data set, it might be referred to as "the middle" value.
So, The variability metrics for each class are listed below:
The further classifications: Class A; Class B;
Range: 30 Range: 30IQR: 12.5 IQR: 20.5MAD: 7.2 MAD: 9.2Greater variability in the data set is suggested by class B's wider interquartile range and mean absolute deviations.
Set A's median will be:
median = (20 + 32+ 36+ 37 + 50) / 5median = 175 / 5median = 35Therefore, the median of the given data set will be 35.
To learn more about Median click on the link
brainly.com/question/26151333
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Which measurement is closest to the shortest distance in miles from Natasha's house to the library?
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Given:
The objective is to find the shortest distance between house and library.
Consider the given triangle as,
Here, A represents the house, B the grocery and C the library.
Since it is a right angled triangle, the distance between the house and the library can be calculated using Pythagoras theorem.
[tex]\text{Hypotenuse}^2=Opposite^2+Adjacent^2[/tex]Apply the given values in the above formula,
[tex]\begin{gathered} AC^2=17^2+0.9^2 \\ AC^2=289+8.1 \\ AC^2=297.1 \\ AC=\sqrt[]{297.1} \\ AC=17.237\text{ miles} \end{gathered}[/tex]If Natasha walks through Grocery store,
[tex]\begin{gathered} AC^{\prime}=AB+BC \\ AC^{\prime}=0.9+17 \\ AC^{\prime}=17.9\text{ miles} \end{gathered}[/tex]By comparing the two ways, ACHence, the hypotenuse distance AC, between house and library is the closest distance.
Professor Ivy’s students have a Mean grade of 69.5 and a Standard Deviation of 6.5.3. If Johnny has an 82 in the class, what would the z-score for Johnny’s grade be? (round to the tenthsplace)4. What percentile does Johnny’s score put him in? (round to the nearest whole number)
Answers
Given:
Mean,ц = 69.5
Standard deviation, σ = 6.5
Let's solve for the following:
• 3. If Johnny has an 82 in the class, what would the z-score for Johnny’s grade be?
Apply the z-score formula:
[tex]z=\frac{x-\mu}{\sigma}[/tex]Where:
x = 82
ц = 69.5
σ = 6.5
Thus, we have:
[tex]\begin{gathered} z=\frac{82-69.5}{6.5} \\ \\ z=\frac{12.5}{6.5} \\ \\ x=1.9 \end{gathered}[/tex]Therefore, the z-score is 1.9
Question 4.
Here, we are to find P(Z<1.9).
Using the standard normal distribution table, we have:
NORMSDIST(1.9) = 0.9712
Now convert to percentage:
0.9712 x 100 = 97.12% = 97%
ANSWER:
3). 1.9
4.) 97%
Triangle XYZ is rotated 90° counterclockwise about the origin.The result is Triangle X'Y'Z', as shown below.
Answers
A shortcut for a 90° counterclockwise rotation:
• If the point (h, k) is rotated 90° counterclockwise rotation, then the final point will be (-k, h).
Answer:
Therefore the coordinates would be:
• X,(-5, 3) → ,X',(-3, -5)
,• Y,(-1, 1) → ,Y',(-1, -1)
,• Z,(-8, -4) → ,Z',(4, -8)
Then, the rule is (x, y) → (-y, x).
Equation of the line that passes through points (8,7) and (0,0)
Answers
Equation of the line:
y = mx+b
where:
m= slope
b= y-intercept
First, we have to find the slope:
m = (y2-y1) / (x2-x1)
Since we have:
(x1,y1) = (8,7)
(x2,y2)= (0,0)
Replacing:
m = (0-7)/ (0-8) = -7/-8 = 7/8
Now, that we have the slope:
y = 7/8 x +b
We can place the point (8,7) in the equation and solve for b:
7 = 7/8 (8) +b
7=7 +b
7-7=b
b=0
Since the y-intercept=0
The final equation is:
y= 7/8x
Find the volume of the cone.9 cmr= 6 cmV = [?] cm3
Answers
The radius of cone is r = 6 cm.
The height of cone is h = 9 cm.
The formula for the volume of cone is,
[tex]V=\frac{1}{3}\pi\cdot r^2\cdot h[/tex]Substitute the values in the formula to determine the volume of cone.
[tex]\begin{gathered} V=\frac{1}{3}\pi\cdot(6)^2\cdot9 \\ =339.29 \\ \approx339.3 \end{gathered}[/tex]Thus, volume of cone is 339.3 cm^3.
P(-3,-5) and Q(1.–3) represent points in a coordinate plane. Find the midpoint of Pe.
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By formula,
Midpoint between two points PQ =
[tex](\frac{x_2+x_1}{2},\text{ }\frac{y_2+y_1}{2})[/tex][tex]\begin{gathered} (\frac{1+-3}{2},\text{ }\frac{-3+\text{ -5}}{2}) \\ \\ \frac{-2}{2},\text{ }\frac{-8}{2}\text{ = (-1,-4)} \\ \\ \end{gathered}[/tex]So, (-1,-4) (option 3)
The Thompson family and the Kim family each used their sprinklers last summer. The Thompson family's sprinkler was used for 25 hours. The Kim family'ssprinkler was used for 35 hours. There was a combined total output of 1075 L of water. What was the water output rate for each sprinkler if the sum of the tworates was 35 L per hour?Thompson family's sprinkler:Kim family's sprinkler:
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Let x be the rate of water output by the Thompson family and let y be the rate of water output by the Kim family.
We know that the Thompson family sprinkler was used for 25 hours, Kim's family sprinkler was used for 35 hours and that there was a combined total output of 1075 L of water; then we have the equation:
[tex]25x+35y=1075[/tex]We also know that the combined water output was 35 L per hour, then:
[tex]x+y=35[/tex]Hence we have the system of equations:
[tex]\begin{gathered} 25x+35y=1075 \\ x+y=35 \end{gathered}[/tex]To solve this system we solve the second equation for y:
[tex]\begin{gathered} x+y=35 \\ y=35-x \end{gathered}[/tex]And we plug this value in the first equation and solve for x:
[tex]\begin{gathered} 25x+35(35-x)=1075 \\ 25x+1225-35x=1075 \\ -10x=1075-1225 \\ -10x=-150 \\ x=\frac{-150}{-10} \\ x=15 \end{gathered}[/tex]Once we have the value of x we plug it in the expression of y:
[tex]\begin{gathered} y=35-15 \\ y=20 \end{gathered}[/tex]Therefore we have that:
[tex]\begin{gathered} x=15 \\ y=20 \end{gathered}[/tex]which means:
Thompson family's sprinkler: 15 L per hour
Kim family's sprinkler: 20 L per hour.