If y varies directly as x, we have that
[tex]y\propto x[/tex]Then
[tex]y=kx(where\text{ k is a constant\rparen}[/tex][tex]\begin{gathered} When\text{ y=84 , x= 6} \\ y=kx \\ 84=6k \\ k=\frac{84}{6}=14 \end{gathered}[/tex]The relationship between x and y is given as
[tex]\begin{gathered} y=kx \\ y=14x \end{gathered}[/tex]Therefore when x= 12, y=?
[tex]\begin{gathered} y=14x \\ y=14\times12=168 \end{gathered}[/tex]Hence, the value of y when x = 12 is 168
Final answer: y = 168
what doesnt belong and why? please someome help me will make brainlist
The one that doesn't belong is 4² = 4² + 4²
Explanation:
2² = 2 × 2
4² = 4 × 4
4² is not equal to 4² + 4²
this is because 4² + 4² = 16 + 16 = 32
while 4² = 4 × 4 = 16
The one that doesn't belong is 4² = 4² + 4²
Find a_1 for the geometric sequence with the given terms. a_3 = 54 and a_5 = 486
ANSWER
[tex]6[/tex]EXPLANATION
We want to find the first term of the sequence.
The general equation for the nth term a geometric sequence is written as:
[tex]a_n=ar^{n-1}[/tex]where a = first term; r = common ratio
Let us use this to write the equations for the third term and the fifth term.
For the third term, n = 3:
[tex]\begin{gathered} a_3=ar^2 \\ \Rightarrow54=ar^2 \end{gathered}[/tex]For the fifth term, n = 5:
[tex]\begin{gathered} a_5=ar^4 \\ \Rightarrow486=ar^4 \end{gathered}[/tex]Let us make a the subject of both formula:
[tex]\begin{gathered} 54=ar^2_{} \\ \Rightarrow a=\frac{54}{r^2} \end{gathered}[/tex]and:
[tex]\begin{gathered} 486_{}=ar^4 \\ a=\frac{486}{r^4} \end{gathered}[/tex]Now, equate both equations above and solve for r:
[tex]\begin{gathered} \frac{54}{r^2}=\frac{486}{r^4} \\ \Rightarrow\frac{r^4}{r^2}=\frac{486}{54} \\ \Rightarrow r^{4-2}=9 \\ \Rightarrow r^2=9 \\ \Rightarrow r=\sqrt[]{9} \\ r=3 \end{gathered}[/tex]Now that we have the common ratio, we can solve for a using the first equation for a:
[tex]\begin{gathered} a=\frac{54}{r^2} \\ \Rightarrow a=\frac{54}{3^2}=\frac{54}{9} \\ a=6 \end{gathered}[/tex]That is the first term.
Write the equation of a circle with a radius of 8 and a center at (4, -9).
The equation of a circle with radius r and center at (h,k) is given by:
(x + h)² + (y + k)² = r²
If r = 8 and (h,k) = (4,-9), we have:
(x-4)² + (y + 9)² = 8²
What would be the transformation of point A when reflected across the x-axis?(1, -2)(-1, 2)(-1, -2)(1, 2)
Given:
The coordinates of point A is (-1, 2).
To find:
The reflection of point A across the x-axis.
Solution:
It is known that the reflection of a point (x, y) about the x-axis is (x, -y).
So, the reflection of point A is (-1, -2).
Thus, option C is correct.
Graph the function by starting with the graph of the basic function and then using the techniques of shifting, compressing, stretching, and/or reflecting.f(x) = -2(x + 1)2 + 2
Okay, here we have this:
Considering the provided function, we are going to graph it, so we obtain the following:
First, we are going to start with the graph of the basic function, so the graph so far is:
Now, we are going to add 1 inside the square, which is reflected in a shift of the graph one unit to the left, so we have:
We continue, multiplying the entire function by -2, which results in a compression of the graph, and in a reflection on the x-axis due to the change of sign, now we have:
And finally we add two units to the whole function, therefore it means that the graph will move up two units, leaving the following graph:
Finally we obtain that the correct answer is the first answer choice.
Write the slope-intercept form of the equation of the line describedthrough: (-5, 4), perp. to y=3/4x+4
Answer:
To find the equation of the line passes through (-5,4) and perpendicular to the line y=3/4 x +4
The slope of the line y=3x/4 +4 is 3/4
Let m be the slope of the required equation of the line,
Since these two lines are perpendicular to each other.
we know that,
Product of the slopes of the perpendicular lines is -1
That is,
[tex]\frac{3}{4}\times m=-1[/tex][tex]m=-\frac{4}{3}[/tex]Slope of the required line is -4/3
Passes through the point (-5,4)
The equation of the line passes through the point (x1,y1) and m as its slope is
[tex](y-y1)=m(x-x1)[/tex]Substitute x1=-5, y1=4 and m=-4/3
we get,
[tex](y-4)=-\frac{4}{3}(x+5)[/tex][tex]y-4=-\frac{4}{3}x-\frac{20}{3}[/tex][tex]y=-\frac{4}{3}x-\frac{20}{3}+4[/tex][tex]y=-\frac{4}{3}x+\frac{(-20)+12}{3}[/tex][tex]y=-\frac{4}{3}x-\frac{8}{3}[/tex]The required slope-intercept form of the equation of the line is y=-4/3 x-8/3.
What is the difference of the complex numbers below? (6+81)-(1-21 ) O A. 7+10) O O O B. 5+10) O C. 5+67 O 0 7 D. 7 +62
we have the expression
[tex](6+8i)-(1-2i)[/tex]Remove parenthesis
[tex]6+8i-1+2i[/tex]Group similar terms
[tex](6-1)+(8i+2i)[/tex]Combine like terms
[tex]5+10i[/tex]The answer is option BHi. I think I am over thinking this question. Can you show me how this works step by step?
We know that:
MN = 7.3
DC = 8.7
M and N are midpoints of AD and BC respectively.
Since DC - MN = 8.7 - 7.3 = 1.4 and M and N are midpoints, we must have:
AB = 7.3 - 1.4 = 5.9
Consider the equation below.x3 – 3x2 – 4 = 1/x-1+ 5The solutions to the equation are approximately x=and x=
Question:
Solution:
Consider the following equation:
[tex]x^3-3x^2-4=\frac{1}{x-1}+5[/tex]this is equivalent to:
[tex]x^3-3x^2=\frac{1}{x-1}+5+4[/tex]that is:
[tex]x^3-3x^2=\frac{1}{x-1}+9[/tex]Multiplying both sides by (x-1), we obtain:
[tex](x-1)(x^3-3x^2)=1+9(x-1)[/tex]this is equivalent to:
[tex](x-1)x^3-3x^2(x-1)=1+9(x-1)[/tex]solving for x, we obtain that the correct solutions are:
[tex]x\text{ }\approx0.90672[/tex]and
[tex]x\approx\: 3.68875[/tex]Approximate the median in each of the three graphs. Explain how you determined the answer.
The median is the centermost data in a set of values.
Since these graphs are for the year 2013, meaning, these graphs contained 365 days of recorded temperature.
For 365 days, the centermost data would be the 183rd day.
Let's take a look at each graph what is the temperature on the 183rd day when data is arranged from lowest to the highest temperature.
As we can see above, the first two bins of the histogram only cover 180 days. Since we are looking for the 183rd day, we moved to the third bin. Hence, the median temperature for City A is between 65 to 75 ℉.
For City B,
The 183rd day in City B is found on the interval 55-75. Hence, the median temperature for City B is between 55-75℉.
Lastly, for City C:
As we can see in the graph above, the 183rd day is approximately closed to the third bin. Hence, for the 183rd day, we can say that the temperature is between 65 to 75℉. The median temperature for City C is between 65 to 75℉.
For which value of x does p(x)=-4 in the graph below
You have to identify which dot in the graph corresponds to p(x)=-4
p(x)=-4 → this expression indicates that the value of the "output" is -4, in the graph, it will correspond to the dot that has y-coordinate= -4
The dots in the graph have the following coordinates:
The coordinates are always given in the following order (x,y), the first coordinate corresponds to the value of x (input) and the second coordinate corresponds to the value of y (output)
From the dots, the only one that has the y-coordinate -4 is the one located in the fourth quadrant with coordinates (2,-4)
How to find slope & y interceptAnd solve for Y-6x+2y=10
Slope intercept form
y= mx + b
Where:
m= slope
b= y-intercept
So, first, we have to solve for y:
-6x +2y = 10
2y = 6x + 10
y = (6x + 10) /2
y = 3x + 5
Slope = 3
y-intercept = 5
The point (-2, 1) is translated 3 units down and 2 units to the left. What are the coordinates of the newpoint?Use the grid below if it helps.-65-43-214-4-3-2-1130-1--2-3-4-5-6(2,-4)o(-4,-2)(0, -4)(0,4)
Given that the point (-2, 1) is translated 3 units down and 2 units to left. The coordinate of the new point would follow the translation rule
[tex]T(x,y)\to(x-2,y-3)[/tex]Therefore, the coordinates of the new point following the translation rule are:
[tex]\begin{gathered} T(-2,1)\to_{}(-2-2,1-3) \\ =(-4,-2) \end{gathered}[/tex]The correct answer is (-4,-2)
According to the graph, what is the solution to this system of equations? O (-4,-3) 0 (-3,-4) O (-5, 5) O (5,-5)
The intersection of the 2 lines is the point of the solution of the graph . Using the graph above , the meeting point of the 2 lines are
[tex](-4,-3)[/tex]There are as many even counting numbers as there are counting numbers. Is this true or false?
The sets of even and odd counting numbers are both infinite in size (number of elements). However, we can map each even counting number to each odd counting number as follows:
[tex]\begin{gathered} 2\rightarrow1 \\ 4\rightarrow3 \\ 6\rightarrow5 \\ 8\rightarrow7 \\ \text{And so on}\ldots \end{gathered}[/tex]So we have the mapping rule:
[tex]\begin{gathered} 2n\rightarrow2n-1 \\ \text{Where }n=1,2,3,\ldots \end{gathered}[/tex]Then, we can say that there are as many even counting numbers as there are counting numbers, or equivalently, that both sets have the same cardinality.
Answer: True
Makayla was scuba diving. She started at at-80 5/9 meters below the surface. She then swam up 20 2/9 meters from her storting location for a break. Alwhat location did she stop for her break compared to sea level?
Answer:
543/9 meters below the surface.
Explanation:
First, we need to transform the mixed numbers into fractions, so 80 5/9 meter and 20 2/9 meters are equivalent to:
[tex]\begin{gathered} A\frac{b}{c}=\frac{A\cdot c+b}{c} \\ 80\frac{5}{9}=\frac{80\cdot9+5}{9}=\frac{725}{9} \\ 20\frac{2}{9}=\frac{20\cdot9+2}{9}=\frac{182}{9} \end{gathered}[/tex]Now, to calculate the location where she stops for a break, we need to take 182/9 and subtract it from 725/9. So:
[tex]\frac{725}{9}-\frac{182}{9}=\frac{725-182}{9}=\frac{543}{9}[/tex]Therefore, Makayla stops for a break at 543/9 meters below the surface.
At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 32 minutes and a standard deviation of 5 minutes. What is the probability that a randomly selected customer will have to wait less than 21 minutes, to the nearest thousandth?
Given:
[tex]\begin{gathered} mean(\mu)=32 \\ Standard-deviation(\sigma)=5 \end{gathered}[/tex]To Determine: The probability that a randomly selected customer will have to wait less than 21 minutes, to the nearest thousandth
Solution
Using normal distribution formula below
[tex]P(Z<\frac{x-\mu}{\sigma})[/tex]Substitute the given into the formula
[tex]\begin{gathered} P(Z<\frac{21-32}{5}) \\ =P(Z<-\frac{11}{5}) \\ =P(Z<-2.2) \end{gathered}[/tex][tex]\begin{gathered} P(X<-2.2)=1-P(X>-2.2) \\ =1-0.986097 \\ =0.01390345 \\ \approx0.014(Nearest\text{ thousandth\rparen} \end{gathered}[/tex]Hence, the probability that a randomly selected customer will have to wait less than 21 minutes, to the nearest thousandth is 0.014
(x²y³z)^1/5Rewrite the rational exponent expression as a radical expression.
Given the following rational exponent:
(x²y³z)^1/5
Putting it into a radical expression will be:
[tex]\text{ \lparen x}^2y^3z)^{\frac{1}{5}}\text{ = }\sqrt[5]{x^2y^3z^}[/tex]Which line is parallel to this one: y=2/3x-9A.y=3/2x+8B.y=2/3x-9C.y=2/3x-1D.y=-3/2x+7
to find the line parallel to th egiven line:
[tex]y=\frac{2}{3}x-9[/tex]the line parallel to the given equation is
[tex]y=\frac{2}{3}x-1[/tex]The graph is,
Graph J(2,-1), K(4,-5), and L(3,1) and reflect across the x=-1. Please draw the line of reflection.
You have the following points:
J(2,-1), K(4,-5), L(3,1)
To reflect the previous points around the line x = -1, consider the horizontal distance of each point to the given line x=-1. The reflected point is obtained by using the same distance to the line but in the other side.
You proceed as follow:
J(2,-1)
the distance of the previous point to the line x=-1 is 2-(-1) = 3. You subtract this value to x = -1. Thus, the x-coordinate of the new point is:
-1-3 = -4
and the new point is:
J(2,-1) => J'(-4,-1)
For K(4,-5) you have:
distance to the line x=-1 is 4-(-1) = 5. Subtract this value to the line.
-1-5 = -6
and the new point is:
K(4,-5) => K'(-6,-5)
For L(3,1):
distance to the line x=-1 is 3-(-1) = 4.
-1-4 = -5
and the new point is:
L(3,1) => L'(-5,1)
A plot of the original and reflected points is given below:
where the figure with black lines is the original figure and the figure with blue lines is the reflected one.
a line intercepts the point (-2, 1) and it has a slope of 2 input the correct values into the point slope formula. y- ? = ? ( x- ?)
We know that the point-slope forumal is given by:
[tex]y-y_1=m(x-x_1)[/tex]where the terms: m, y₁ and x₁ are numbers.
Given that the line passes through a given points we have that:
m is the slope of the line
y₁ is the y value of that given point
x₁ is the x value of that given point
In this case we have that
1. It has a slope of 2, then m = 2
2. The line intercepts the point (-2, 1). Since (-2, 1) = (x₁, y₁) then
x₁ = -2 and y₁ = 1
We have that
m = 2
x₁ = -2
y₁ = 1
Let's replace those values in the first formula:
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \downarrow \\ y-1=2(x-(-2)) \\ y-1=2(x+2) \end{gathered}[/tex]Answer- the point-slope formula of this line is: y - 1 = 2(x + 2)7. Given an arithmetic sequence, find a26if a4 = 71 and a32 = 1. 8. Given a geometric sequence, find az ifa1 = 729 and a2 = – 243.
Arithmetic sequence
a= 26
a4 = 71
a32 = 1.
an = a1 + (n-1)d
a4 = a1 + (3)*d
71 = a1 + (3)*d (I)
a32 = a1 + (31)*d
1 = a1 + 31 d (II)
to find a1 and d we subtract (II)- (I)
1 = a1 + 31 d (II)
-( 71 = a1 + (3)*d) (I)
________________
-70 = 0 28 d
-70 = 28 d
d= -70/ 28
d= -5/2
Replacing d in (I)
71 = a1 - (3)(5/2)
a1= 71 + (3)(5/2)
a1= 78.5
The arithmetic sequence is
an = a1 + (n-1)d
an = 78.5 + 2.5 (n-1)
Verifying
a4 = 78.5 - 2.5 (3) = 78.5 -7.5= 71
a32 = 78.5 +2.5 (31) =78.5 -77.5 = 1
a26 = 78.5 + 2.5 (25) = 78.5 - 62.5 = 16
Give me some minutes
______________________
Answer
a26 = 16
Problem 12a) solve for x, 3(-2x +5) = 9(x-5), x=b) x = –27, x =
The given equation is
[tex]3(-2x+5)=9(x-5)[/tex]Dividing by 3 both sides, we get
[tex]\frac{3\mleft(-2x+5\mright)}{3}=\frac{9\mleft(x-5\mright)}{3}[/tex][tex]-2x+5=3(x-5)[/tex]Multiplying 3 and (x-5) as follows.
[tex]-2x+5=3\times x-3\times5[/tex][tex]-2x+5=3x-15[/tex]Adding 15 on both sides, we get
[tex]-2x+5+15=3x-15+15[/tex][tex]-2x+20=3x[/tex]Adding 2x on both sides, we get
[tex]-2x+20+2x=3x+2x[/tex][tex]20=5x[/tex]Dividing by 5, we get
[tex]\frac{20}{5}=\frac{5x}{5}[/tex][tex]4=x[/tex]Hence the value of x is 4.
what digit is in the
7 more than t
The algebraic expression is:
[tex]t+7[/tex]Answer: t + 7
Demetrius is single but would like to take care of his mother in the event of his death. He would like to replace her income of $25,000 a year with a multiple of 30. How much life insurance should he buy?
Answer:
750000
Step-by-step explanation:
QuestionFind the equation of the line with slope -2 which goes through the point (7, -1).Give your answer in slope-intercept form y = mx +b.Provide your answer below:
From the question, we can deduce the following:
• slope, m = -2
,• The line goes through the points: (7, -1)
Let's find the equation of the line in slope-intercept form.
Apply the slope-intercept formula of a line:
y = mx + b
Where m is the slope and b is the y-intercept.
Here, the slope, m is = -2
We also have the point:
(x, y) ==> (7, -1)
Substitue -2 form, 7 for x, and -1 for y to find b.
y = mx + b
-1 = -2(7) + b
-1 = -14 + b
Add 14 to both sides:
-1 + 14 = -1 + 14 + b
13 = b
The y-intercept is 13.
Therefore, the equation of the line in slope-intercept form is:
y = -2x + 13
ANSWER:
y = -2x + 13
A baseball player has a batting average of 0.33. What is the probability that he has exactly 4 hits in his next 7 atbats? Round to 3 decimal places.The probability is
Given that the player can or cannot hit the ball, then this situation can be modeled with the binomial distribution.
Binomial distribution formula
[tex]P=_nC_xp^x(1-p)^{n-x}^{}[/tex]where
• P: binomial probability
,• nCx: number of combinations
,• p: probability of success in a single trial
,• x: number of times for a specific outcome within n trials
,• n: number of trials
Substituting with n = 7, x = 4, and p = 0.33, we get:
[tex]\begin{gathered} P=_7C_4(0.33)^4(1-0.33)^{7-4} \\ P=35(0.33)^4(0.67)^3 \\ P\approx0.125 \end{gathered}[/tex]The probability is 0.125
if D equals 6x - 7 + 8 equals 4x + 9 find the DB
If DE = 6x - 7 and AE = 4x + 9, find DB.
DE and AE are line segments. We need to equal DE = AE
[tex]DE=AE[/tex][tex]6x-7=4x+9[/tex][tex]6x-4x=9+7[/tex][tex]2x=16[/tex][tex]x=8[/tex]Now, DB = DE + EB
[tex]DB=6x-7+4x+9[/tex][tex]DB=10x+2[/tex]x = 8
[tex]DB=10(8)+2\rightarrow DB=82[/tex]DB is 82
9) write in standard formthrough: (4,4), parallel to y=-6x + 5
Data
Point (4, 4)
Equation
y = -6x + 5
Procedure
As the straight line is parallel we use the same slope of the original straight line.
m = -6
Now we will calculate y-intercept
[tex]\begin{gathered} b=y-mx \\ b=4-(-6)\cdot4 \\ b=4+24 \\ b=28 \end{gathered}[/tex]The equation would be:
[tex]y=-6x+28[/tex][tex]6x+y=28[/tex]Now in the standard form: 6x+y=28
B. The perimeter of this rectangle is 20 centimeters. What is the value of X
Statement Problem: Find the value of x in the diagram below, given the perimeter of a rectangle as 20centimeters.
Solution:
The perimeter of a rectangle is;
[tex]P=2(l+w)[/tex]Where the length and width of the given rectangle is;
[tex]\begin{gathered} l=(x+3)cm \\ w=(x+1)cm \end{gathered}[/tex]Thus, the value of x is;
[tex]\begin{gathered} 2(l+w)=20 \\ 2(x+3+x+1)=20 \\ \text{Divide both sides by 2},\text{ we have;} \\ \frac{2\mleft(x+3+x+1\mright)}{2}=\frac{20}{2} \\ x+3+x+1=10 \\ \text{Collect like terms, we have;} \\ 2x+4=10 \\ \end{gathered}[/tex]Then, we subtract 4 from both sides of the equation, we have;
[tex]\begin{gathered} 2x+4-4=10-4 \\ 2x=6 \\ \text{Divide both sides by 2, we have;} \\ \frac{2x}{2}=\frac{6}{2} \\ x=3 \end{gathered}[/tex]The value of x is 3