As they are in the right ratio, option D, "Renting a movie for $2 for the first day and $1 for each day after the first day," demonstrates a proportionate relationship.
Proportional Relationship
When two variables' ratios are equivalent, proportional relationships between them emerge. The fact that one variable is consistently equal to the constant value of the other in a proportional connection is another way to think of them. This constant is known as the "constant of proportionality."
Ratio
In mathematics, a ratio shows how frequently one number appears in another. For instance, the ratio of apples to mangos in a bowl of fruit would be nine to six if there were nine apples and six mangos. Apples make up 9:15 of the entire fruit, whereas Mangos make up 6:9 of the total fruit.
Option D, "Renting a movie for $2 for the first day and $1 for each day after the first day," illustrates a proportional relationship because the ratio is correct
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Choose the scenarios that demonstrate a proportional relationship for each person's income.
Millie works at a car wash and earns $17.00 per car she washes.
Bryce has a cleaning service and charges $25.00 plus $12.50 per hour.
Carla makes sandwiches at her job and earns $7.85 per hour.
Tino is a waiter and makes $3.98 per hour plus tips.
Choose the scenarios that demonstrate a proportional relationship for each person's income.
Millie works at a car wash and earns $17.00 per car she washes.
Bryce has a cleaning service and charges $25.00 plus $12.50 per hour.
Carla makes sandwiches at her job and earns $7.85 per hour.
Tino is a waiter and makes $3.98 per hour plus tips.
Quadrilateral HGEF is a scaled copy of quadrilateral DCAB. What is themeasurement of lin EG?
Answer:
14 units
Explanation:
If quadrilaterals HGEF and DCAB are similar, then the ratio of some corresponding sides is:
[tex]\frac{FH}{BD}=\frac{EG}{AC}[/tex]Substitute the given side lengths:
[tex]\begin{gathered} \frac{6}{3}=\frac{EG}{7} \\ 2=\frac{EG}{7} \\ \implies EG=2\times7 \\ EG=14 \end{gathered}[/tex]The measurement of line EG is 14 units.
Using the GCF you found in Part B, rewrite 72 + 81 as two factors. One factor is the GCF and the other is the sum of two numbers that do not have a common factor. Show your work.
The factors of 72 and 81 are
[tex]\begin{gathered} 72=2^3\cdot3^2 \\ 81=3^4 \end{gathered}[/tex]Therefore, their GCF is equal to 3^2=9
Then,
[tex]72+81=9\cdot8+9\cdot9=9(8+9)[/tex]The answer is 9(8+9).
The factors of 8 and 9 are
[tex]\begin{gathered} 8\to2,4,8 \\ 9\to3,9 \end{gathered}[/tex]A factory makes car batteries. The probability that a battery is defective is1/6 If 400 batteries are tested, about how many are expected to be defective?A. 40 B. 25C. 16D. 375
Since there are 400 batteries are tested
Since the probability of the defective batteries is 1/6
The number of defective batteries =
[tex]\frac{1}{16}\times400=25[/tex]The answer is B
entionaction f(x) = 4.12x +12. If f(x) = -2(5)*, what is f(2)?A100B.20fC227-2050C. -20D. -50boioht of 144I
Problem
We have the following expression given:
f(x)= -2(5)^x
And we want to find f(2)
Solution
so we can do the following:
f(2)= -2 (5)^2 = -2*25 = -50
Use slope to determine if lines AB and CD are parallel, perpendicular, or neither 6. A(-3, 8), B(3, 2), C(7,1), D(5,-1)m(AB) m(CD) Types of Lines
What is the value of x? 20/72=x/360
The value of x = 100
Question 6 What is the factored form of the expression below? 7 - 16 O OD (x-8)(x - 8) (x - 4)(x + 4) (x - 4)(x - 4) (x-8)(x + 8) Oo
If :
[tex]x^2-16[/tex][tex]\begin{gathered} \sqrt[]{x^2}=x \\ \sqrt[]{16}=4 \end{gathered}[/tex]Then:
[tex]x^2-16\text{ =(x-4)(x+4)}[/tex]Answer: ( x - 4 ) ( x + 4 )
5 cm3 cm3 cm5 cm3 cmPrisma5 cmPrism BWhich of the following statements are true about the solids shown above?Check all that apply.A. Prisms A and B have different values for lateral surface area.O B. Prism B has a total surface area of 110 cm?O C. Prism A has a lateral surface area of 60 cm?D D. Prism B has a larger surface area.
Note that the lateral surface area is the area of the faces of the solid, excluding the cross-sectional faces i.e. faces which are perpendicular to the longitudinal axis.
The lateral surface area of prism A is calculated as,
[tex]\begin{gathered} LSA_A=2(5\times3)+2(5\times3)_{} \\ LSA_A=30+30 \\ LSA_A=60 \end{gathered}[/tex]Similarly, the lateral surface area of prism A is calculated as,
[tex]\begin{gathered} LSA_B=2(3\times5)+2(5\times5)_{} \\ LSA_B=30+50 \\ LSA_B=80 \end{gathered}[/tex]Clearly, prisms A and B have different values of lateral surface area.
So option A is the correct statement.
The total surface area is the sum of all the faces of the solid.
Since we have already calculated the LSA i.e. sum of area of 4 faces of the prism, we can add the area of the two remaining cross sectional faces to get the total area.
The total cross section area of prism B is calculated as,
[tex]\begin{gathered} A_B=2(5\times3) \\ A_B=30 \end{gathered}[/tex]So the total surface area of prism B becomes,
[tex]\begin{gathered} TSA_B=LSA_B+A_B_{} \\ TSA_B=80+30 \\ TSA_B=110 \end{gathered}[/tex]The total surface area of prism B is 110 sq. cm.
So option B is also correct.
Note that we have already found that the lateral surface area of prism A is 60 sq. cm.
Therefore, option C is also correct.
The total cross section area of prism A is calculated as,
[tex]\begin{gathered} A_A=2(3\times5) \\ A_A=30 \end{gathered}[/tex]So the total surface area of prism A becomes,
[tex]\begin{gathered} TSA_A=LSA_A+A_A \\ TSA_A=60+30 \\ TSA_A=90 \end{gathered}[/tex]The total surface area of prism A is 90 sq. cm.
It is oberved that prism B has a larger surface area.
So, option D is also correct.
Hence, we can conclude that all the given statements are correct.
10.1.3The hour hand of a clock moves from 5 to 9 o'clock. Through how many degrees does it move?
Step 1: Lets calculate angle on each hour hand
since the wall clock takes the shape of a cirle
Therefore,
The total angles in a walk clock is 360°
Angle on each hour hand is
There are 12 hour hands on the clock ,
Therefore,
[tex]\begin{gathered} \text{Angle on each hour hand is =}\frac{360^0}{hands\text{ on the clock}}^{} \\ \text{Angle on each hour hand =}\frac{360^0}{12}=30^0 \end{gathered}[/tex]Since the hour hand moved from 5 o'clock to 9 o'clock
It has moved a distance of (9 - 5)= 4 hands on the clock
If each hand on the clock=30°
Therefore,
The angle in degrees moved through 4 hour hands on the clock will be calculated as,
[tex]\begin{gathered} \text{Angle moved = angle on each hand}\times no\text{ of hands moved} \\ \text{Angle moved=30}^0\times4=120^0 \end{gathered}[/tex]The hour hand of the clock moved from 5 o'clock to 9 o'clock through an angle of 120°
Round 7488 to the nearest thousand
The thousand place value is the 4th digit to the left of the decimal point. This means that the digit is 7.
If the first digit after 7 is greater than or equal to 5, 7 would increase by 1. If it is less than 5, 7 remains the same. Since 4 is less than 5, 7 remains. The rest digits turns to 0. Thus, the answer is
7000
Swine Flu is attacking Springfield. The function below determines how many people have swine where t=time in days and S=the number of people in thousands.
A.find s(4)
[tex]\begin{gathered} s(4)=9(4)-4 \\ s(4)=36-4 \\ s(4)=32 \end{gathered}[/tex]B. means that in 4 days there will be 32000 infected people
C. find t to S(t)=23
[tex]\begin{gathered} 23=9t-4 \\ 9t=23+4 \\ t=\frac{27}{9} \\ t=3 \end{gathered}[/tex]D. means there will be 23,000 infected people after 3 days
E. Graph
to draw the line we need two points which we already have but we will add another to make a table of 3 values the new value is t=1
[tex]\begin{gathered} s(1)=9(1)-4 \\ s(1)=5 \end{gathered}[/tex]table
graph
8.1 km to miles and feet
Given
[tex]8.1\operatorname{km}[/tex]It should be noted that
[tex]\begin{gathered} 1\operatorname{km}=0.621371miles \\ 1\text{mile}=5280\text{feet} \end{gathered}[/tex][tex]\begin{gathered} \text{convert 8.1km to miles} \\ 1\operatorname{km}=0.621371\text{miles} \\ 8.1\operatorname{km}=8.1\times0.621371 \\ 8.1\operatorname{km}=5.0331051\text{miles} \end{gathered}[/tex][tex]\begin{gathered} 8.1\operatorname{km}=5\text{miles}+0.0331051\text{miles} \\ \text{convert 0.0331051miles to fe}et \\ 1\text{miles}=5280ft \\ 0.0331051\text{miles}=0.0331051\times5280feet \\ 0.0331051\text{miles}=174.79feet \end{gathered}[/tex]Hence, 8.1km is 5 miles and 174.79 feet
find the exact values of the six trigonometric functions of the angle 0 shown in the figure(Use the Pythagorean theorem to find the third side of the triangle)
The right angled triangle is given with reference angle theta.
The opposite side (facing the reference angle) is 3, while the hypotenuse (facing the right angle) is 5. The adjacent shall be calculated using the Pythagoras' theorem as follows;
[tex]\begin{gathered} \text{Adj}^2+3^2=5^2 \\ \text{Adj}^2=5^2-3^2 \\ \text{Adj}^2=25-9 \\ \text{Adj}^2=16 \\ \text{Adj}=\sqrt[]{16} \\ \text{Adj}=4 \end{gathered}[/tex]Therefore, the trigonometric functions of angle theta are shown as follows;
[tex]\begin{gathered} \sin \theta=\frac{opp}{hyp}=\frac{3}{5} \\ \cos \theta=\frac{adj}{hyp}=\frac{4}{5} \\ \tan \theta=\frac{opp}{adj}=\frac{3}{4} \\ \csc \theta=\frac{hyp}{opp}=\frac{5}{3} \\ \sec \theta=\frac{hyp}{adj}=\frac{5}{4} \\ \cot \theta=\frac{adj}{opp}=\frac{4}{3} \end{gathered}[/tex]Solve the equation 2x^3 – 5x² + x + 2 = 0 given that 2 is a zero of f (x) = 2x^3 – 5x^2 + x +2.
Hence, 2 is a zero of f(x). That is x - 2 is a factor of f(x).
So we can find
[tex]\frac{2x^3-5x^2+x+2}{x-2}[/tex][tex]\Rightarrow\frac{2x^3-5x^2+x+2}{x-2}=2x^2-x-1[/tex][tex]\text{Next we solve }2x^2-x-1=0[/tex][tex]\begin{gathered} \Rightarrow2x^2-x-1=0 \\ 2x^2+x-2x-1=0 \\ x(2x+1)-1(2x+1)=0 \\ \Rightarrow(x-1)(2x+1)=0 \\ \Rightarrow x-1=0\text{ or 2x+1=0} \\ \Rightarrow x=1\text{ or 2x=-1} \\ x=1\text{ or x =-}\frac{1}{2} \end{gathered}[/tex]Hence,
[tex]x=2,1,\text{ or -}\frac{1}{2}[/tex]Larry says all numbers that have a 2 in the one's place are composite numbers. Explain if Larry is correct or incorrect.
A composite number is defined as a whole number that have more than two factors; from this definition we conclude that all whole numbers that are not prime are composite numbers.
Since all even numbers are not prime we conclude that Larry is correct; all numbers that have a 2 in the one's place are composite. In fact all even numbers are composite with exception of 2 itself.
State 3 solutions to the inequality: (1 Point) 3−4>5
We have the inequality:
[tex]3x\text{ - 4 }>5[/tex]Let's find out 3 solutions, as follows:
3x - 4 > 5
Adding 4 at both sides of the inequality:
3x - 4 + 4 > 5 + 4
3x > 9
Dividing by 3 at both sides, we have:
3x/3 > 9/3
x > 3
Now, we can find 3 solutions that fulfill the condition of the inequality:
x = 4, x = 5, x = 6
These three solutions are > 3
may ou solve the system of linear equations by substitution
y= 11 + 4x
3x +2y = 0
Put the first equation into the second one. (replace the value of y)
3x +2 (11 + 4x) = 0
Solve for x:
3x + 22 + 8x = 0
3x+8x = -22
11x = -22
x = -22/11
x = -2
Replace x=-2 in the first equation and solve for y
y= 11 + 4 (-2)
y= 11-8
y= 3
Solution:
x= -2 , y=3
If f(x) = x + 1, find f(x + 7). Hint: Replace x in the formula by x+7.f(x + 7) =
The original function is:
[tex]f(x)\text{ = x+1}[/tex]We want to find the value of the function when the input is "x + 7". So in the place of the original "x" we will add "x+7".
[tex]\begin{gathered} f(x+7)\text{ = (x+7)+1} \\ f(x+7)\text{ = x+7+1} \\ f(x+7)\text{ = x+8} \end{gathered}[/tex]The value of the expression is "x + 8"
Ready for me for a quadratic function with vertex (3,9)
We have to write an equation of a quadratic function that has a vertex at (3,9) and pass through the origin.
We can use the vertex form of the quadratic equation:
[tex]y=a(x-h)^2+k[/tex]where the vertex has coordinates (h,k).
In this case, (h,k) = (3,9).
From the formula we can see that the parameter a that can take any value and still have the same vertex. We will use the parameter "a" to make it pass through the origin.
The vertex form of the equation is then:
[tex]y=a(x-3)^2+9[/tex]As it pass through the origin, then the equation should be satisfied when x = 0 and y = 0:
[tex]\begin{gathered} 0=a(0-3)^2+9 \\ 0=a(-3)^2+9 \\ 0=a\cdot9+9 \\ -9=9a \\ -\frac{9}{9}=a \\ a=-1 \end{gathered}[/tex]Then, as a = -1, we can write the equation as:
[tex]y=-(x-3)^2+9[/tex]Answer: An example of quadratic function with vertex (3,9) that pass through the origin is y = -(x-3)² + 9.
in which quadrant is the given point located (2,-4)
Answer: 4th Quadrant
Step-by-step explanation:
When plotted, the point (2, -4) lies in the 4th quadrant.
price of gas at store was 4.29 per gallon the next week it went up .55 and down .25 and back up 8.30 and finally it went down$.15 what is the price per gallon now
The final price of gas per gallon after number of reduction and increment is $12.74.
Given,
The price of gas at store = 4.29 per gallon
The increased amount of gas = 0.55
The new price of gas = 4.29 + 0.55 = 4.84 per gallon
Then decreased 0.25. So, the price of gas = 4.84 - 0.25 = 4.59 per gallon
Again the price increases 8.30 and the new price become, 4.59 + 8.30 = 12.89 per gallon
Then, the price of gas finally went down to .15.
Therefore, the price of gas now is:
12.89 - 0.15 = 12.74
That is, the final price of gas per gallon after number of reduction and increment is $12.74.
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Passes through (8,8) with slope 11/6
Given:
point (8,8).
slope 11/6
The slope intercept form is,
[tex]y=mx+b[/tex]where m is the slope and b is the y-intercept.
we know that m=11/6 so subistute in the equation.
[tex]y=\frac{11}{6}x+b[/tex]Now, let us plug in the point in the equation to find the value of b that is the y-intercept.
[tex]undefined[/tex]Triangle HFG is similar to triangle RPQ. Find the value of x. Find the length of HG.
Answer:
• x=1
,• HG=8 units
Explanation:
If triangles HFG and RPQ are similar, the ratios of their corresponding sides are:
[tex]\frac{HF}{RP}=\frac{HG}{RQ}=\frac{FG}{PQ}[/tex]Substitute the given values:
[tex]\frac{4}{2}=\frac{6x+2}{x+3}=\frac{6}{3}[/tex]First, we solve for x:
[tex]\begin{gathered} \frac{4}{2}=\frac{6x+2}{x+3} \\ 2=\frac{6x+2}{x+3} \\ 2(x+3)=6x+2 \\ 2x+6=6x+2 \\ 6-2=6x-2x \\ 4=4x \\ x=1 \end{gathered}[/tex]Finally, calculate the length of HG.
[tex]\begin{gathered} HG=6x+2 \\ =6(1)+2 \\ =8\text{ units} \end{gathered}[/tex]The Terrell Middle School wants to plant a community garden. They plan togrow and harvest vegetables, which will then be sold to raise funds for futuregardening.1. The science teacher, Ms. Maeda, wants the school to start composting.She borrows $392 from a school fund for supplies to make thecompost bins.Part AStudents plan to pay back half the debt now through fundraising,and the rest after the harvest. Write and solve an equation to representthe debt they will repay through fundraising. Use a negative integer toshow debt.
Total money $392
half of $392 is 196
one half would be paid through fundraising
The debt would be the other half
[tex]\begin{gathered} The\text{ debt} \\ x=\frac{-1}{2}(392) \\ x=-196\text{ dollars} \end{gathered}[/tex]THE FINAL ANSWER
x=-196 dollars
5x + 4 = x + 8. What is the solution for 'x'?
Given the equation
[tex]5x+4=x+8[/tex]To solve this first pass all x-related terms to the left side of the equation and all other terms to the right side:
[tex]5x-x=8-4[/tex]And solve
[tex]\begin{gathered} 4x=4 \\ x=\frac{4}{4} \\ x=1 \end{gathered}[/tex]Which of the following are mathematical sentences? Check all that apply. A. 34 B. r + 7 = 4 C. 5g = 9 D. 4x E. 8r = 12 F. x = 1
The mathematical sentences that can be found in the sentence are:
B. r + 7 = 4 C. 5g = 9 D. 4x E. 8r = 12 F. x = 1What are mathematical sentences?A mathematical sentence can be described as the statement that comprises the two expressions nor more than two expression.
It should be noted that these two expressions can make use of the numbers as well as the variables and in some of the cases combination of them however the mathematical sentence do encompass the symbols which could be inform of equals, greater than, as well as less than.
Therefore, the options that are examples of mathematical sentences are option B C D E F.
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The mathematical sentences which can be found in the sentence are;
B. r + 7 = 4, C. 5g = 9, D. 4x, E. 8r = 12 and F. x = 1
What are mathematical sentences?A mathematical sentence can be described as a statement that comprises two expressions or more than two expressions.
Expression can be defined as the collection of numbers variables and functions by using signs like addition, subtraction, multiplication, and division.
Remember that these two expressions can make use of the numbers as well as the variables and in some cases a combination of them however the mathematical sentence does encompass the symbols which could be in form of equals, greater than, as well as less than.
Hence, the options that are examples of mathematical sentences are ; B. r + 7 = 4, C. 5g = 9, D. 4x, E. 8r = 12 and F. x = 1
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Probability knowledge check (this is math not chemistry I am looking at the tab correctly)
Given: The odds in favor of receiving a gift are 4/19.
Required: To determine the probability of receiving a gift.
Explanation: The probability of an event A that has an odd of happening as A/B can be calculated as
[tex]P(A)=\frac{A}{A+B}[/tex]Here A=4 and B=19. Putting the values, we get,
[tex]\begin{gathered} P(A)=\frac{4}{4+19} \\ =\frac{4}{23} \\ \end{gathered}[/tex]Final Answer: The probability of Brian receiving a gift is 4/23.
Which number is greater in each set?
We have three set of numbers and we must choose the greater value in each set
1.
[tex]\frac{1}{3}or\frac{1}{4}or\frac{1}{5}[/tex]When the numerator is 1, the greater fraction is the one that has the small denominator.
So, in this case the greater number is
[tex]\frac{1}{3}[/tex]2.
[tex]\frac{1}{4}or\frac{4}{3}or\frac{5}{6}[/tex]In this case we can rewrite the fractions as fractions with the same denominator
[tex]\frac{1}{4}=\frac{3}{12}[/tex][tex]\frac{4}{3}=\frac{16}{12}[/tex][tex]\frac{5}{6}=\frac{10}{12}[/tex]Then, the greater number is the one that has the greater numarator
So, it is
[tex]\frac{16}{12}=\frac{4}{3}[/tex]in this case the greater number is
[tex]\frac{4}{3}[/tex]3.
[tex]\frac{16}{5}or3\frac{2}{5}or3.25[/tex]In this case we can rewrite the numbers as decimal numbers
[tex]\frac{16}{5}=3.2[/tex][tex]3\frac{2}{5}=3.4[/tex][tex]3.25=3.25[/tex]In this case the greater number is
[tex]3\frac{2}{5}[/tex]X -5x+4 =015.x+6=0A rectangular painting is 3 feet shorter in length than it is tall (height).12. Write a polynomial to represent the area of the painting.13. Write a polynomial to represent the perimeter of the painting.14. The painting has the unique quality of having an area that has a value that is equal to the value of theperimeter. Find the height of the painting.15. What is the extraneous solution to the polynomial created when the area is set equal to the perimeter?
The rectangular painting is 3 feet shorter in length than in height.
Let "x" represent the painting height, then its length can be expressed as "x-3"
12.
The area of the rectangular painting can be calculated by multiplying its length by its height.
[tex]A=l\cdot h[/tex]For the painting
h=x
l=x-3
-Replace the formula with the expressions for both measurements:
[tex]A=(x-3)x[/tex]-Distribute the multiplication on the parentheses term:
[tex]\begin{gathered} A=x\cdot x-3\cdot x \\ A=x^2-3x \end{gathered}[/tex]The polynomial that represents the area of the painting is:
[tex]A=x^2-3x[/tex]13.
The perimeter of a rectangle is calculated by adding all of its sides, or two times its length and two times its height. You can calculate the perimeter of the painting as follows:
[tex]P=2l+2h[/tex]We know that
h=x
l=x-3
Then the perimeter can be expressed as follows:
[tex]P=2(x-3)+2(x)[/tex]-Distribute the multiplication on the parentheses term:
[tex]\begin{gathered} P=2\cdot x-2\cdot3+2x \\ P=2x-6+2x \end{gathered}[/tex]-Order the like terms together and simplify them to reach the polynomial:
[tex]\begin{gathered} P=2x+2x-6 \\ P=4x-6 \end{gathered}[/tex]14.
The painting has the unique quality of having an area that is equal to the value of the perimeter, then we can say that:
[tex]A=P[/tex]-Replace this expression with the polynomials that represent both measures:
[tex]x^2-3x=4x-6[/tex]To solve the expression for x, first, you have to zero the equation, which means that you have to pass all therm to the left side of the equation. Do so by applying the opposite operation to both sides of the equal sign.
[tex]\begin{gathered} x^2-3x-4x=4x-4x-6 \\ x^2-7x=-6 \\ x^2-7x+6=-6+6 \\ x^2-7x+6=0 \end{gathered}[/tex]We have determined the following quadratic equation:
[tex]x^2-7x+6=0[/tex]Using the quadratic formula we can calculate the possible values for x. The formula is:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where
a is the coefficient that multiplies the quadratic term
b is the coefficient that multiplies the x term
c is the constant of the quadratic equation
For our equation the coefficients have the following values:
a=1
b=-7
c=6
Replace these values in the formula and simplify:
[tex]\begin{gathered} x=\frac{-(-7)\pm\sqrt[]{(-7)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{7\pm\sqrt[]{49-24}}{2} \\ x=\frac{7\pm\sqrt[]{25}}{2} \\ x=\frac{7\pm5}{2} \end{gathered}[/tex]Next is to calculate the addition and subtraction separately:
-Addition
[tex]\begin{gathered} x=\frac{7+5}{2} \\ x=\frac{12}{2} \\ x=6 \end{gathered}[/tex]-Subtraction
[tex]\begin{gathered} x=\frac{7-5}{2} \\ x=\frac{2}{2} \\ x=1 \end{gathered}[/tex]The possible values of x, i.e., the possible heights of the painting are:
x= 6ft
x=1 ft
15.
To determine the extraneous solution created when the area was set equal to the perimeter you have to calculate the corresponding length for both possible values of the height:
For the height x= 1ft, the corresponding length of the painting would be -2ft. This value, although mathematically correct, is not a possible measurement for the painting's length since these types of measures cannot be negative.
