What we need to do is follow up with Sondra.
Starts at 12.15 yd
Ascendes 38.281, therefore:
[tex]12.15+38.5=50.65[/tex]A total 50.65 yd
Then, descends 15.25, therefore:
[tex]50.65-12.25=38.5[/tex]This means that the relative 38.5 - (15.25 - 12.25) = 38.5 - 3 = 35.5, it is means 35 1/2
The answer is 35 1/2
Harry took a loan from the bank.D represents Harry's remaining debit (In dollars) after t months.D = -200t + 9000
D = His remaining debt in dollars
t = number of months
[tex]\begin{gathered} D=-200t+9000 \\ D=9000-200t \end{gathered}[/tex]The size of the loan will be $9000 . From the equation you can notice the product of 200 and t is always subtracted at t months from the amount he borrowed($9000)
if the great circle circumference of a sphere is 16 pi yards, find its surface area.
Next, find the surface area.
[tex]\begin{gathered} \text{Surface area = 4}\pi r^2 \\ \text{ = 4}\pi\text{ }\times8^2 \\ \text{ = 4}\pi\text{ x 64} \\ \text{ = 256}\pi yd^2 \end{gathered}[/tex]Elena is conduction a study about the effects of toxins in the water on the hormones of fish. Elena surveys 350 male fish in a river and finds that 150 of the male fish have egg cells growing inside them. According to Elena’s survey, what is the ratio of male fish with egg cells to male cells in the river?
Answer:
3:7
Explanation:
We know that Elena surveys 350 male fish in a river.
Out of these 350 fish, 150 have egg cells growing inside them.
Therefore,
male fish with egg cells: male fish in the river = 150: 350
The next step is to write this ratio as a fraction and simplify it.
Writing the ratio as a fraction gives
[tex]\frac{150}{350}[/tex]dividing both the numerator and the denominator by 50 gives
[tex]\frac{150\div50}{350\div50}[/tex][tex]=\frac{3}{7}[/tex]Hence, the ratio of the male fish with egg cells to total male fish in the river is 3:7.
In a coordinate plane, quadlateral PQRS has vertices P(0,7), Q(4,6), R(2,3), S(-1,3). Find the coordinates of the vertices of the image after each reflection.Reflection across the line y = x
First we need to draw the graph
Then reflect with respect to the X axis
which is basically changing the sign of the Y values for each point
So, we can calculate the new points
[tex]\begin{gathered} P^{\prime}(0,-7) \\ Q^{\prime}(4,-6) \\ R^{\prime}(2,-3) \\ S^{\prime}(-1,-3) \end{gathered}[/tex]Can you please help me with the answers, I don’t need the work I just need the answer please
Given that
The angles of the triangle are
30, 60 and 90
Consider the sine law formula
[tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}[/tex]Let A=30, B=60, and C=90, and substitute these values in the sine law, we get
[tex]\frac{a}{\sin 30^o}=\frac{b}{\sin 60^o}=\frac{c}{\sin 90^o}[/tex][tex]\frac{a}{\frac{1}{2}}^{}=\frac{b}{\frac{\sqrt[]{3}}{2}}=\frac{c}{1}[/tex]The ratio of the legs a and b is
[tex]\frac{a}{b}=\frac{\frac{1}{2}}{\frac{\sqrt[]{3}}{2}}=\frac{1}{2}\times\frac{2}{\sqrt[]{3}}=\frac{1}{\sqrt[]{3}}[/tex][tex]1\colon\sqrt[]{3}[/tex]The ratio of the legs b and c is
[tex]\frac{b}{c}=\frac{\frac{\sqrt[]{3}}{2}}{1}=\frac{\sqrt[]{3}}{2}[/tex][tex]\sqrt[]{3}\colon2[/tex]The ratio of the legs a and c is
[tex]\frac{a}{c}=\frac{\frac{1}{2}}{1}=\frac{1}{2}[/tex][tex]1\colon2[/tex]Hence the answer is
[tex]1\colon\sqrt[]{3}[/tex]Option C is correct.
The box-and-whisker plot below represents some data set. What percentage of the data values are greater than 69?
Given:
The box-and-whisker plot below represents some data set.
As shown:
The median = 65
The value of Q1 = 61
The value of Q3 = 69
The percentage of the data values that are greater than 69 will be the percentage of the data that are greater than Q3
So, the answer will be 25%
Can you please help me out with a question
As you can see in the given figure, there are two intersecting chords inside the circle.
Recall that the "Intersecting Chords Theorem" is given by
[tex]AE\cdot EC=BE\cdot DE[/tex]For the given case, we have
AE = 7
BE = 6
EC = 9
Let us substitute these values into the above equation and solve for DE
[tex]\begin{gathered} AE\cdot EC=BE\cdot DE \\ 7\cdot9=6\cdot DE \\ 63=6\cdot DE \\ \frac{63}{6}=DE \\ 10.5=DE \\ DE=10.5 \end{gathered}[/tex]Therefore, the length of DE is 10.5 units.
A table of values of a linear function is shown below. Find the output when the input is n. input: 1 2 3 4 n output: 3 1 -1 -3
We have the points of a linear function and need to find the equation that represent.
Because it is a linear function, we can find its equation with two points.
We get the points (1,3) and (2,1):
[tex]\begin{gathered} We\text{ call input as x and output as y:} \\ P_1=(x_1,y_1)=(1,3),P_2=(x_2,y_2)=(2,1) \\ y-y_1=\frac{(y_2-y_1)}{(x_2-x_1)}(x-x_1) \\ y-3=\frac{(1-3)}{(2-1)}(x-1) \\ y-3=-\frac{2}{1}(x-1)=-2(x-1) \\ y=-2x-2\cdot(-1)+3 \\ y=-2x+2+3 \\ y=-2x+5 \end{gathered}[/tex]We can check that the points (3,-1) and (4,-3) also satisfy the equation that we found above:
[tex]\begin{gathered} \text{For point (3,-1):} \\ y=-2\cdot3+5=-6+5=-1 \\ \text{For point (4,-3):} \\ y=-2\cdot4+5=-8+5=-3 \end{gathered}[/tex]The above shows that the points satisfy the equation.
So, for input=n the output is:
[tex]\text{output}=-2\cdot n+5[/tex]The table summarizes results from 977 pedestrian deaths that were caused by automobile accidents.Pedestrian DeathsDriverPedestrian Intoxicated?Intoxicated?YesNoYes4885No260584If one of the pedestrian deaths is randomly selected, find the probability that neither the pedestrian nor the driver wasintoxicated. Please enter a decimal to 4 places.Probability
We are asked to find the probability that neither the pedestrian nor the driver was intoxicated.
Probability is given by
P = number of desired outcomes/total number of outcomes
From the given table we know that
Total number of pedestrian deaths = 977
The desired outcomes are
Driver intoxicated = No and Pedestrian intoxicated = No (584)
So the probability is
P = 584/977
P = 0.5977
Therefore, the probability that neither the pedestrian nor the driver was intoxicated is 0.5977
What is the domain of h?y7h---7 6 5 4 3 276+5+4+3+21++ +++++1 2 3 4 5 6 72+-3+-4+-5 .-6+-7
The domain of a function is the set of all possible x-values.
In this case, the function h(x) is discrete, then the domain is the set of x-values where h(x) is defined, that is,
Domain: {-2, -1, 1, 5, 6}
i need help with this problem Find rate of change.
Step 1:
First, pick two points where the line intercept with both the horizontal and the vertical axis.
( 0 , 300 ) and (
Rotation of 270° (x,y) or (a,b) becomes
A totation of 270º will chncge the sign of the x and y coordinate so the condinates will change:
[tex]\begin{gathered} (x,y)\to(-x,-y) \\ (a,b)\to(-a,-b) \end{gathered}[/tex]I need the answers please show work so I don’t fail
Solution
- The way to solve the question is that we should substitute the values of x and y given into the formula given to us.
- The formula given to us is:
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ where, \\ (x_1,y_1)\text{ are the points given to us} \\ m\text{ is the slope} \end{gathered}[/tex]- Thus, we can solve the question as follows:
Question 1:
[tex]\begin{gathered} m=5 \\ x_1=3,y_1=6 \\ \\ \text{ Thus, the equation is:} \\ y-6=5(x-3) \end{gathered}[/tex]Question 2:
[tex]\begin{gathered} m=\frac{2}{7} \\ x_1=-5,y_1=4 \\ \\ \text{ Thus, the equation is:} \\ y-4=\frac{2}{7}(x-(-6)) \\ \\ y-4=\frac{2}{7}(x+6) \end{gathered}[/tex]Question 3:
[tex]\begin{gathered} m=-\frac{3}{2} \\ x_1=-7,y_1=-10 \\ \\ \text{ Thus, the equation is:} \\ y-(-10)=-\frac{3}{2}(x-(-7)) \\ \\ y+10=-\frac{3}{2}(x+7) \end{gathered}[/tex]Final Answer
Question 1:
[tex]y-6=5(x-3)[/tex]Question 2:
[tex]y-4=\frac{2}{7}(x+6)[/tex]Question 3:
[tex]y+10=-\frac{3}{2}(x+7)[/tex]You Try! As an Equation Example B: A rectangle has a width that is 5 feet less than the length. The area of the rectangle is 126 square feet. Find the dimensions of the rectangle. Write an equation to model this problem and solve it in the sketch area below. 市里。 Entrega
Let the width be w and the length be l
Since the width is 5ft less than the length,
then
w=l-5
The area of the rectangle is 126 square feet
Then
l times w = 126
therefore
l(l-5) = 126
[tex]\begin{gathered} \Rightarrow l^2-5l=126 \\ \Rightarrow l^2-5l-126=0 \end{gathered}[/tex]Factorising the equation, we have:
[tex]\begin{gathered} l^2+9l-14l-126=0 \\ \Rightarrow(l+9)(l-14)=0 \\ \Rightarrow l=-9\text{ or 14} \end{gathered}[/tex]Since l cannot be negative
then the only correct option is l=14
Since w = l - 5
then
w = 14 - 5 = 9
Therefore the length is 14ft and the width is 9ft
The equation of the line of best fit is y= 25x+7.5. What does the y-intercept represent?
y= 25x+7.5
The general equation of a line is given as
y = mx + c
where m is the slope and c is the y-intercept
Comparing with the equation
the y intercept is 7.5
12. One year ago the median price for a home was $342,000. Now the current median price for a home is $215,000. What was the percent decrease in the median price of a home over the last year? Give your answer to at least one decimal place.%
One year ago the median price for a home was $342,000. Now the current median price for a home is $215,000. What was the percent decrease in the median price of a home over the last year? Give your answer to at least one decimal place.
In this problem
$342,000 ------> represents 100%
so
Applying proportion
Find out how much percentage represents the difference (342,000-215,000=127,000)
100/342,000=x/127,000
solve for x
x=(100/342,000)*127,000
x=37.1%If 3 girls can decorate 2 holiday cards in 1.5 minutes, how many holiday cards can 6 girls decorate in 1 hour if they work at the same speed?
Notice that 2*3=6.
On the other hand, according to the question, 3 girls can decorate a total of
[tex]\begin{gathered} \frac{60}{1.5}=40 \\ \Rightarrow40\cdot2=80 \end{gathered}[/tex]80 holiday cards in one hour. Therefore, 6 girls can decorate 2*80=160 holiday cards in the same amount of time.
The answer is 160 holiday cards per hour
Tracey paid $145 for an item that was originally priced at $550. What percent of the original price did Tracey pay?
Tracey paid $145 for an item that was originally priced at $550. What percent of the original price did Tracey pay?
we know that
$550 represent 100%
so
Applying proportion
100/550=x/145
solve for x
x=(100/550)*145
x=26.36%Susan monitors the number of strep infections reported in a certain neighborhood in a given week.The recent numbers are shown in this table:WeekNumber of People020126234344According to her reports, the reported infections are growing at a rate of 30%.If the number of infections continues to grow exponentially, what will the number of infections bein week 10?
SOLUTION
The formula for growth rate is given below;
[tex]y=a(1+r)^x[/tex]Where;
y = number of infections,
a = initial number of infections (20),
r = rate (30% or 0.3) and
x = the week to find (10)
[tex]\begin{gathered} y=a(1+r)^x \\ y=20(1+0.3)^{10} \\ y\text{ = 20 (13.7858)} \\ y\text{ = 275.72} \end{gathered}[/tex]The number of infections in week 10 will be 276.
The function shown by the graph is non-differentiable at x = 4. Why?
A. It has a vertical tangent line at x = 4.
B. It has a jump discontinuity at x = 4.
C. It has a corner at x = 4.
D. It has a vertical asymptote at x = 4.
Answer: In photo below
You're Welcome :)
The reason for the function being non-differentiable at x = 4 is given as follows:
C. It has a corner at x = 4.
What are the conditions for differentiability at a point?A function is differentiable at a point x = a if:
The function is continuous at x = a.The first derivative of the function is continuous at x = a.The continuity of a function at a point x = a is defined as follows:
Same lateral limits.Lateral limits are equal to the numeric value, that is, equal to f(a).Analyzing the graphed function at point x = 4, we have that:
The function is continuous, as the lateral limits are equal to the numeric value.The first derivative of the function is not continuous.The reason for the first derivative being not continuous is that:
To the left of x = 4, the function is increasing, hence it has a positive derivative.To the right of x = 4, the function is constant, hence it has a derivative of zero.Due to the different lateral limits, the first derivative is not continuous and then the function is not differentiable, thus option c is correct.
More can be learned about differentiable functions at brainly.com/question/27864122
#SPJ1
Please help me with this problem I am trying to help my son to understand I have attached what I have helped him with so far just need to be sure i am correct:Solve the system of equations.13x−y=90y=x^2−x−42 Enter your answers in the boxes. ( __,__) and (__,__)
y=xTo solve the system of equations, follow the steps below.
Step 01: Substitute the value of y from equation 2 in equation 1.
In the second equation:
[tex]y=x^2-x-42[/tex]In the first equation:
[tex]13x-y=90[/tex]So, let's substitute y by x² - x - 42.
[tex]\begin{gathered} 13x-y=90 \\ 13x-(x^2-x-42)=90 \\ 13x-x^2+x+42=90 \end{gathered}[/tex]Adding the like terms:
[tex]-x^2+14x+42=90[/tex]Subtracting 90 from both sides:
[tex]\begin{gathered} -x^2+14x+42-90=90-90 \\ -x^2+14x-48=0 \end{gathered}[/tex]Step 02: Use the quadratic formula to solve the equation.
For a quadratic equation ax² + bx + c = 0, the quadratic formula is:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \end{gathered}[/tex]In this question, the equation is -1x² + 14x + -48 = 0, then, teh coeffitients are:
a = -1
b = 14
c = -48
Substituting the values and solving the equation:
[tex]\begin{gathered} x=\frac{-14\pm\sqrt{14^2-4*(-1)*(-48)}}{2*(-1)} \\ x=\frac{-14\pm\sqrt{196-192}}{-2} \\ x=\frac{-14\pm\sqrt{4}}{-2}=\frac{-14\pm2}{-2} \\ x_1=\frac{-14-2}{-2}=\frac{-16}{-2}=8 \\ x_2=\frac{-14+2}{-2}=\frac{-12}{-2}=6 \end{gathered}[/tex]Step 03: Substitute the values of x in one equation and find y.
Knowing that:
[tex]y=x^2-x-42[/tex]Let's substitute x by 6 and 8 and find the ordered pairs that are the solution of the system.
First, for x = 8:
[tex]\begin{gathered} y=8^2-8-42 \\ y=64-8-42 \\ y=14 \end{gathered}[/tex]Second, for x = 6:
[tex]\begin{gathered} y=6^2-6-42 \\ y=36-48 \\ y=-12 \end{gathered}[/tex]So, the solutions for the system of equations are (8, 14) and (6, -12).
Answer: (8, 14) and (6, -12).
Show that (fof-1) (x)=x and (F-1 of)(x) = x for the following pair of functions.f(x) = 5-4x, f'(x) =5-754Show that (fof-1)(x) = x.(fof- ')(x) = 0Write the expression for the composition.= x(Do not simplify. Type an exact answer, using radicals as needed.)
To obtain the expression for the composition of the function of the inverse function of x, the following steps are necessary:
Step 1: Write out the expression for the function of x and the inverse function of x, as below:
[tex]\begin{gathered} f(x)=\sqrt[5]{5-4x} \\ ^{}f^{-1}(x)=\frac{5-x^5}{4} \end{gathered}[/tex]Step 2: Write out the expression for the composition of the function of the inverse function of x, as below:
[tex](f^{-1}Of))(x)=^{}\frac{5-(\sqrt[5]{5-4x})^5}{4}[/tex]Thus, the above is how the expression for the composition of the function of the inverse function of x is to be written out
y=-2/3x +7y = 2x - 3For each equation, write the slope and y-intercept.Graph both equations on the same graph (do this on paper)
The given equations of the lines are in their slope-intercept form, that is:
[tex]\begin{gathered} y=mx+b \\ \text{ Where m is the slope and} \\ b\text{ is the y-intercept of the line} \end{gathered}[/tex]Then, we have:
• First equation
[tex]\begin{gathered} y=-\frac{2}{3}x+7 \\ \boldsymbol{m=-\frac{2}{3}} \\ \boldsymbol{b=7} \end{gathered}[/tex]• Second equation
[tex]\begin{gathered} y=2x-3 \\ \boldsymbol{m=2} \\ \boldsymbol{b=-3} \end{gathered}[/tex]Now, to graph the first equation, we can find two points through which the line passes:
• First point
If x = 3, then we have:
[tex]\begin{gathered} y=-\frac{2}{3}x+7 \\ y=-\frac{2}{3}\cdot3+7 \\ y=-2+7 \\ y=5 \end{gathered}[/tex]That means that the line passes through the point (3,5).
• Second point
If x = 6, then we have:
[tex]\begin{gathered} y=-\frac{2}{3}x+7 \\ y=-\frac{2}{3}\cdot6+7 \\ y=-2\cdot2+7 \\ y=-4+7 \\ y=3 \end{gathered}[/tex]That means that the line passes through the point (6,3).
To graph the second equation, we can find two points through which the line passes:
• First point
If x = 3, then we have:
[tex]\begin{gathered} y=2x-3 \\ y=2\cdot3-3 \\ y=6-3 \\ y=3 \end{gathered}[/tex]That means that the line passes through the point (3,3).
• Second point
If x = 4, then we have:
[tex]\begin{gathered} y=2x-3 \\ y=2\cdot4-3 \\ y=8-3 \\ y=5 \end{gathered}[/tex]That means that the line passes through the point (4,5).
Now that we know two points through which each line passes, we can graph them and then join them to obtain the graph of both equations:
For the following exercise, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptotes of the function. use that information to sketch graph.
Answer:
The expression is given below as
[tex]a(x)=\frac{x^2+2x-3}{x^2-1}[/tex]The horizontal intercepts will be at y=0
[tex]\begin{gathered} a(x)=\frac{x^2+2x-3}{x^2-1} \\ \frac{x^2+2x-3}{x^2-1}=0 \\ x^2+2x-3=0 \\ x^2+3x-x-3=0 \\ x(x+3)-1(x+3)=0 \\ (x-1)(x+3)=0 \\ x-1=0,x+3=0 \\ x=1,x=-3 \end{gathered}[/tex][tex]\begin{gathered} x^2-1=0 \\ x^2=1 \\ x=\pm1 \\ x=1,x=-1 \end{gathered}[/tex]Hence,
The horizontal intercepts is at x = -3
The vertical intercept is at x=0
[tex]\begin{gathered} a(x)=\frac{x^2+2x-3}{x^2-1} \\ y=\frac{0^2+2(0)-3}{0^2-1} \\ y=\frac{-3}{-1} \\ y=3 \end{gathered}[/tex]Hence,
The vertical intercept is at y=3
A vertical asymptote is a vertical line that guides the graph of the function but is not part of it. It can never be crossed by the graph because it occurs at the x-value that is not in the domain of the function. A function may have more than one vertical asymptote.
[tex]\begin{gathered} a(x)=\frac{x^2+2x-3}{x^2-1} \\ a(x)=\frac{(x-1)(x+3)}{(x-1)(x+1)_{}} \\ \text{hence, the vertical aymspote will be at} \\ x+1=0 \\ x=-1 \end{gathered}[/tex]Hence,
The vertical asymptotes is at x= -1
The horizontal asymptotes will be calculated using the image below
[tex]\begin{gathered} a(x)=\frac{x^2+2x-3}{x^2-1} \\ a=1,b=1,n=m=1 \\ y=\frac{a}{b} \\ y=\frac{1}{1} \\ y=1 \end{gathered}[/tex]Hence,Do you have any questions about the steps to solve your question?
The horizontal asymptotes is y=1
The graph is represented below as
Convert the following angle from degrees to radians. Express your answer in simplestform.150°
Recall that:
[tex]2\pi radians=360^{\circ}.[/tex]Therefore:
[tex]150^{\circ}=\frac{150*2\pi}{360^}\text{ radians.}[/tex]Simplifying the above result, we get:
[tex]150^{\circ}=\frac{5}{6}\pi.[/tex]Answer: [tex]\begin{equation*} \frac{5}{6}\pi. \end{equation*}[/tex]D=2730 mi t=9.75 h find r
In order to calculate the rate r (in this case, the speed), we can use the following formula:
[tex]\begin{gathered} \text{distance}=\text{speed}\cdot\text{time} \\ d=r\cdot t \end{gathered}[/tex]So, using the values of d = 2730 and t = 9.75, we have:
[tex]\begin{gathered} 2730=r\cdot9.75 \\ r=\frac{2730}{9.75} \\ r=280 \end{gathered}[/tex]So the value of r is equal to 280 miles per hour.
can you please solve this problem then tell me what was wrong with the answer
Given:
[tex]64^{\circ\text{ }}and\text{ }x^{\circ}[/tex]Sum of the angles of Same side interior angles is 180 degree.
[tex]\begin{gathered} 64+x=180 \\ x=180-64 \\ x=116^{\circ} \end{gathered}[/tex]The volume of this cone is 53,851 cubic yards. What is the height of this cone? Use pi = 3.14 and round your answer to the nearest hundredth.
Volume of a cone = 1/3 π r^2 h
Where:
r= radius = 35 yd
h= height
π = 3.14
Replacing with the values given:
53,851 = 1/3 (3.14) 35^2 h
Solve for h
53,851 / (1/3 (3.14) 35^2)= h
h= 42
Analyze the data in the line plot "attached "Use the data to construct a line plot.Number of students in a classroom:22, 28, 31, 33, 28, 29, 31, 28, 29, 32, 27, 18, 29, 31, 30, 31, 32, 27, 29, 33
Answer:
Most classrooms have between 29-31 students.
Step-by-step explanation:
To construct a line plot, create a number line that includes all the numbers or values in the data set. Then place an x over each data value on the number line, if a value occurs more than one place x's as necessary:
Write the following fractions in thesimplest form:1. 20/52. 12/963. 100/2
So the first fraction is:
1)
[tex]\frac{20}{5}[/tex]So we have to find a number that is multiple of 20 and 5, in this case is the number 5. the we have to divide the two numbers in 5:
[tex]\frac{5}{5}=1\to\text{ }\frac{20}{5}=4[/tex]So at the end the fraction is going to be equal to:
[tex]\frac{20}{5}=4[/tex]2)
[tex]\frac{12}{96}[/tex]and 12 and 96 are multiples of two, so we can divide by 2
[tex]\frac{12}{2}=6\to\frac{96}{2}=48[/tex]now we need to find other number that is multiple of 6 and 48, and agins is the number 2:
[tex]\frac{6}{2}=3\to\frac{48}{2}=24[/tex]Now 3 and 24 are multiples of 3, so we make the same procedure:
[tex]\frac{3}{3}=1\to\frac{24}{3}=8[/tex]So at the end is going to be:
[tex]\frac{12}{96}=\frac{1}{8}[/tex]3)
[tex]\frac{100}{2}[/tex]100 and 2 are multiples of 2, so we divide them by 2:
[tex]\frac{100}{2}=50\to\frac{2}{2}=1[/tex]So the final Resold will be:
[tex]\frac{100}{2}=\frac{50}{1}=50[/tex]