ANSWER:
[tex]\begin{gathered} \text{Length = 21}\frac{2}{5}\text{ centimeters} \\ \\ \text{Perimeter = 76 centimeters} \end{gathered}[/tex]EXPLANATION:
Since the length is 4⅘ longer than the width, the length will be:
Length = 4⅘ + 16⅗
[tex]\begin{gathered} W\text{ = 16}\frac{3}{5}\text{ centimeters} \\ Length\text{ = 4}\frac{4}{5}\text{ }_{}+\text{ 16 }\frac{3}{5} \end{gathered}[/tex]Let's solve for length:
[tex]\begin{gathered} Length\text{ = 4}\frac{4}{5}\text{ }_{}+\text{ 16 }\frac{3}{5}\text{ = }\frac{24}{5}\text{ + }\frac{83}{5}\text{ = }\frac{107}{5}\text{ = 21}\frac{2}{5}\text{ centimeters} \\ \\ \text{Length = 21}\frac{2}{5}\text{ centimeters} \end{gathered}[/tex]From the given dimensions, it shows the picture frame is a rectangle.
To find the perimeter, use the perimeter of a rectangle formula:
Perimeter of a rectangle = 2(Length + Width)
[tex]\begin{gathered} \text{Perimeter = 2(}\frac{107}{5}+\frac{83}{5}) \\ \\ \text{ = 2(}\frac{190}{5}) \\ \\ \text{ = }2(38) \\ \\ \text{ = 76 centimeters} \end{gathered}[/tex]This question is very complicated which is something we are barely learning. I hope you can help and I appreciate the help.
There are four walls, one roof and one floor.
Mr. Smith will only paint the walls, but in one of them there is a door not to be painted.
The two walls with no doors have dimensions of 6 feet x 5 feet.
Their individual area is 6 * 5 = 30 square feet.
Their combined area is 2 * 30 = 60 square feet.
The back wall and the front wall have dimensions of 12 feet x 6 feet.
Their individual area is 12 * 6 = 72 square feet.
The front wall has a door of dimensions of 3 feet x 5 feet.
The area of the door is 3 * 5 = 15 square feet.
This area must be subtracted from the area of the fron wall.
Area of the front wall = 72 - 15 = 57 square feet.
The total area to be painted in blue is:
60 + 72 + 57 = 189 square feet
In the diagram, GH bisects ZFGI.Solve for x and find mZFGH,b. Find mZHGL.Find mZFGI.a. X(Simplify your answer.)
As shown in the diagram:
GH bisects the angle FGI
So, the measure of the angle FGH = measure of the angle HGI
so,
2x - 9 = 3x - 28
solve for x
2x - 3x = -28 + 9
-x = -19
x = 19
So, mand m
Hi there… I need some help help with this question.
ANSWERS
a. 1/2
b. 1001
c. 20
d. 8
e. 0.16
EXPLANATION
a. There are 4 women and 4 men on the hiring committee, which is a total of 8 people. The probability that a randomly selected person is a woman is,
[tex]P(W)=\frac{4}{8}=\frac{1}{2}[/tex]Hence, the probability that the person drawing the names from the hat is a woman is 1/2.
b. The applicant pool consists of 6 database administrators and 8 network engineers, which is a total of 14 applicants. We want to choose 4 applicants,
[tex]_4C_{14}=\frac{14!}{4!(14-4)!}=\frac{14\cdot13\cdot12\cdot11\cdot10!}{4!\cdot10!}=\frac{14\cdot13\cdot12\cdot11}{4!}=1001[/tex]Hence, there are 1001 ways to choose the group to be hired.
c. There is a total of 6 database administrators, and we want to choose 3,
[tex]_3C_6=\frac{6!}{3!(6-3)!}=\frac{6!}{3!\cdot3!}=20[/tex]Hence, there are 20 ways of choosing 3 database administrators.
d. There is a total of 8 network engineers, and we want to choose 1,
[tex]_1C_8=\frac{8!}{1!\cdot(8-1)!}=\frac{8\cdot7!}{1\cdot7!}=8[/tex]Hence, there are 8 ways of choosing 1 network engineer.
e. In part b, we found that there is a total of 1001 ways of choosing the 4 people to be hired. Also, in parts c and d, we found that there are 20 ways of choosing 3 database administrators and 8 ways of choosing 1 network engineer. The probability that this is the combination of people hired is,
[tex]P(3DA+1NE)=\frac{20\cdot8}{1001}=\frac{160}{1001}\approx0.16[/tex]Hence, the probability that the random selection of four persons to be hired will result in 3 database administrators and 1 network engineer is approximately 0.16.
*16. What is the leading coefficient of the polynomial function f(x) = 9-2x + 6x² + 5x³?A. 9 B. 3 C. 5 D. 4
ANSWER :
C. 5
EXPLANATION :
The Leading coefficient is the coefficient of the leading term.
The leading term is the term with the highest degree.
From the problem, we have the polynomial :
[tex]f(x)=9-2x+6x^2+5x^3[/tex]The term with the highest degree is 5x^3
Therefore, the leading coefficient is 5
Relative error as a percent rounded to the nearest tenth of a percent
Answer:
Explanation:
Given:
Expected value of the measurement = 15.75 cm
Actual value of the measurement = 15.71 cm
To find:
The relative error
Relative error formula is given as:
[tex]Relative\text{ error = \mid}\frac{Actual\text{ - expected}}{expected}|\text{ }\times100\text{ \%}[/tex][tex]\begin{gathered} Relative\text{ error = \mid}\frac{15.71\text{ - 15.75}}{15.75}|\times100\text{ \%} \\ \\ Relative\text{ error = \mid}\frac{-0.04}{15.75}|\text{ }\times100\text{ \%} \\ \\ Relative\text{ error = \mid-0.00254\mid }\times\text{ 100\%} \end{gathered}[/tex][tex]\begin{gathered} Absolute\text{ value of a negative number gives a positive number} \\ \\ Relative\text{ error = 0.00254 }\times100\text{ = 0.254} \\ \\ Relative\text{ error = 0.3 \%} \end{gathered}[/tex]At a coffee shop, the first 100customers' orders were as follows.SmallMediumLargeHot54822Cold8125What is the probability that a customerordered a small given that he or sheordered a hot drink?P(Small | Hot ) = [?]Round to the nearest hundredth.
Explanation
The probability of P(Small | Hot ) is easily observable from the table. This is given as
[tex]\begin{gathered} P(Small|Hot)=\frac{5}{5+48+22}= \\ =\frac{5}{75} \\ =0.07 \end{gathered}[/tex]The final answer is 0.07
Three slices of cheese pizza and four slices of pepperoni pizza cost $12.50. Twoslices of cheese pizza and one slice of pepperoni pizza cost $5.00. What is the priceof one slice of pepperoni pizza?
Price of one slice = ?
Then write
3X + 4Y = 12.50
2X + 1Y = 5.00
Then now find Y
Multiply by 4, and substract 2X + Y = 5
4• ( 2X + Y ) = 4• 5.00
8X + 4Y = 20
now substract 3X + 4Y = 12.5
(8X + 4Y)- ( 3X + 4Y) = 20 - 12.5
(8X - 3X )+ 4Y - 4Y = 7.5
5X + 0 = 7.5
. X = 7.5/5 = 1+ 1/2 = 1.5
Then ANSWER IS
Price of 1 slice of pepperoni = $1.5 dollars
I need help with homework question and please help with plotting the points on the graph please it’s highly important for the equation. I have the answer already I just need help plotting the dots on the line. It’s two lines. One line has two points and the second line has two points as well and I already have the outcome but really I stress on placing the coordinates on the line
Given the set of inequalities:
-2x - 2y > 1
y ≥ -2
Let's graph the system of linear inequalities and shade the solution set.
For the first inequality, rewrite in slope-intercept form:
y = mx + b
Add 2x to both sides:
-2x + 2x - 2y > 2x + 1
-2y > 2x + 1
Divide through by 2:
[tex]\begin{gathered} \frac{-2y}{-2}>\frac{2x}{-2}+\frac{1}{-2} \\ \\ y<-x-\frac{1}{2} \end{gathered}[/tex]Now, let's get two points from this inequality.
When: x = 1.5:
[tex]\begin{gathered} x=1.5 \\ y<-1.5-\frac{1}{2} \\ y<-2 \\ \\ \\ \text{WHen x = 0} \\ y<-0-\frac{1}{2} \\ y<-0.5 \end{gathered}[/tex]For the first inequality, we have the points:
(x, y) ==> (1.5, -2), (0, -0.5)
Plot the points and connect the points using a dashed line.
Shade the area below the boundary region since y is less than.
• For the second inequality:
[tex]y\ge-2[/tex]
This inequality is a horizontal line at y = -2.
We can get any two points on the line:
(x, y) ==> (4, -2), (1.5, -2)
Draw a dashed line at y = 2.
feet by You are preparing to tile the backsplash in a kitchen. The area you are tiling measures 8 1/2 feet. The tiles you plan to use are sold in boxes that have enough tiles to cover 10 square feet. What is the minimum number of boxes of tiles you should order to complete the job?A.1B. 2C. 12D. 13E. 20
hello
the area of the room is given by
[tex]8\frac{1}{2}by1\frac{1}{2}[/tex]let's convert the mixed fraction to improper fraction
[tex]\begin{gathered} 8\frac{1}{2}=\frac{17}{2} \\ 1\frac{1}{2}=\frac{3}{2} \end{gathered}[/tex]now, let's multiply the two dimensions given to find the area in squared feet.
[tex]\frac{17}{2}\times\frac{3}{2}=\frac{51}{4}[/tex]the area of the room is 51/4 ft^2
we can now find how many boxes of tiles will cover the room
1 box covers 10ft^2
let the number of boxes of tiles to cover 51/4ft^2 be represented by x
1 box = 10
x box = 51/4
[tex]\begin{gathered} 1=10 \\ x=\frac{51}{4} \\ \text{cross multiply both sides and solve for x} \\ 1\times\frac{51}{4}=10\times x \\ \frac{51}{4}=10x \\ \text{divide both sides by 10} \\ \frac{\frac{51}{4}}{10}=\frac{10x}{10} \\ x=\frac{51}{4}\times\frac{1}{10}=\frac{51}{40}=1.275 \end{gathered}[/tex]the number of boxes required to cover the room is 1.275 boxes and he'll need a minimum of 2 boxes to do so.
the answer is option B
How many ways can a person toss a coin 14 times so that the number of heads is between 6 and 9 inclusive?
How many ways can a person toss a coin 14 times so that the number of heads is between 6 and 9 inclusive?
the formula of combination is equal to
[tex]\text{nCr}=\frac{n!}{r!(n-r)!}[/tex]For r between 6 and 9
For r=6
n=14
substitute
[tex]14\text{C6}=\frac{14!}{6!(14-6)!}=\frac{14!}{6!(8)!}=\frac{14\cdot13\cdot12\cdot11\cdot10\cdot9}{6\cdot5\cdot4\cdot3\cdot2\cdot1}[/tex]14C6=3,003
For r=7
n=14
substitute
[tex]14\text{C7}=\frac{14!}{7!(14-7)!}=\frac{14!}{7!(7)!}=\frac{14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}[/tex]14C7=3,432
For r=8
n=14
substitute
[tex]14\text{C8}=\frac{14!}{8!(14-8)!}=\frac{14!}{8!(6)!}=\frac{14\cdot13\cdot12\cdot11\cdot10\cdot9}{6\cdot5\cdot4\cdot3\cdot2\cdot1}[/tex]14C8=3,003
For r=9
n=14
substitute
[tex]14\text{C9}=\frac{14!}{9!(14-9)!}=\frac{14!}{9!(5)!}=\frac{14\cdot13\cdot12\cdot11\cdot10}{5\cdot4\cdot3\cdot2\cdot1}[/tex]14C9=2,002
adds the combinations
3,003+3,432+3,003+2,002=11,440
11,440 waysorder for least to greatest 93.389 0.28 0.0043 0.002 30.59 1.49
From least to greatest, we have
0.002 0.0043 0.28 1.49 30.59 93.389
Write the slope-intercept form of the eqı 1) Slope = -7, y-intercept = -4
We want to write the slope-intercept form of
Please help :( It’s my study guide for my upcoming test
Let x the number of quartes and y the number of nickels
So (1) x + y = 47
Solve for x
x = 47 - y
Then .25x +.05 =4.95
It is better if you multiply both sides by 100 to get rid of the decimal
100(.25x +.05) = 100(4.95)
(2) 25x + 5y = 495
Replace the first x value in the second equation
(2) 25x + 5y = 495
25(47 - y ) + 5y = 495
Then solve the equation for y
1175 - 25y + 5y = 495
-25y + 5y = 495 - 1175
-20y = -680
y = -680/ -20
y = 34 nickels
Replace this y value in the x equation
x = 47 - y
x = 47 - 34
x = 13 quarters
4. Solve for the variable in the following proportion: 36/c = 45/10
We need to solve for "c"in the proportion:
36/c = 45/10
so we cross multiply:
36 * 10 = 45 * c
operate
360 = 45 * c
divide by 45 on both sides to isolate "c"
360 / 45 = c
c = 8
I need help with this question please. Just ignore the wording below it
f(x) = x² + 5x - 66
Explanation:The zeros of the quadratic equation are -11 and 6
Thsi means that:
x + 11 = 0
x - 6 = 0
The function will therefore be found as:
f(x) = (x + 11)(x - 6)
Expanding the function above
f(x) = x² - 6x + 11x - 66
f(x) = x² + 5x - 66
Therefore, the quadratic function that is in standard form and has zeros -11 and 6 is:
f(x) = x² + 5x - 66
Solve the quadratic equation by factoring.2x^2+24x+22=0
Solution
[tex]\begin{gathered} 2x^2+24x+22=0 \\ Divide\text{ through by 2} \\ x^2+12x+11=0 \\ x^2+11x+x+11=0 \\ x(x+11)+1(x_+11)=0 \\ (x+11)(x+1)=0 \\ x+11=0\text{ or x+1=0} \\ x=-11\text{ or x = -1} \end{gathered}[/tex]Write a quadratic that represents the table . please Explain how you created your equation
a quadratic equation is of the form
[tex]y=ax^2+bx+c[/tex]then. if x = 0, y = 7:
[tex]\begin{gathered} 7=a(0)^2+b(0)+c \\ c=7 \end{gathered}[/tex]and, if x = 1, y = 16
[tex]\begin{gathered} 0=a(1)^2+b(1)+7 \\ 0=a+b+7 \\ a+b=-7\text{ eq 1 } \end{gathered}[/tex]and if x = 2, y = 27
[tex]\begin{gathered} 27=a(2)^2+b(2)+7 \\ 27=4a+2b+7 \\ 27-7=4a+2b+7-7 \\ 4a+2b=20\text{ } \\ 2a+b=10\text{ eq2} \end{gathered}[/tex]then solve for a and b with the equations 1 and 2
[tex]\begin{gathered} \begin{bmatrix}a+b=-7 \\ 2a+b=10\end{bmatrix} \\ a+b=-7 \\ a+b-b=-7-b \\ a=-7-b \\ 2\mleft(-7-b\mright)+b=10 \\ -14-2b+b=10 \\ -14-b=10 \\ -14-b+14=10+14 \\ -b=24 \\ \frac{-b}{-1}=\frac{24}{-1} \\ b=-24 \end{gathered}[/tex]for a
[tex]a=-7-b=-7-(-24)=-7+24=17[/tex]answer, the equation is:
[tex]y=17x^2-24x+7[/tex]The radius of circle O (not shown) is 4, and the radian measure of central angle AOB is between 3pi/4 and 5pi/4. which could be the length of arc AB?
SOLUTION
Write out the formula for the length of an arc
[tex]\begin{gathered} \text{length of Arc=}\theta\times r \\ \text{Where }\theta\text{ is in radians } \\ r=4 \end{gathered}[/tex]Angle given is between
[tex]\frac{3\pi}{4}\text{ and }\frac{\text{5}\pi}{4}[/tex]Substitute each of the value for Θ in the formula above
[tex]\begin{gathered} \text{When }\theta=\frac{3\pi}{4} \\ \text{Then} \\ \text{Length of Arc=}\theta\times r=\frac{3\pi}{4}\times4=3\pi \end{gathered}[/tex]Also
[tex]\begin{gathered} \text{when }\theta=\frac{5\pi}{4} \\ \text{Then} \\ \text{Length of Arc=}\frac{5\pi}{4}\times4=5\pi \end{gathered}[/tex]Hence
The length of the Arc is between
[tex]\begin{gathered} 5\pi\text{ } \\ \text{and } \\ 3\pi \end{gathered}[/tex]Therefore
The length of the Arc AB could be 4π
Answer :Option B
i wasn’t sure what the real answer was i did l x w and got 168
Solution
The area of the parallelogram is
[tex]\begin{gathered} A=bh \\ A=21\times8 \\ A=168in^2 \end{gathered}[/tex]Therefore the area of the figure = 168in²
2 numbers whose product is -84 and whose sum is -17
Solution
- We are asked to find two numbers with a product of -84 and a sum of -17.
- Let the two numbers be x and y.
- We can form equations using the above statement. These equations are formed below
[tex]\begin{gathered} x\times y=-84 \\ xy=-84\text{ (Equation 1)} \\ \\ x+y=-17\text{ (Equation 2)} \end{gathered}[/tex]- Now that we have the two equations, we can proceed to solve them simultaneously.
- This is done using substitution as shown below
[tex]undefined[/tex]I am in 9th grade learning Algebra 1 and I need help to understand it. Can you please help me?
1) Considering that we have the statement "A number and -5 has a result of 2".
2) We can rewrite it as a Linear Equation, calling this number by x we can write it out:
[tex]x-5=2[/tex]Then we have a One step equation. The first thing to do is to isolate the x variable on the left side. So let's manipulate this equation by adding 5 to both sides:
[tex]\begin{gathered} x-5=2 \\ x-5{\textcolor{blue}{+5}}=2{\textcolor{blue}{+5}} \\ x+0=7 \\ x=7 \end{gathered}[/tex]By adding 5 to the left side we get rid of that -5 on the left side, and since it is an equality, we have to add 5 to the right side as well
3) Hence, to solve Step equations we need to manipulate the equation to isolate the variable on one side.
if g(y) = 5, then solve for g(-1)
We have the following:
[tex]undefined[/tex]2) y = x + 3 + 3 A) Domain: x 2-3 Range: y = 3 B) Domain: x 2-3 Range: y s3 C) Domain: x 2 3 Range: y 2-3 D) Domain: x 2-3 Range: y 2-3
Domain : Domain of a function is the set of input values for which the function is real and defined.
Given function is :
[tex]y=\sqrt[]{x+3}+3[/tex]Since if x less than - 3 then the square root will be into the form of complex number i,
So x ≥ -3
So Domain will be : x ≥ -3
Interval notation : [ -3, infinity)
Range : Range is the set of all the output values of the function :
The range of the funtion is:
[tex]f(x)\ge3[/tex]Intervale notation : [3, infinity)
Domain = x ≥ -3, [-3, inifinity)
Range : f(x)≥3, Interval notation [ 3, infinity)
Answer : A)
Domain x ≥- 3
Range y ≥ 3
Lyndie is making reduced copies of a photo 25 centimeters in height. She sets the copy machine to an 80% size reduction.
PART A
Write a percent equation that represents the relationship of the height of the first copy to the height of the original photo. 38 3-3 Represent and Use the Percent Equation
PART B
Lyndie wants to make another copy that will have a height of 17 cm. The copy machine settings increase or decrease in increments of 5%. Which photo should she make her copy from, the original or her first copy? Explain.
The succession time should be atleast t=9 to get a final copy that is less than 15% of the original size.
Lyndie is making reduced copies of a photo 25 centimeters in height. She sets the copy machine to an 80% size reduction.
Part a
Let the size of the page be q, when it is reduced to 80%, its size becomes
= 80%*q
= 0.80(q)
= 0.80q
When you want to return it into its original size q, you need to multiply the page by x
such that
x(0.80q) = q
[tex]x = \frac{q}{(0.80q)}[/tex]
[tex]x = \frac{1}{0.80}[/tex]
x = 1.25
x = 125%
Hence, the enlargement needed to be done is 25%.
Part b
The size of the page after t number of copying done is given by
[tex]C(t) = C_{0}(0.80)^{t}[/tex]
where [tex]C_{0}[/tex] is the original size of the page.
We want to find a value for t ∈ Ζ such that
[tex]\frac{C(t)}{C_{0} } = (0.80)^{t}[/tex]
[tex]0.15 \leq (0.80)^{t}[/tex]
To solve this equation, we can apply natural logarithm.
≅ [tex]In(0.15) \leq In(0.80)^{t} \\\\In(0.15) \leq tIn(0.80)\\\\\frac{In(0.15)}{In(0.80)} \leq t\\ \\0.80 \leq t[/tex]
Hence the answer is the succession time should be atleast t = 9 to get a final copy that is less than 15% of the original size.
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Divide using the long division method.x^2+ 6x + 4/x + 5
ANSWER
[tex]x+1-\frac{1}{x+5}[/tex]EXPLANATION
We want to divide the given polynomial by long division:
[tex]\frac{x^2+6x+4}{x+5}[/tex]To do this, we have to divide each term in the numerator by the first term in the denominator and multiply by the second term.
This is repeated until the last term is divided. That is:
Since we cannot divide further, the remainder is written as a fraction of the divisor.
In other words, the solution to the division is:
[tex]x+1-\frac{1}{x+5}[/tex]Simplify:(2+i)-(2+3i)
Answer:
-2i
Explanation:
Given the expression:
[tex]\mleft(2+i\mright)-\mleft(2+3i\mright)[/tex]To simplify, first, we remove the brackets.
[tex]=2+i-2-3i[/tex]Next, we collect like terms and simplify.
[tex]\begin{gathered} =2-2+i-3i \\ =-2i \end{gathered}[/tex]
Suppose that three geological study areas are set up on a map at points please check photo
So we must find the center of the earthquake. We have three points and we know the distances from each of these points to the earthquake. In order to find the center we just need to make three circles, each centered in one of the three points and its radius must be the distance to the center of the earthquake. If we do this correctly then the three circles will meet in a given point D which is the center of the earthquake.
In order to draw a circle using the tool given by the question you'll need its center and a point in the circumference. So let's construct each of the circles:
First we have point A=(-15,2) which is at a distance of 13mi from the earthquake. So we must construct a circle centered around A with a radius of 13 units. Any point at a distance of 13 units from A will be useful, for example a point that has a horizontal distance of 13 units from A. We'll name this point E and we have:
[tex]E=(-15+13,2)=(-2,2)[/tex]So the first circle is the one that passes through (-2,2) and is centered around (-15,2).
Now we repeat this process with the other circles. We have B=(-11,1) and its distance to the earthquake is 10 miles so we can add 10 to its x-value to find a point that is at a distance of 10 units from it:
[tex]F=(-11+10,1)=(-1,1)[/tex]So this circle is centered around (-11,1) and it passes through (-1,1).
For the third circle we have C=(-6,3) and its distance to the earthquake is 5 miles. Then a point located at 5 miles from C could be:
[tex]G=(-6+5,3)=(-1,3)[/tex]So the third circle is centered around (-6,3) and passes through (-1,3).
With all this information we can graph the three circles:
As you can see these three circles intercept each other at (-3,7). Then the earth quake is located at (-3,7).
AnswerThe graphs are displayed in the picture above. The center of the earthquake is located at (-3,7).
I dont know how to do number 18 on my homework
Beth walked 3 blocks in 15 minutes.
Then, we have that:
[tex]\frac{3}{15}\cdot\frac{3}{3}=\frac{9}{45}[/tex]We have that multiplying the rate (ratio) by the same number in the numerator and in the denominator, we will have equivalent fractions (and the same ratio).
Option a is true. We have the result above.
Then, for option b, we cannot obtain an equivalent fraction. It is false.
For option c, we have the same as for option b. It is false.
For option d:
[tex]\frac{3}{15}\cdot\frac{4}{4}=\frac{12}{60}[/tex]Then, option d is true.
6(k-8)=96k=?help please!
answer: k = 24
6). A movie theater sold twenty-five tickets on Saturday and five tickets on Thursday. They soldhow many times as many tickets on Saturday as they sold on Thursday?
A movie theater sold 25 tickets on Saturday and 5 tickets on Thursday.
If we compare 25 and 5, we can see that,
[tex]25=5\cdot5[/tex]In other words, they sold 5 more times on Saturday than on Thursday