Solution
For this case we know that
m m < DBC = 3x+2
So then we can do the following:
4x -5 = 3x+2
4x-3x = 5+2= 7
x = 7
A dilation with a scale factor of 2 maps ....
The measure of the angles is the same. After a dilation what changes is the length of the sides, but the angles are the same of
[tex]\measuredangle B\text{ = }\measuredangle E[/tex]The second choice is correct.
Given the line segment with points P (1,6) andQ (9,-4). What is the length of PQ?(2 Points)
When 2 coordinate points are given, we can find its length by using the distance formula. Which is
[tex]\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]We
• take differences in y coordinates and x coordinates
,• square them
,• take their sum
,• take square root of the answer
Tha's all.
So, let's do the steps:
y diff: -4 -6 = -10
x diff: 9-1 = 8
Now, it becomes:
[tex]\begin{gathered} \sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2} \\ =\sqrt[]{(-10)^2+(8)^2} \\ =\sqrt[]{164} \\ =2\sqrt[]{41} \end{gathered}[/tex]The length of PQ (exact) is:
[tex]2\sqrt[]{41}[/tex]In decimal: 12.81
Answer this question based on the knowledge of angle in a Circle.
From the given diagram, PQ is a diameter and since the triangle inside a semicircle is right angled, hence triangle PQR is a right triangle.
Also since the side QR is parallel to OS, hence the line PR is perpendicular to the lines QR and OS
Hence;
PR ⊥ QR and OS ⊥ PR
Proof that From △PRQ,
Since OS ⊥ PR, hence △PSR is divided into two similar triangles by the line OS showing that the base angles (
Recall that for an isosceles triangle, the base angles and two opposite sides are equal. Based on the proof above, we can conclude that;△PSR is isosceles triangle showing that SP = SRSimplify the expression `2\sqrt{a^{2}b^{8}}\left(ab^{3}\right)^{-1}`You may type many lines to show your work. Enter equations inside the text using the square-root button below.
ANSWER
2b
EXPLANATION
To simplify this expression, we have to apply some of the exponents' properties. First, the square root is a fractional exponent,
[tex]\sqrt{x}=x^{1/2}[/tex]So we can rewrite the expression as,
[tex]2\sqrt{a^2b^8}(ab^3)^{-1}=2(a^2b^8)^{1/2}(ab^3)^{-1}[/tex]Then, we can distribute the exponents into the multiplication,
[tex](xy)^z=x^zy^z[/tex]In this problem,
[tex]2(a^2b^8)^{1/2}(ab^3)^{-1}=2(a^2)^{1/2}(b^8)^{1/2}(a)^{-1}(b^3)^{-1}[/tex]Exponents of exponents are multiplied,
[tex](x^y)^z=x^{yz}[/tex]In this problem,
[tex]2(a^2)^{1/2}(b^8)^{1/2}(a)^{-1}(b^3)^{-1}=2\cdot a^{2\cdot1/2}\operatorname{\cdot}b^{8\operatorname{\cdot}1/2}\operatorname{\cdot}a^{-1}\operatorname{\cdot}b^{3\operatorname{\cdot}(-1)}[/tex]Simplify if possible,
[tex]2\cdot a^{2\cdot1/2}\operatorname{\cdot}b^{8\operatorname{\cdot}1/2}\operatorname{\cdot}a^{-1}\operatorname{\cdot}b^{3\operatorname{\cdot}(-1)}=2\cdot a^1\operatorname{\cdot}b^4\operatorname{\cdot}a^{-1}\operatorname{\cdot}b^{-3}[/tex]Now, the product of two powers with the same base is equal to the base raised to the sum of the exponents,
[tex]x^y\cdot x^z=x^{y+z}[/tex]In this problem,
[tex]2\cdot a^1\operatorname{\cdot}b^4\operatorname{\cdot}a^{-1}\operatorname{\cdot}b^{-3}=2\cdot a^{1-1}\operatorname{\cdot}b^{4-3}[/tex]Solve the subtractions,
[tex]2\cdot a^{1-1}\operatorname{\cdot}b^{4-3}=2\cdot a^0\cdot b^1=2b[/tex]Hence, the simplified expression is 2b.
Anna is 10 years older than Brad. The sum of their ages is 26. The system of equations representing their ages is _ Solve the system using inverses. Then select the correct calculation and the ages of Anna and Brad.
Answer:
The correct answer is option B.
[tex]undefined[/tex]Find the solution for the given the system of equations:Y= (1/2)x - 1/2 and y=2^(x+3)
Answer:
This system has no solution.
Step-by-step explanation:
The solution of this system is the ordered pair that is the solution to both equations, we can solve this using the graphical method, which consists of graphing both equations in the same coordinate system.
The solution to the system will be at the point where the two functions intersect.
Since the functions do not intersect, this system has no solution.
Ayana drew a scale drawing of a house and its lot. The backyard, which is 70 feet long in real life, is 203 inches long in the drawing. What scale did Ayana use for the drawing?29 inches : [ ] feet
Let m be scale used by individual for drawing.
Then the product of scale factor and original length is equal to the length in drawing. So,
[tex]\begin{gathered} 70\times m=203 \\ m=\frac{203}{70} \\ =\frac{29}{10} \end{gathered}[/tex]So, 29 inches of drawing is corresponding to 10 feet of house.
What is the solution of the system?{4x−y=−38x+y=3Enter your answer in the boxes.
Solution:
Given:
[tex]\begin{gathered} 4x-y=-38 \\ x+y=3 \end{gathered}[/tex]Solving the system simultaneously by substitution method;
[tex]\begin{gathered} 4x-y=-38\ldots\ldots\ldots\ldots\ldots(1) \\ x+y=3\ldots\ldots\ldots\ldots\ldots\ldots.(2) \\ \\ \text{making y the subject of the formula from equation (2);} \\ x+y=3 \\ y=3-x\ldots\ldots\ldots\text{.}\mathrm{}(3) \\ \\ \text{Substituting equation (3) into equation (1);} \\ 4x-y=-38 \\ 4x-(3-x)=-38 \\ 4x-3+x=-38 \\ 4x+x=-38+3 \\ 5x=-35 \\ \text{Dividing both sides by 5;} \\ x=-\frac{35}{5} \\ x=-7 \\ \\ \text{Substituting the value of x into equation (3) to get y;} \\ y=3-x \\ y=3-(-7) \\ y=3+7 \\ y=10 \end{gathered}[/tex]Therefore, the solution of the system is;
[tex](x,y)=(-7,10)[/tex]Distance-Time Grapfoss Object ng Constant Speed - 9) Achillwit and string a speed The graph above shows the Giant the ball towed from starting pucat in 5 seconds.
In the given graph, the following are the records of the distance of the tennis ball at a certain time (seconds) after being hit:
Time (seconds) Distance (meters)
1 0.5
2 1
3 1.5
4 2
5 2.5
To be able to get the speed of the tennis ball, let's use any of the data (Time and Distance Covered) in the graph, and use the formula in calculating the speed.
[tex]\text{ Speed = }\frac{Dis\tan ce}{Time}[/tex]Let's use 1 second = 0.5 meter. We get,
[tex]\text{ Speed = }\frac{0.5\text{ meter}}{1\text{ second}}[/tex][tex]\text{Speed = 0.5 meter/second}[/tex]Therefore, the speed of the tennis ball is 0.5 m/s.
The answer is letter A.
Find the point on the curve y=5x+1 closest to the point (0,4).
Given the coordinates of two points P and Q, we can calculate the distance between them using the formula:
[tex]\begin{gathered} \begin{cases}P={(x_P},y_P) \\ Q={(x_Q},y_Q)\end{cases} \\ . \\ d=\sqrt{(x_P-x_Q)^2+(y_P-y_Q)^2} \end{gathered}[/tex]In this case, we want the smallest distance between a point in the curve and the point (0, 4)
Then, we know that there is a point that we can call Q = (x, y) that is the closest to the point (0, 4). We can write, using the distance formula:
[tex]d=\sqrt{(x-0)^2+(y-4)^2}=\sqrt{x^2+(y-4)^2}[/tex]The equation given is:
[tex]y=5x+1[/tex]We want to rewrite the distance formula to include the equation of the curve. Since there is a term 'x²', we can solve the equation for x and square on both sides:
[tex]\begin{gathered} y=5x+1 \\ . \\ y-1=5x \\ , \\ x=\frac{y}{5}-\frac{1}{5} \\ . \\ x^2=(\frac{y}{5}-\frac{1}{5})^2 \end{gathered}[/tex]Now we can substitute in the distance equation:
[tex]d=\sqrt{(\frac{y}{5}-\frac{1}{5})^2+(y-4)^2}[/tex]We can see that this is a distance function for any point of the curve to the point (0, 4). This is actually a function of y.
[tex]d(y)=\sqrt{(\frac{y}{5}-\frac{1}{5})^2+(y-4)^2}[/tex]Now, we can apply calculus to find the minimum of this function. Let's take the first derivative:
[tex]d^{\prime}(y)=\frac{\frac{2}{5}(\frac{y}{5}-\frac{1}{5})+2(y-4)}{2\sqrt{(\frac{y}{5}-\frac{1}{5})^2+(y-4)^2}}[/tex]Simplify:
[tex]d^{\prime}^(y)=\frac{26y-101}{25\sqrt{(\frac{y}{5}-\frac{1}{5})^2+(y-4)^2}}[/tex]And since we want to find a minimum, we need to also calculate the second derivative:
[tex]d^{\prime}^{\prime}(y)=\frac{26\sqrt{(y-4)^2+(\frac{y}{5}-\frac{1}{5})^2}-\frac{(2(y-4)+\frac{2}{5}(\frac{y}{5}-\frac{1}{5})(26y-101)}{2\sqrt{(y-4)^2+(\frac{y}{5}-\frac{1}{5})^2}}}{25((y-4)^2+(\frac{y}{5}-\frac{1}{5})^2)}[/tex]Simplify:
[tex]d^{\prime}^{\prime}(y)=\frac{9}{25((y-4)^2+(\frac{y}{5}-\frac{1}{5})^2)^{\frac{3}{2}}}[/tex]Now, we need to find the critical points of the function. The critical points are the x values where the first derivative is 0.
Then:
[tex]d^{\prime}(y)=\frac{26y-101}{25\sqrt{(y-4)^2+(\frac{y}{5}-\frac{1}{5})^2}}[/tex]Is a quotient, For a quotient to be 0, the only way this is possible is for the numerator to be 0. Then:
[tex]\begin{gathered} 26y-101=0 \\ . \\ y=\frac{101}{26} \end{gathered}[/tex]And now, to see if this critical point is a minimum, we evaluate it in the second derivative, if the second derivative is positive in this critical point, the function has a minimum at that point:
[tex]d^{\prime}^{\prime}(\frac{10}{26})=\frac{9}{25((\frac{101}{26}-4)^2+(\frac{101}{26}-\frac{1}{5})^2)^{\frac{3}{2}}}\approx1.76746[/tex]Then, the function d(y) has a minimum at y = 101/26
Now, we need to find the x coordinate of this point. We use the equation of the curve:
[tex]\begin{gathered} \frac{101}{26}=5x+1 \\ . \\ x=(\frac{101}{26}+1)\cdot\frac{1}{5}=\frac{15}{26} \end{gathered}[/tex]Thus, the answer to the point in the curve that is the closest to (0, 4) is:
[tex](\frac{15}{26},\frac{101}{26})[/tex]Please solve equation for maximum and minimum
Answer:
17^(1/11), which occurs at x=9
Step-by-step explanation:
To find the Absolute Extrema in a set of points, you need to evaluate (plug in) the endpoints, and maxima/minima of the equations and figure out the greatest and lowest ones.
1.) By using this method, the first step is to find the Relative Maximums/Minimums of these areas. We can do this by finding the derivative of the equation, and setting that equal to 0 and solving. [tex]\frac{d}{dx} (x^2-64)x^{\frac{1}{11}} = (x^2-64)^{-10/11} * 2x[/tex]. If we set this equal to 0, we will find that x = 0. Therefore, x=0 is a minimum. Since this point belongs to the interval of [-8, 9], we can use it.
2.) Plug the endpoints of the interval and the result from our calculations. If we do this, we get f(-8)=0, f(9) = 17^(1/11), f(0)=0
3.) Since we are finding the Maxima, we look for the greatest value, which is 17^(1/11), which occurs at x=9
Find the volume of the sphere. Round your answer to the nearest tenth.A) 2,289.1 m^3B) 3,052.1 m^3C) 24,416.6 m^3D) 12,437.4 m^3
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
sphere:
diameter = 18 m
Step 02:
geometry:
volume of the sphere:
v = 4/3 π r³
π = 3.14
r = d / 2 = 18 m / 2 = 9 m
v = 4/3 π (9 m)³ = 3052.1 m³
The answer is:
3052.1 m³
4.
The value of a truck decreases exponentially since its purchase. The two points on the
graph shows the truck's initial
value and its value a decade afterward.
[6040,000)
a) Express the car's value, in dollars, as a function of time
d, in decades, since purchase.
(1 24,000)
b) Write an expression to represent the car's value 4 years
after purchase.
c) By what factor is the value of the car changing each year? Show your reasoning.
Answer:
a. v = 40 000 (3/ 5)^d
b. v = 40 000 (3/5)^(4/10)
c. 0.95
Explanation:
The exponential growth is modelled by
[tex]v=A(b)^d[/tex]We know that points (0, 40 000) and (1, 24 000) lie on the curve. This means, the above equation must be satsifed for v = 40 000 and d = 0. Putting v = 40 000 and d = 0 into the above equation gives
[tex]40\; 000=Ab^0[/tex][tex]40\; 000=A[/tex]Therefore, we have
[tex]v=40\; 000b^d[/tex]Similarly, from the second point (1, 24 000) we put v = 24 000 and d = 1 to get
[tex]24\; 000=40\; 000b^1[/tex][tex]24\; 000=40\; 000b^{}[/tex]dividing both sides by 40 000 gives
[tex]b=\frac{24\; 000}{40\; 000}[/tex][tex]b=\frac{3}{5}[/tex]Hence, our equation that models the situation is
[tex]\boxed{v=40\; 000(\frac{3}{5})^d\text{.}}[/tex]Part B.
Remember that the d in the equation we found in part A is decades. Since there are 10 years in a decade, we can write
t = 10d
or
d = t/10
Where t = number of years
Making the above substitution into our equation gives
[tex]v=40\; 000(\frac{3}{5})^{\frac{t}{10}}[/tex]Therefore, the car's value at t = 4 is
[tex]\boxed{v=40\; 000(\frac{3}{5})^{\frac{4}{10}}}[/tex]Part C:
The equation that gives the car's value after t years is
[tex]v=40\; 000(\frac{3}{5})^{\frac{t}{10}}[/tex]which using the exponent property that x^ab = (x^a)^b we can rewrite as
[tex]v=40\; 000\lbrack(\frac{3}{5})^{\frac{1}{10}}\rbrack^t[/tex]Since
[tex](\frac{3}{5})^{\frac{1}{10}}=0.95[/tex]Therefore, our equation becomes
[tex]v=40\; 000\lbrack0.95\rbrack^t[/tex]This tells us that the car's value is changing by a factor of 0.95 each year.
graph g(x) where f(x) = 2x-5 and g(x) = f(x+1)
The graph therefore is shown below;
Robbie ran 21 miles less than O'neill lastweek. Robbie ran 17 miles. How manymiles did O'neill run?
We need to find the distance, in miles, that O'Neill ran.
We know that Robbie ran 21 miles less than O'Neill. This is the same as saying that O'Neill ran 21 miles more than Robbie.
Also, we know that Robbie ran 17 miles.
Then, to find the distance O'Neill ran, we need to add 21 miles to the 17 miles Robbie ran.
We obtain:
[tex]17\text{miles}+21\text{miles}=38\text{miles}[/tex](6-5g)(-3g+9)Multiply polynominals
The two polynomials are multiplied as follows:
[tex](6-5g)(-3g+9)[/tex]Open the first bracket as follows:
[tex](6-5g)(-3g+9)=6(-3g+9)-5g(-3g+9)[/tex]Open the two brackets as follows:
[tex]\begin{gathered} 6(-3g+9)-5g(-3g+9)=-18g+54+15g^2-45g \\ =15g^2-63g+54 \end{gathered}[/tex]Hence the multiplication of the polynomials results in:
[tex](6-5g)(-3g+9)=15g^2-63g+54[/tex]Line segments, AB, BC, CD, DA create the quadrilateral graphed on the coordinate grid above. The equations for two of the four line segments are given below. Use the equations of the line segments to answer the questions that follow. AB: y = -x + 1 1
Given the equations of the side length as shown;
AB y = 1/3 x + 1
CD = y = -3x+11
Before we determine whether they are parallel or perpendicular, we must first know that;
Two parallel lines have the same slope i.e Mab = Mcd
For two lines to be perpendicular then the product of the slopes must give -1 i.e MabMcd = -1
Comapring both equations with the general equation of a line y = mx+c;
For line AB: y = 1/3 x + 1
Mab = 1/3
For line CD: y = -3x+11
Mcd = -3
Taking the product of the slope;
MabMcd = 1/3 * -3
MabMcd = -1
Is AB perpendicular or parallel to CD? Since the product of their slope gives -1, hence the lines AB and CD are PERPENDICULAR to each other.
Find and graph the intercepts of the following linear equation:5x-2y=-10
x-intercept = -2
y-intercept = 5
Explanation:[tex]\begin{gathered} Given: \\ 5x\text{ - 2y = -10} \end{gathered}[/tex]x-intercept is the value of x when y = 0
To get the x-intercept, we will substitute y with zero:
[tex]\begin{gathered} 5x\text{ - 2\lparen0\rparen = -10} \\ 5x\text{ = -10} \\ divide\text{ both sides by 5:} \\ x\text{ = -10/5} \\ \text{x = -2} \\ So,\text{ the x-intercept = -2} \end{gathered}[/tex]y-intercept is the value of y when x = 0
To get the y-intercept, we will substitute x with zero:
[tex]\begin{gathered} 5(0)\text{ - 2y = -10} \\ -2y\text{ = -10} \\ divide\text{ both sides by -2:} \\ \frac{-2y}{-2}\text{ = }\frac{-10}{-2} \\ division\text{ of same signs give positive sign} \\ y\text{ = 5} \\ So,\text{ the y-intercept = 5} \end{gathered}[/tex]Plotting the graph:
Two cards are drawn from a standard deck without replacement. What is the probability that the first card is a diamond and the second card is red
The probability that the first card drawn without replacement is a diamond and the second card drawn is red is approximately 0.12
Probability of an event without replacementThe probability for an event without replacement implies that once an item is drawn, then we do not replace it back to the sample space before drawing another item.
In a standard deck of cards, there is a total of 52 cards with 13 diamonds and 26 reds (including diamonds)
probability that the first card drawn is diamond = 13/52
probability that the first card drawn is red = 25/51 {because diamond is also a red and was drawn first without replacement}
So;
probability that the first card is a diamond and the second card is red = (13/52) × (25/51)
probability that the first card is a diamond and the second card is red = 325/2652
probability that the first card is a diamond and the second card is red = 0.1225
Therefore, the proprobability that the first card drawn without replacement a diamond and the second card is red is approximately 0.12
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How many yards are there in 72 miles? Round answer to nearest 100th (2-decimal places).
Given,
The number of total miles are 72.
As know that,
There are 1760 yards in one mile.
The number of yards in 72 mile is,
[tex]\text{Number of yards=72}\times1760=126720\text{ miles}[/tex]Hence, the nnumber of yards in one mile is 126720.
Find the equation of the circle having the given properties:
The equation of a circle is given by:
[tex](x-h)^2+(y-k)^2=r^2[/tex]3.
Using the points (4, 0), (0, 2) and (0, 0), we have:
[tex]\begin{gathered} (0,0)\colon \\ h^2+k^2=r^2 \\ (0,2)\colon \\ h^2+(2-k)^2=r^2 \\ h^2+4-2k+k^2=r^2 \\ \text{Comparing both equations:} \\ h^2+k^2=h^2+4-2k+k^2 \\ 4-2k=0 \\ k=2 \\ (4,0)\colon \\ (4-h)^2+k^2=r^2 \\ 16-8h+h^2+4=r^2 \\ \text{Comparing this last equation with the first one:} \\ 16-8h+h^2+4=h^2+4 \\ 16-8h=0 \\ h=2 \\ \\ h^2+k^2=r^2 \\ r^2=4+4=8 \end{gathered}[/tex]So the equation of this circle is:
[tex](x-2)^2+(y-2)^2=8[/tex]function "p" is in the form y = ax² + c if the values of "a" and "c" are both less than 0, which graph could represent "p" ?A) graph AB) graph BC) graph CD) graph D
INFORMATION:
We know that:
- function "p" is in the form y = ax² + c
- the values of "a" and "c" are both less than 0
And we must select the graph that can represent p.
STEP BY STEP EXPLANATION:
To select the correct option we must know that:
- the form y = ax² + c is the form of a parabola
- the sign of "a" determines whether the parabola opens up or down
- "c" will move the function c units up or down about the origin.
Now, since "a" and "c" are both less than 0 we can conclude that:
- the parabola opens down because "a" is negative (less than 0)
- the parabola will be c units down about the origin because "c" is negative
Finally, we can see that the option which met the conditions is graph B.
ANSWER:
B) graph B
Find the value of tan X rounded to the nearest hundredth, if necessary.
5
W
1
√26
X
The value of tan x is 5
We need to find the value of tan x
Tan is one of the trignometric functions and the range of a tan function varies from 0 ≤ tan x ≤ 2π
In the triangle vwx, the perpendicular of x is VW and the base is WX
tan x = perpendicular / base
Here, the perependicular is 5 and base is 1
tan x = 5/1
tan x = 5
Therefore, the value of tan x is 5
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Identify how many solutions there are to the system of equations represented on the following graph. Treat the red and black graphs as one circle. H H
The solution of two or more equations is the point where the equation intersects. The more that the graph of the equation intersects, the more its solutions.
From the given graph of circle and parabola, they intersect three times. Therefore, there are 3 solutions in this given system of equations.
which operation is applied to 3 and ×+5 in the expression 3(x+5) over 0.2
In the expression " 3(x+5) over 0.2" the word "over" indicates that 0.2 is dividing the first term 3(x+5), you can write the calculation as follows:
[tex]\begin{gathered} \frac{3(x+5)}{0.2} \\ \cdot-\cdot or\cdot-\cdot \\ 3(x+5)\div0.2 \end{gathered}[/tex]The operation is a division.
45 pointsSolve the logarithmic equation below. All work must be shown to earn full credit and
We know that the substraction of two logarithm of the same base is related to a division:
[tex]\log _460-\log _44=\log _4(\frac{60}{4})[/tex]Since 60/4 = 15, then
[tex]\log _4(k^2+2k)=\log _415[/tex]Then, the expressions in the parenthesis are equal:
k² + 2k = 15
Factoring the expression
Now, we can solve for k:
k² + 2k = 15
↓ substracting 15 both sides
k² + 2k - 15 = 0
Since
5 · (-3) = -15 [third term]
and
5 - 3 = 2 [second term]
we are going to use 5 and -3 to factor the expression:
k² + 2k - 15 = (k -3) (k +5) = 0
We want to find what values should have k so
(k -3) (k +5) = 0
if k -3 = 0 or if k +5 = 0, the expression will be 0
So
k - 3 = 0 → k = 3
k +5 = 0 → k = -5
Answer: k = 3 or k = -5A certain strain of bacteria is growing at a rate of 44% per hour, and with 2,000 bacteria initially, this event can be modeled by the equation B(t) = 2,000(1.44)t. With this fast growth rate, scientists want to know what the equivalent growth rate is per minute. Using rational exponents, what is an equivalent expression for this bacterial growth, expressed as a growth rate per minute?
The given equation for the growth rate per hour is:
[tex]B(t)=2,000(1.44)^t[/tex]Where t is the time in hours.
The equivalent growth rate per minute would be the equivalent in minutes for hours, then:
[tex]1\min \cdot\frac{t\text{ hours}}{60\min }=\frac{t}{60}[/tex]Where t is the time in minutes, then the answer is:
[tex]B(t)=2,000(1.44)^{\frac{t}{60}}[/tex]Each participant must pay $14 to enter the race. Each runner will be given a T-shirt that cost race organizers $3.50. If the T-shirt was the only expense for the race organizers, which of the following expressions represents the proportion of the entry fee paid by each runner that would be donated to charity?
The total amount paid by each participant is $14, from which $3.5 will be used for the T-shirt. As that is the only expense, all the remaining money will be available to donate to charity.
[tex]14-3.5=10.5[/tex]Then, for each participant, $10.5 of the $14 paid will be donated to charity.
To find the proportion of the fee paid that would be donated we just need to divide those values:
[tex]\frac{\text{Amount of money donated}}{\text{Total fee paid}}[/tex][tex]\frac{10.5}{14}=\frac{3}{4}[/tex]Then, 3/4 is the proportion of the entry fee paid that will be donated to charity.
What is the radius of a circle with circumferenceC= 40 CM
Solution:
The circumference C, of a circle is;
[tex]C=\pi r^2[/tex]Given;
[tex]C=40cm,\pi=3.14[/tex]The radius r, is;
[tex]\begin{gathered} 40=3.14(r^2) \\ \text{Divide both sides by 3.14} \\ \frac{40}{3.14}=\frac{3.14(r^2)}{3.14} \\ r^2=12.74 \\ \text{Take the square root of both sides;} \\ \sqrt[]{r^2}=\sqrt[]{12.74} \\ r=3.57 \end{gathered}[/tex]The radius of the circle is 3.57cm
Solve the inequalities. State whether the inequalities have no solutions or real solutions.4( 3 - 2x) > 2( 6 - 4x)
Given an inequalities show if it have no solutions or real solutions:
[tex]4(3-2x)>2(6-4x)[/tex]Step 1: Expand the inequalities
[tex]\begin{gathered} 4(3-2x)>2(6-4x) \\ 12-8x>12-8x \end{gathered}[/tex]Step 2: Subtract 12 from both side of the inequalities
[tex]\begin{gathered} 12-8x>12-8x \\ \text{subtract 12 from both side} \\ 12-8x-12>12-8x-12 \\ -8x>-8x \end{gathered}[/tex]Step 3: Add 8x to both side of the inequalities
[tex]\begin{gathered} -8x>-8x \\ \text{add 8x to both side} \\ -8x+8x>-8x+8x \\ 0>0 \end{gathered}[/tex]Therefore the solution for the above inequalities is No solution