Answer:
One-third of the difference of h and 1
Two functions, function A and function B, are shown below:Function Axy714816918Which statement best compares the rate of change of the two functions?The rate of change of both functions is 2.The rate of change of both functions is 3.The rate of change of function A is greater than the rate of change of function B.The rate of change of function B is greater than the rate of change of function A.
Answer
The rate of change of both functions is 2.
Explanation
To know the statement that best compares the rate of change of the two functions, we need to first calculate the rate of change for each function.
Rate of change of function A
Using x₁ = 7, y₁ = 14, x₂ = 8 and y₂ = 16
Rate of change = Δy/Δx
Δy = (y₂ - y₁) = 16 - 14 = 2
Δx = (x₂ - x₁) = 8 - 7 = 1
⇒ Rate of change = 2/1 = 2
Rate of change of function B
From the graph
Using coordinate x₁ = 2, y₁ = 4, x₂ = 3 and y₂ = 6
Rate of change = Δy/Δx
Δy = (y₂ - y₁) = 6 - 4 = 2
Δx = (x₂ - x₁) = 3 - 2 = 1
⇒ Rate of change = 2/1 = 2
Since the rate of both functions are the same (2), then the statement that best compares the rate of change of the two functions in the options given is "The rate of change of both functions is 2"
what are three requirements for fully defining a reference point?
1 - reference point should consist of abstract coordinates.
2- it should be stationary
3- it should be related to all the variables in the frame.
segment C prime D prime has endpoints located at C′(0, 0) and D′(4, 0). It was dilated at a scale factor of one half from center (4, 0). Which statement describes the pre-image?A-segment CD is located at C(2, 0) and D(6, 0) and is half the length of segment C prime D prime periodB- segment CD is located at C(2, 0) and D(6, 0) and is twice the length of segment C prime D prime periodC- segment CD is located at C(−4, 0) and D(4, 0) and is twice the length of segment C prime D prime periodD-segment CD is located at C(−4, 0) and D(4, 0) and is half the length of segment C prime D prime period
Segment C prime D prime has endpoints located at C′(0, 0) and D′(4, 0). It was dilated at a scale factor of one half from centre (4, 0). the pre-image
B- segment CD is located at C(2, 0) and D(6, 0) and is twice the length of segment C prime D prime period
According to the question,
Segment C prime D prime has endpoints located at C' (0, 0) and D' (2, 0).
The coordinates are given as:
C' (0, 0) and D' (4, 0).
Since,
Centre of dilation = D = (4,0)
Here, CD seems to be the dilated image of CD by something like a factor of two. It follows that M must have been at (0,0).
It's one-half units left from the centre of dilated.
Then, C` = 1/2 x 4 = 2
Since the dilation is (4, 0),
C = (2+4, 0) = (6,0)
Hence,
segment CD is located at C(2, 0) and D(6, 0) and is twice the length of segment C prime D prime period
What is segment?Segment simplifies data collection and integrates new tools, allowing you to spend more time using data and less time collecting it. A segment allows you to track events that occur when a user interacts with user interfaces. "Interfaces" is the segment's umbrella term for all the digital real estate you own: your website, mobile apps and processes running on a server or OTT device.
When you capture interaction data in a segment, you can send it (often in real time) to your marketing, product and analytics tools and data warehouses. In most cases, you don't even need to touch the tracking code to connect to the new tools.
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Use a calculator to evaluate the expression. (Do not round until the final answer. Then round to three decimal places as needed.)
2.303
1) For the following expression:
[tex]\frac{\ln30+\ln15}{\log_{10}30+\log_{10}15}[/tex]We can simplify that and then round it off to the nearest thousandth:
2) Let's rewrite them simplifying using the logarithm property of multiplication:
[tex]\begin{gathered} \frac{\ln30+\ln15}{\log_{10}30+\log_{10}15}= \\ \frac{\ln(30\cdot15)}{\log_{10}30+\log_{10}15}= \\ \frac{\ln(30\cdot15)}{\log_{10}(30\cdot15)}= \\ \frac{\ln(450)}{\log_{10}(450)}= \end{gathered}[/tex]Note that the base of the Natural Log is the Euler's number "e" so let's move on now using the calculator, finally:
[tex]\frac{\ln(450)}{\log_{10}(450)}=\frac{6.10924}{2.65321}=2.30258\ldots\approx2.303[/tex]Note that only at the last step we have rounded it off. And that's the
answer
Melina made a scale drawing of a building.She used a scale in which 0.5 inch represents 1 foot. Which graph represents this relationship?
From the graph, the y - axis 10 uints while the x - axis is 5 units
The x - axis is labeled inches and its half of the feet
For every half inch on x - axis you have 1 feet
The graph that displays the scale is graph D
The answer is OPTION D
it says find x 110° x and 25° in a triangle
There are two known angles in such a manner:
We know that the sum of internal angles of a triangle is equal to 180 degrees. This means that we can find the missing angle by adding all the internal angles and making it equal to 180
[tex]\begin{gathered} 25+110+x=180 \\ 135+x=180 \\ x=180-135 \\ x=45 \end{gathered}[/tex]The missing angle is 45 degrees.
[tex]f(x) = (x - 2) ^{2}(x + 3)(x + 1)^{2} [/tex]the multiplicity of the root x=2 is...?
The solution of the factor with power 2 in the function f(x) can be found as:
(x-2)=0
x=2.
So, the root is x=2.
The multiplicity is the power of the factor (x-2) with its root given as x=2.
So, the multiplicity of the root x=2 is 2.
The wholesale price for a bookcase is 152$. A certain furniture marks up the wholesale price by 36%. find the price of the bookcase in the furniture store. round answer by the nearest cent, as necessary
The price of the bookcase in the funiture store is:
$206.72
Explanation:Given that the markup is 36% of $152
This is:
0.36 * 152 = $54.72
Therefore, the price of the bookcase in the funiture store is:
$152 + $54.72
= $206.72
In an all boys school, the heights of the student body are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. What is the probability that a randomly selected student will be taller than 71 inches tall, to the nearest thousandth?
The probability that a randomly selected student will be taller than 71 inches tall is 0.010.
We use z score formula to calculate :
z = (x-μ)/σ
where,
z = standard score
x = observed value
μ = mean of students height
σ = standard deviation of students height
x = 63 inches
μ = 70 inches
σ = 3 inches
For x shorter than 63 inches we calculate
Z = (x - μ)/σ
then put the given values in above equation.
= (63 - 70)/3
= -2.33333
Probability value is :
P(x<63) = 0.0098153
Approximately to the nearest thousandth = 0.010
The probability that a randomly selected student will be taller than 71 inches tall is 0.010.
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the volume v of a fixed amount of a gas variety directly as the temperature T and inversely as the pressure P. suppose that V =42cm3 when T=84 kelvin and P=8kg/cm2 find the temperature when V =74cm3 and P=10 kg/cm2
ok
I'll use the law of gases to solve this problem
V1 = 42 cm^3
T1 = 84 °K
P = 8 Kg/cm^2
T2 = x
V2 = 74 cm^3
P2 = 10 kg/cm^2
Equation
P1V1/T1 = P2V2/T2
Solve for T2
T2 = P2V2T1 / P1V1
Substitution
T2 = (10*74*84) / (8*42)
Simplification
T2 = 62160 / 336
Result
T2 = 185°K
Find the equation for the line that passes through the point (1,0), and that is perpendicular to the line with the
step 1
Find out the slope of the given line
we have
-(4/3)x+2y=4/3
isolate the variable y
2y=(4/3)x+(4/3)
Divide both sides by 2
y=(4/6)x+(4/6)
simplify
y=(2/3)x+(2/3)
the slope is m=2/3
Remember that
If two lines are perpendicular, then their slopes are negative reciprocal
that means
the slope of the perpendicular line to the given line is
m=-3/2
step 2
Find out the equation in slope-intercept form of the perpendicular line
y=mx+b
we have
m=-3/2
point ( 1,0)
substitute and solve for b
0=-(3/2)(1)+b
0=-(3/2)+b
b=3/2
therefore
the equation is
y=-(3/2)x+(3/2)ory=-1.5x+1.5Two planes, which are 2320 miles apart, fly toward each other. Their speeds differ by 80 mph. If they pass each other in 4 hours,what is the speed of each?Step 1 of 2: Use the variable x to set up an equation to solve the given problem. Set up the equation, but do not take steps to solve it.
Given the word problem, we can deduce the following information.
1. Two planes, which are 2320 miles apart, fly toward each other.
2. Their speeds differ by 80 mph.
3. They pass each other in 4 hours.
To find the speed of each plane, we use the formula:
distance = (rate)(time)
Since they are flying towards each other, the sum of both speeds is 2x+80. So,
distance = (rate)(time)
2320 miles = (2x+80 mph)(4 hrs)
Thus, the equation to solve this is:
2320 = (2x+80)(4)
Company A has a monthly budget of 2 x 10^4 dollars. Company B has
a monthly budget of 5 x 10^8 dollars. How many times greater is the
monthly budget for company B than for company A?
The budget is 20000 times greater.
What are basic arithmetic?Mathematics' fundamentals are arithmetic operations. Addition, subtraction, multiplication, and division are the main operations that make up this concept. The phrase "mathematical operations" also refers to these.
The math operation of subtracting two integers reveals the difference between them. The '-' sign is used to indicate it. In math, subtraction is the process of taking one number away from another to determine what is left over after something has been taken away. Rational number operations are equivalent to those performed on whole numbers. The main distinction is that rational numbers take the form p/q, where p and q are integers and q is not equal to 0. It is necessary to take the LCM of the numerators when adding or subtracting two rational integers.
Here we are discussing the four basic rules of arithmetic operations for all real numbers.
Addition (sum; ‘+’)Subtraction (difference; ‘-’)Multiplication (product; ‘×’ )Division (÷)Company A = $2 × [tex]10^{4}[/tex]e
Company B = $ 5 × [tex]10^{8}[/tex]
The difference = $2 × [tex]10^{4}[/tex]
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You are selling drinks at the carnival to raise money for your club. You sell lemonadefor $6 for 2 cups and orange drinks for $9 for 3 cups. Your sales totaled $240. Let xbe the number of cups of lemonade and y be the number of orange drinks. Write anyequation in standard form for the relationship above.
Let x be the number of cups of lemonade sold, and y the number of cups of orange drinks sold, then we can set the following equation:
[tex]6(\frac{x}{2})+9(\frac{y}{3})=240.[/tex]Now, recall that the standard form of a linear equation is:
[tex]Ax+By=C,[/tex]Where, A≥0, B and C are integers.
Simplifying the first equation, we get:
[tex]3x+3y=240.[/tex]Answer:
[tex]3x+3y=240.[/tex]Let f(x)=5x.Let g(x)=5x−7.Which statement describes the graph of g(x)with respect to the graph of f(x)? g(x)is translated 7 units down fromf(x).g(x)is translated 7 units left fromf(x).g(x)is translated 7 units right from f(x).g(x)is translated 7 units up fromf(x).
Given
[tex]\begin{gathered} f(x)=5x \\ g(x)=5x-7 \end{gathered}[/tex]According to rules of transformation:
f(x)+c shift c units up and f(x)-c shift c units down.
For the given function g(x) = 5x-7, 7 is being subtracted from 5x.
Where 5x is represented by f function.
Therefore, we could apply the rules of transformation f(x)-c shift c units down.
Here the value of c is 7.
Answer: g(x) is translated 7 units down from f(x)
Jina spends $16 each time she travels the toll roads. She started the month with $240 in her toll road account. The amount, A (in dollars), that she has left in the account after t trips on the toll roads is given by the following function.=A(t)=240-16tAnswer the following questions.(a)How much money does Jina have left in the account after 11 trips on the toll roads?$(b)How many trips on the toll roads can she take until her account is empty?trips
GIVEN:
We are told that Jina had an opening balance of $240 in her toll road account.
Also, we are told that the amount left in the toll road account is given by the function;
[tex]A(t)=240-16t[/tex]Required;
(a) To find how much money she has left in her acount after 11 trips.
(b) To find out how many trips she can take until her account is empty.
Step-by-step solution;
We first take note of the variable t, which represent the number of trips taken. Also, the function shows how many trips multiplied by 16 would be subtracted from the opening balance. The result would be how much amount (variable A) would be left in her account.
Therefore;
(a) After 11 trips, Jina would have;
[tex]\begin{gathered} A(t)=240-16t \\ \\ A(11)=240-16(11) \\ \\ A(11)=240-176 \\ A(11)=64 \\ \end{gathered}[/tex]For the (A) part, the answer is $64.
(b) For her account to be empty, then the function given would be equal to zero. That is, after an unknown number of trips, the balance would be zero. We can now re-write the function as follows;
[tex]\begin{gathered} A(t)=240-16t \\ \\ 0=240-16t \end{gathered}[/tex]Add 16t to both sides of the equation;
[tex]\begin{gathered} 16t=240-16t+16t \\ \\ 16t=240 \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }16: \\ \\ \frac{16t}{16}=\frac{240}{16} \\ \\ t=15 \end{gathered}[/tex]This means after 15 trips she would have emptied her toll road account.
ANSWER:
[tex]\begin{gathered} (A)=\text{\$64} \\ \\ (B)=15\text{ }trips \end{gathered}[/tex]A class had a quiz where scores ranged from 0 to 10.N(s) models the number of students whose score on the quiz was s.What does the statement N(8) > N(5) mean?Group of answer choicesA score of 8 is greater than a score of 5.There are more students who scored 8 than students who scored 5.There are 8 students who scored higher than 5.
The expression N(s) models the number of students that got a score "s" on the quiz.
Then the expression N(8) represents the number of students that scored 8 on the quiz.
And N(5) represents the number of students that scored 5 on the quiz.
[tex]N(8)>N(5)[/tex]Can be read as: "The number of students whose score on the quiz was 8 is greater than the number of students whose score on the quiz was 5"
Therefore, you can conclude that there were more students who scored 8 than students who scored 5. (option 2)
6. Write a quadratic function whose graph has a vertex of (-4,-2) and passes through the point (-3,1).
(h,k) are the coordinates of the vertex.
Use the given point (x,y) and the vertex (h,k) in the equation above to find the value of a:
Point: (-3 ,1) x = -3 y=1
Vertex: (-4 , -2) h= -4 k=-2
[tex]\begin{gathered} 1=a(-3-(-4))^2+(-2) \\ 1=a(-3+4)^2-2 \\ 1=a(1)^2_{}-2 \\ 1=a-2 \\ 1+2=a \\ 3=a \end{gathered}[/tex]Use the vertex and a to write the equation:
[tex]\begin{gathered} y=3(x-(-4))^2+(-2) \\ \\ \\ y=3(x+4)^2-2 \end{gathered}[/tex]2) Add or subtract the following polynomials: (5pts each) 1) (98-7x' +5x-3)+(2x* +4x'-6x-8) = ii) (8x* +6x - 4x2 -2)-(3x* – 5x – 7x+9)=
When we are adding/subtracting polynomials, we add or subtract like terms.
For example,
x^2 added with x^2 terms
x^4 added with x^4 terms
numbers (constants) added with numbers etc.
2 i)[tex](9x^5-7x^2+5x-3)+(2x^4+4x^3-6x-8)[/tex]Since we are "adding" the 2nd parenthesis polynomial, we can take out the parenthesis and put them in order and them simply add/subtract(!) The steps are shown below:
[tex]\begin{gathered} (9x^5-7x^2+5x-3)+(2x^4+4x^3-6x-8) \\ =9x^5-7x^2+5x-3+2x^4+4x^3-6x-8 \\ =9x^5+2x^4+4x^3-7x^2+5x-6x-3-8 \\ =9x^5+2x^4+4x^3-7x^2-x-11 \end{gathered}[/tex]Note: there were like terms with "x's" and "constants". We added/subtracted them only.
Hello I need help completing this practice math problem, I will include a picture. Thank you so much!
To answer this question we will set and solve a system of equations.
Let k be the number of orders that Kala served on Monday, a be the number of orders that Abdul served, and j be the number of orders that Joe served.
Since they served a total of 71 orders, Kala served 5 fewer orders than Abdul, and Joe served 2 times as many orders as Abdul, then we can set the following system of equations:
[tex]\begin{gathered} k+a+j=71, \\ k=a-5, \\ j=2a\text{.} \end{gathered}[/tex]Substituting the second and third equation in the first one we get:
[tex]a-5+a+2a=71.[/tex]Adding like terms we get:
[tex]4a-5=71.[/tex]Adding 5 to the above equation we get:
[tex]\begin{gathered} 4a-5+5=71+5, \\ 4a=76. \end{gathered}[/tex]Dividing the above equation by 4 we get:
[tex]\begin{gathered} \frac{4a}{4}=\frac{76}{4}, \\ a=19. \end{gathered}[/tex]Finally, substituting a=19 in the second and third equation we get:
[tex]\begin{gathered} k=19-5=14, \\ j=2\cdot19=38. \end{gathered}[/tex]Answer:
Number of orders Kala served: 14.
Number of orders Abdul served: 38.
Number of orders Joe served: 19.
ive tried to do this question multiple times but i just cant seem to understand it
The domain of the function which is the entire x values during the strike is
[tex]0\leq x\leq230[/tex]A rectangular garden plot measure 3.1 meters by 5.6 meters as shown Find the area of the garden in square meters
Given:
Length(l) of the garden is 3.1 meters
Width(w) of the rectangular garden is 5.6 meters
[tex]\begin{gathered} \text{Area of the garden=}l\times w \\ =3.1\times5.6 \\ =17.36 \end{gathered}[/tex]Area of the garden is 17.36 square meters.
The area in square millimeters of a wound has decreased by the same percentage every day since it began to heal. The table shows the wound's area at the end of each day.
Given the table showing the number of days since wound began to heal and area of wound in square millimeters
To determine the statement that are correct from the option provided
From the table shown it can be seen that as the day increases by 1, the area of wound in square millimeters decreases by a common ratio of
[tex]\frac{20}{25}=\frac{16}{20}=\frac{12.8}{16}=\frac{10.24}{12.8}=0.8[/tex]Suppose that an expression to represent the area of wound is
[tex]ab^c[/tex]The modelled expression from the table is
[tex]\begin{gathered} a=25 \\ b=0.8 \\ c=n-1 \\ \text{Therefore, we have} \\ 25(0.8^{n-1}) \end{gathered}[/tex]Let us use the modelled expression to verify each of the given conditions
The modelled expression can be simplified as shown below:
[tex]\begin{gathered} 25(0.8^{n-1}) \\ \text{Note},\text{ using indices rule} \\ \frac{a^n}{a}=a^{n-1} \\ \text{Therefore:} \\ 0.8^{n-1}=\frac{0.8^n}{0.8} \end{gathered}[/tex]Then, we have the modelled expression becomes
[tex]25(0.8^{n-1})=25\times\frac{0.8^n}{0.8}=\frac{25}{0.8}\times0.8^n=31.25(0.8^n)[/tex]From the two modelled expression we can see that
[tex]\begin{gathered} \text{when:} \\ c=n-1,a=25,b=0.8 \\ c=n,a=31.25,b=0.8 \end{gathered}[/tex]Then we can conclude that the two conditions that are true from the options are
If the value of c = n, the value of a is 31.25, and
If the value of c = n, the value of b is 0.8
Leo is constructing a tangent line from point Q to circle P. What is his next step? Mark the point of intersection of circle P and segment PQ. Construct arcs from point P that are greater than half the length of segment PQ. Construct a circle from point Q with the radius PQ. Plot a new point R and create and line perpendicular to segment PQ from point R
The next step to constructing a tangent line from Q to circle P is to construct the perpendicular bisector of the segment PQ.
For this, Leo can construct arcs from point P and from point Q that are greater than half the length of segment PQ.
AnswerThe next step is to construct arcs from point P that are greater than half the length of segment PQ.
can somone hep me please
Hi
a) = (8x2) x (10 ‐³ x10 ‐⁴)
= 8 x 2 you get 16 then 10‐³-⁴
16 x 10 ‐⁷
= 1.6 x 10¹ x 10 ‐⁷
= 1.6 x 10 ‐⁶
final answer
1.6 x 10 ‐⁶
I need problem C solved and for the work to be shown, Solve for the variable(s) in each triangle
Given:
Given that a right triangles.
Required:
To find the value of variables in each triangle.
Explanation:
In right triangles,
[tex]hup^2=opp^2+adj^2[/tex](C)
Here,
[tex]undefined[/tex]Finding supplementary and complementary angles (a) An angle measures 50°. What is the measure of its complement? (b) An angle measures 135°. What is the measure of its supplement? measure of the complement: measure of the supplement: 0 0 O X ?
SOLUTION
(a) Complementary angles are angles that add up to 90 degrees. So the angle that will complement 50 degrees will add to it to get 90. Let the angle be x, we have
[tex]\begin{gathered} 50\degree+x\degree=90\degree \\ 50+x=90 \\ x=90-50 \\ x=40\degree \end{gathered}[/tex]Hence the measure of the compelement is 40 degrees
(b) Supplementary angles are angles that add up to 180 degrees. So the angle that will supplement 135 degree will add to it to make it 180 degrees. Let this angle be y, so we have
[tex]\begin{gathered} 135\degree+y\degree=180\degree \\ y=180-135 \\ y=45\degree \end{gathered}[/tex]Hence measure of the supplement is 45 degrees
Write the equation in point slope and slope intercept form of a line that passes through the given point and has given slope m.(5,-6);m=-1
Given:
A line passes through the point,
[tex](x_1,y_1)=(5,-6)[/tex]The slope of the line is m = -1.
The objective is to find the equation of the line in point-slope and slope-intercept form.
Explanation:
To find equation in point-slope form:
The general formula of point-slope form is,
[tex]y-y_1=m(x-x_1)\text{ . . . . . . ..(1)}[/tex]On plugging the given values in equation (1),
[tex]\begin{gathered} y-(-6)=-1(x-5) \\ y+6=-x+5\text{ . . . . . .(2)} \end{gathered}[/tex]To find the equation in slope-intercept form,
The general formula of slope-intercept form is,
[tex]y=mx+b\text{ . . . . (3)}[/tex]On further solving the equation (2),
[tex]\begin{gathered} y+6=-x+5 \\ y=-x+5-6 \\ y=-x-1 \end{gathered}[/tex]Hence,
The equation of the line in point-slope form is y+6 = -x+5.
The equation of the line in slope-intercept form is y = -x-1.
ava's family drove to disneyland for spring break. Her mom and dad shared the driving duties for a total of 24 hours. Her mom drove 75 miles per hour, and her dad drove 60 miles per hour. If they drove a total of 1,710 miles, how many hours did each person drive for?
Total driving time =24
Mom drove =75 mile per hours
Dad drove = 60 miles per hours
Total distance =1710
Let
[tex]\begin{gathered} \text{ mom driving time =}^{}t_1 \\ \text{dad driviving time=}^{}t_2 \\ \text{Mom driving distance =}x \\ \text{ So dad driving distance=}^{}1710-x \end{gathered}[/tex]Total time:
[tex]t_1+t_2=24[/tex]Formula:
[tex]\text{ Spe}ed=\frac{\text{ Distance}}{\text{ Time}}[/tex]For Ava's mom:
[tex]\begin{gathered} \text{Speed}=\frac{\text{ Distance}}{\text{ time}} \\ 75=\frac{x}{t_1} \\ x=75t_1^{} \end{gathered}[/tex]For Ava's dad:
[tex]\begin{gathered} \text{ Spe}ed=\frac{\text{ Distance}}{\text{ Time}} \\ 60=\frac{1710-x}{t_2} \\ 60t_2=1710-x \\ x=1710-60t_2 \end{gathered}[/tex]Put the value of "x" then:
[tex]\begin{gathered} x=75t_1 \\ x=1710-60t_2 \\ so\colon \\ 75t_1=1710-60t_2 \\ 75t_1+60t_2=1710 \\ 15(5t_1+4t_2)=15\times114 \\ 5t_1+4t_2=114 \end{gathered}[/tex]Solve the both eq then:
[tex]\begin{gathered} t_1+t_2=24 \\ 4t_1+4t_2=96 \\ 5t_1+4t_2=114 \\ \text{then:} \\ 5t_1-4t_1+4t_2-4t_2=114-96 \\ t_1=18 \\ \end{gathered}[/tex]So Ava's mom drive 18 hours
[tex]\begin{gathered} t_1+t_2=24 \\ 18+t_2=24 \\ t_2=24-18 \\ t_2=6 \end{gathered}[/tex]Ava's dad driving 6 houras
Hello I I am confused because their are two different letters.
Let's begin by listing out the information given to us:
Line AB is parallel to Line CD; this implies that the angle formed by the two lines are right angles (90 degrees)
E is the intersecting point of both lines AB & CD (figure attached)
Let us put this into its mathematical form:
[tex]\begin{gathered} m\angle AED=(6x-24)=90^{\circ} \\ 6x-24=90\Rightarrow6x=90+24 \\ 6x=114\Rightarrow x=19 \\ x=19 \\ m\angle CEB=(4y+32)=90^{\circ} \\ 4y+32=90\Rightarrow4y=90-32 \\ 4y=58\Rightarrow y=17 \\ y=17 \end{gathered}[/tex]