Given:
Equation of a circle on a graph with center(3, -2).
To find:
Equation of a circle.
Explanation:
General eqution of a circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex]Solution:
From the graph, we can see that center is (3, -2) and radius equal 3.
So, equation of a circle is
[tex](x-3)^2+(y+2)^2=3^2[/tex]Hence, this is the equation of a circle.
Find the interval in which the following quadratic is decreasing.
The quadratic is decreasing in the interval in which the y values decrease with the increase in x values.
In the interval, (-∞, 0), the y values decrease with increase in x values.
Hence, the quadratic is decreasing in the interval (-∞, 0),
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
Maria is at the top of a cliff and sees a seal in the water. If the cliff is 40 feet above the water, Marla's eye-level is 5.5 feet, and the angle of depression is 52°, what is the horizontal distance from the seal to the cliff, tothe nearest foot?
SOLUTION
Let us make a diagram to interpret the question
from the diagram above, we can make the right-angle triangle as follows
So we can use SOHCAHTOA to solve this. The opposite side to the angle 52 degrees is 45.5 ft, this is gotten by adding the height of the cliff to Maria's height from her feet to her eyes.
The adjacent side is d, that is the distance from the seal to the cliff, so we have
[tex]\begin{gathered} TOA\text{ tan}\theta\text{ = }\frac{opposite}{adjacent} \\ tan52\degree=\frac{45.5}{d} \\ cross\text{ multiply, we have } \\ tan52\degree d=45.5 \\ d=\frac{45.5}{tan52} \\ d=35.54849 \end{gathered}[/tex]Hence the answer is 36 foot to the nearest foot
solve 2x^2+5x-3>0 quadratic inequalities
The solution set of the inequality 2 · x² + 5 · x - 3 > 0 is (- ∞, - 3) ∪ (1 / 2, + ∞).
How to solve a quadratic inequality
Herein we find a quadratic inequality, whose solution set can be found by factoring the expression and determine the interval where the expression is greater than zero. Initially, we use the quadratic formula to determine the roots of the quadratic function:
2 · x² + 5 · x - 3 = 0
x₁₂ = [- 5 ± √[5² - 4 · 2 · (- 3)]] / (2 · 2)
x₁₂ = (- 5 ± 7) / 4
x₁ = 1 / 2, x₂ = - 3
Then, the factored form of the inequality is:
(x - 1 / 2) · (x + 3) > 0
In accordance with the law of signs, we must look for that intervals such that: (i) (x - 1 / 2) > 0, (ii) (x + 3) > 0, (ii) (x - 1 / 2) < 0, (x + 3) < 0. Then, the solution set of the quadratic inequality is:
Inequality form - x > 1 / 2 ∨ x < - 3
Interval form - (- ∞, - 3) ∪ (1 / 2, + ∞)
The solution set of the inequality is (- ∞, - 3) ∪ (1 / 2, + ∞).
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The displacement (in meters) of a particle moving in a straight line is given by s = t^2 - 9t + 15,where t is measured in seconds.(A)(i) Find the average velocity over the time interval [3,4].Average Velocity = ___ meters per second(ii) Find the average velocity over the time interval [3.5,4].Average Velocity=____meters per second(iii) Find the average velocity over the time interval [4,5].Average Velocity= ____meters per second(iv) Find the average velocity over the time interval (4,4.5] Average Velocity = ____meters per.(B) Find the instantaneous velocity when t=4.Instantaneous velocity= ____ meters per second.
Given
The displacement (in meters) of a particle moving in a straight line is given by s = t^2 - 9t + 15,
!!PLEASE HELP IMMEDIATELY!!
Solve the inequality
-1/3x - 12 > 21 or -6x + 10 < -2
x < ? or x > ?
solve for both
Answer:
x < 2 or x > -11
Step-by-step explanation:
b. 1. add 12 to both sides to get -1/3x > 33
2. multiply by -3/1 to both sides to get x > -11
a. 1) subtract 10 to both sides
2) divide by -6 to both sides
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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Find the minimum weight resistance possible for A 230 pound man
Hello there. To find this minimum weight resistance, we need to convert the percentage value to decimals and multiply it by the weight of the person.
8% converted to decimals is equal to 0.08.
Now, multiply it by the weight of the 230 pound man
0.08 * 230 = 18.4 pounds
This is the minimum weight resistance this U gym offers to the customers.
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
What is the slope of the line that passes through the points (6,-10) and (3,-13)? Write in simplist form
Use the slope formula to find the slope of a line that goes through two points:
[tex]\begin{gathered} \text{Coordinates of two points}\rightarrow\text{ }(x_1,y_1),(x_2,y_2) \\ \text{Slope of a line through those points}\rightarrow m=\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]Substitute the coordinates (6,-10) and (3,-13) into the slope formula:
[tex]\begin{gathered} m=\frac{(-13)-(-10)}{(3)-(6)} \\ =\frac{-13+10}{3-6} \\ =\frac{-3}{-3} \\ =1 \end{gathered}[/tex]Therefore, the slope of a line that passes through those points, is 1.
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
An 80% confidence interval for a proportion is found to be (0.27, 0.33). Whatis the sample proportion?
Step 1
Given;
Step 2
When repeated random samples of a certain size n are taken from a population of values for a categorical variable, the mean of all sample proportions equals the population percentage (p).
[tex]\begin{gathered} Sample\text{ proportion=}\hat{p} \\ \hat{p}\pm margin\text{ error=cofidence interval} \end{gathered}[/tex]Thus;
[tex]\begin{gathered} Let\text{ }\hat{p}=x \\ Margin\text{ of error=y} \\ x-y=0.27 \\ x+y=0.33 \end{gathered}[/tex]checking properly, the sample proportion =0.30, because
[tex]\begin{gathered} 0.30-0.03=0.27 \\ 0.30+0.03=0.33 \end{gathered}[/tex]Answer; Option D
[tex]0.30[/tex]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+1.5Identify the values of a, b, and c for the quadratic equation given:y=-x2 +9a =b =C=
Question:
Solution:
A quadratic Equation in Standard Form is given by the following formula:
[tex]ax^2+bx\text{ + c = 0}[/tex]now, the given equation is
[tex]y=-x^2+9[/tex]this is equivalent to:
[tex]f(x)=-x^2+9[/tex]According to the Quadratic Equation in Standard Form, we can conclude that
[tex]a\text{ = -1}[/tex][tex]b\text{ = 0}[/tex]
and
[tex]c\text{ = 9}[/tex]3. An equation that crosses the y-axis at -5 and crosses the x-axis at 24. An equation that crosses the y-axis at -5 and crosses the x-axis at -65. An equation that crosses the y-axis at -5 and crosses the point (2,3)
We need to find the equation of the line which:
• crosses the y-axis at -5
,• crosses the x-axis at 2
The y-axis cutting point is (0,-5)
The x-axis cutting point is (2,0)
The equation of line is:
[tex]y=mx+b[/tex]Where m is the slope and b is the y-axis cutting point
m is given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Where
y_2 = 0
y_1 = -5
x_2 = 2
x_1 = 0
So, slope is:
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{0--5}{2-0}=\frac{0+5}{2}=\frac{5}{2}[/tex]We got m, we also know b.
The y cutting point is -5, so b = -5
The equation is:
[tex]y=\frac{5}{2}x-5[/tex]The graph would look like:
More clear version:
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.
One angle measures 140°, and another angle measures (5k + 85)°. If the angles are vertical angles, determine the value of k.
The value of k when one angle measures 140°, and another angle measures (5k + 85)° and if the angles are vertical angles is 11.
What is vertical angles?
Vertical angles are angles opposite each other where two lines cross.
Note: Vertical angles are equal.
To calculate the value of k, we use the principle of vertical angle
From the question,
140 = (5k+85)°Solve for k
5k = (140-85)5k = 55Divide both side by the coefficient of k (5)
5k/5 = 55/5k = 11Hence, the value of k is 11.
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Convert the function p(x) = 2(x – 4)(x + 3)
Expanding the expression,
[tex]\begin{gathered} p(x)=2(x-4)(x+3) \\ \rightarrow p(x)=2(x^2+3x-4x-12) \\ \rightarrow p(x)=2(x^2-x-12) \\ \rightarrow p(x)=2x^2-2x-24 \end{gathered}[/tex]We get that:
[tex]p(x)=2x^2-2x-24[/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.
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"
Graph the line with the given slope m and y-intercept b.
m = 4, b = -5
Answer:
Step-by-step explanation:
What we know:
m = 4, b = -5
y = mx + b where m is the gradient/slope and b is the y-intercept
Substitute m and b values:
y = 4x + -5 which is the same as y = 4x - 5
Substitute all x values to find y coordinate:
When x = -7, y = (4 x -7) - 5 = -33
When x = -6, y = (4 x -6) - 5 = -29
When x = -5, y = …
Continue for all x values
15 points?Solve for A 5/A = P A = ???? That’s all it saysPlease state what A is.
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]Find the sum of the arithmetic series 31+37 +43 +49 +... where n=8,OA. 416B. 1668OC. 832D. 834Reset Selection
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given details
[tex]\begin{gathered} a_1=31 \\ n=8 \\ d=37-31=6 \end{gathered}[/tex]STEP 2: Write the formula for finding sum of arithmetic series
STEP 3: Find the sum of the series
By substitution,
[tex]\begin{gathered} S_8=\frac{8}{2}[2(31)+(8-1)6] \\ S_8=4(62+42) \\ S_8=4(104)=416 \end{gathered}[/tex]Hence, the sum is 416
5 Which equations have the same value of x as 6 2 3 -9? Select three options. -9(6) 5x+4=-54 5x+4=-9 5x=-13 5X=-58
The given equation is-
[tex]\frac{5}{6}x+\frac{2}{3}=-9[/tex]If we multiply the equation by 6, we would have the same value for the variable x since we are multiplying the same number on each side. So, the second choice is an equivalent equation to the given one.
Let's multiply by 6.
[tex]\begin{gathered} 6\cdot\frac{5}{6}x+6\cdot\frac{2}{3}=-9\cdot6 \\ 5x+4=-54 \end{gathered}[/tex]So, the third expression is also an equivalent expression.
Then, let's subtract 4 on each side.
[tex]\begin{gathered} 5x+4-4=-54-4 \\ 5x=-58 \end{gathered}[/tex]The last choice is also an equivalent expression.
Therefore, the right choices are 2, 3, and 6.2. The product of two consecutive odd numbers is 143. Find the numbers. (Hint: If the first odd number is x, what is the next odd number?)
Step-by-step explanation:
we have the 2 numbers x and (x+2).
x × (x + 2) = 143
x² + 2x = 143
x² + 2x - 143 = 0
the general solution to such a quadratic equation
ax² + bx + c = 0
is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case this is
x = (-2 ± sqrt(2² - 4×1×-143))/(2×1) =
= (-2 ± sqrt(4 + 572))/2 = (-2 ± sqrt(576))/2 =
= (-2 ± 24)/2 = (-1 ± 12)
x1 = -1 + 12 = 11
x2 = -1 - 12 = -13
so, we have 2 solutions : 11 and 13, -13 and -11
11× 13 = 143
-11×-13 = 143
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]Exercise 2 Find a formula for Y in terms of X
Given:
y is inversely proportional to square of x.
The equation is written as,
[tex]\begin{gathered} y\propto\frac{1}{x^2} \\ y=\frac{c}{x^2}\ldots\ldots\ldots c\text{ is constant} \end{gathered}[/tex]Also y = 0.25 when x = 5.
[tex]\begin{gathered} y=\frac{c}{x^2} \\ 0.25=\frac{c}{5^2} \\ 25\times0.25=c \\ c=\frac{25}{4} \end{gathered}[/tex]So, the equation of y interms of x is,
[tex]y=\frac{25}{4x^2}[/tex]When x increases,
[tex]\begin{gathered} \lim _{x\to\infty}y=\lim _{x\to\infty}(\frac{25}{4x^2}) \\ =\frac{25}{4}\lim _{x\to\infty}(\frac{1}{x^2}) \\ =0 \end{gathered}[/tex]Hence, the value of x increases then y decreases.
You roll a die. What is the probability that you’ll get a number less than 3?0.3330.50.6670.75
Recall that the numbers in a die are 1,2,3,4,5,6.
[tex]S=\mleft\lbrace1,2,3,4,5,6\mright\rbrace[/tex]Hence the number of possible outcomes is 6.
[tex]n(S)=6[/tex]We need a number less than 3. Let A be this event.
[tex]A=\mleft\lbrace1,2\mright\rbrace[/tex]The favorable outcome is 2.
[tex]n(A)=\mleft\lbrace1,2\mright\rbrace[/tex]Since there are 1,2 less than 3 in a die.
[tex]P(A)=\frac{Favourable\text{ outcomes}}{\text{Total outcomes}}=\frac{n(A)}{n(S)}[/tex]Substitute n(A)=2 and n(S)=6, we get
[tex]P(A)=\frac{2}{6}=\frac{1}{3}=0.333[/tex]Hence the required probability is 0.333.
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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