A.
Overall it has a relation that there are more sold cups when the temperature is lower. On the other hand, based on the 40 degrees part, that have to different values of two different days, we can say is not the only factor.
B.
The best lineal approach is the line created with the points at 20 and 80 degrees. First the slope:
[tex]m=\frac{y1-y2}{x1-x2}=\frac{20-10}{20-80}=\frac{10}{-60}=-\frac{1}{6}[/tex]Now the intercept with y axis, b:
[tex]\begin{gathered} y=mx+b \\ 20=20(-\frac{1}{6})+b \\ 20+\frac{20}{6}=b=23.33=\frac{70}{3} \end{gathered}[/tex]The final line formula is:
[tex]y=-\frac{x}{6}+\frac{70}{3}[/tex]Exercise 1: What's In2.Mark’s temperature goes 1.5°C higher from the normal body temperature. What is Marks temperature now?A. 38.5°CB. 37.5°CC. 36.5°CD. 36.5C
The normal body temperature of a human is 37°C.
If Mark's temperature goes 1.5°C higher than that temperature, his new temperature will be:
[tex]\Rightarrow37+1.5=38.5°C[/tex]OPTION A is the correct option.
Question 17
2(h - 6) + 20 = -4
A school bus with the football team left Jefferson HighSchool and drove at an average speed of 48 mph. A schoolbus with the cheerleading squad left 2 hours later and wasable to catch up to the football team after 6 hours. Whatwas the speed of the bus carrying the school's cheerleadingsquad?
Given data
A school bus with the football team from Jefferson High School drove at an average speed of 48mph
Another school bus with the cheerleading squad left 2 hours later and caught with the football team after 6 hours.
Required
To find the speed of the bus carrying the Cheerleading squad.
Step 1
Determine the distance the bus carrying the football team had travelled in the first 2 hours
Speed is given as
[tex]\begin{gathered} \text{speed =}\frac{dis\tan ce}{time} \\ \text{where sp}eed\text{ = 48mph} \\ \text{time = 2 hours} \\ \text{distance = sp}eed\text{ x time} \\ \text{distance = 48 x 2 =96miles} \end{gathered}[/tex]Step 2
Determine of the distance covered by the bus with the football team in the next 6 hours and find the total distance in 8 hours
[tex]\begin{gathered} \text{Distance = sp}eed\text{ }\times\text{ time} \\ where \\ \text{speed = 48mph} \\ \text{time = 6 hours} \\ \text{Distance = 48 }\times\text{ 6 = 288 miles} \end{gathered}[/tex]The total distance in 8 hours covered by the bus = 288 + 96 = 384miles
Step 3
Determine the speed of the bus carrying the Cheerleaders
The total distance to be covered by the Cheerleaders is 384 miles
The total time of their journey to catch with the bus carrying the Football team is 6hours
Hence the speed of the bus required is given as
[tex]\begin{gathered} \text{Speed = }\frac{dis\tan ce}{time} \\ \text{speed = }\frac{384}{6} \\ \text{speed = 64mph} \end{gathered}[/tex]Therefore, the speed of the bus carrying the school's Cheerleaders squad is 64mph
Each n in the model represents the same value. 136.5 What is the value of n? there are 7 N's in the problem
The value of n must be 136.5/7
n = 136.5/7
n = 19.5
Result n = 19.5
Of the twenty-two students in a classroom, ten are transfer students, seven are nursing students, four are AAS students and one student is undecided.If three students are chose randomly, without replacement, find the probability that all three students are nursing students.
Given that:
• There are a total number of 22 students in the classroom.
,• 10 of them are transfer students.
,• 7 are nursing students.
,• 4 are AAS students.
,• 1 student is undecided.
,• Three students are chosen randomly.
Since you need to find the probability that all three students that are chosen randomly are nursing students, you need to set up that this is:
[tex]P(A)[/tex]Where Event A is that one of the students is a nursing student.
Therefore, the probability that three of the chosen students are nursing students can be set up as:
[tex]\begin{gathered} P=P(A)\cdot P(A)\cdot P(A)=P(A)^3 \\ \\ P=P(A)^3 \end{gathered}[/tex]Knowing that the total number of students is 22 and 7 of them are nursing students, you know that:
[tex]P(A)=\frac{7}{22}[/tex]Therefore:
[tex]P=(\frac{7}{22})^3[/tex][tex]P=0.0322[/tex]Hence, the answer is:
[tex]P=0.0322[/tex]What is the equation of the line below in slope-intercept form?(4 Points)x-3y = 6y =- 2y = 3x - 2y = - ** - 2y = -3x - 2
Let's make y the subject of the equation.
[tex]\begin{gathered} x-6=3y \\ y=\frac{x-6}{3} \\ y=\frac{1}{3}x-\frac{6}{3} \\ y=\frac{1}{3}x-2 \end{gathered}[/tex]The correct option is A
NEED ASAP ILL GIVE BRAINLIEST IF CORRECT
A reflection across which line(s) carries the trapezoid onto itself?
If we reflect about x =2, it is a mirror image on each side ( left and right). There is no other line where we can have a mirror image on each side.
Find LM if LN = 137mm.
What is the first step for finding the quotient of 3x^3 z^5/5y * x^2 z^6/20y^3
The initial expression is:
[tex]\frac{3x^3z^5}{5y}\text{ / }\frac{x^2z^6}{20y^3}[/tex]So the first step is to multiply the numerator of the second fraction with the denominator of the first franction and the denominator of the second fraction by the numerator of the first fraction so:
[tex]\frac{3x^3z^6}{5y}(\frac{20y^3}{x^2z^6})[/tex]So is option C)
Please help and answer this question ASAP! :)
Answer:
Odd, Even, Even, Neither=========================
The difference between odd and even functions is that:
f(-x) = f(x) for even functions,f(-x) = - f(x) for odd functions.Let's test this property for the given functions.
Function f(x)f(-4) = - f(4) = 8 and f(-2) = - f(2) = 1, so this is an odd functionFunction g(x)g(4) = g(-4) = -4 and g(2) = g(-2) = 2, so this is an even functionFunction j(x)j(2) = j(-2) = 2 and j(1) = (j-1) = - 4, so this is an even functionFunction k(x)k(-4) = 9, k(4) = 1 and k(-2) = 4, k(2) = 0, since each value is different this is neither odd nor even functionwhich ordered pair is a solution of 6X + 7 < 21
Substitute 2 for x and 1 for in the inequality to verify that ordered pair satisfy the inequality or not.
[tex]\begin{gathered} 6\cdot2+7\cdot1<21 \\ 12+7<21 \\ 19<21 \end{gathered}[/tex]The inequality is trus so point (2,1) satisfy the inequality.
Substitute 4 for x and 1 for y in the inequality to verify that ordered pair satisfy the inequality or not.
[tex]undefined[/tex]At a birthday party, guests ate 452 plates ofchocolate cupcakes and 2/3 plates of cherrycupcakes. How many did the guests eat altogether?If 5 plates of chocolate cupcakes and 5 plates ofcherry were made, how much of each are left?
Given
[tex]\begin{gathered} \text{Ate 4}\frac{5}{12}\text{Plates of chocolate cupcakes} \\ \text{And} \\ \text{Ate 2}\frac{1}{3}\text{plates of cherry} \end{gathered}[/tex]We are to add them together to know the total
[tex]4\frac{5}{12}+2\frac{1}{3}=6\frac{5+4}{12}=6\frac{9}{12}=6\frac{3}{4}[/tex]The final answer
[tex]6\frac{3}{4}[/tex]A bakery makes and sells hot cocoa bombs during the holidays. The first 12 hot cocoa bombs of an order cost is $4.00 each. Each of the next 6 hot cocoa bombs cost $3.50 each. For orders exceeding 18, the cost drops to $3 each. The function C(x) represents the bakery's pricing.
Solution
Step 1
Given data for C(x), the bakery's pricing
[tex]\begin{gathered} F\text{or this range 0}\leq x\leq12ofhotcocoabombs\text{ we use C(x) =4x} \\ \text{For this range }1218,ofhotcocoabombs\text{ we useC(x) = }3x+15 \end{gathered}[/tex]Required
Step 1
To find the cost of 8 hot cocoa bombs
[tex]\begin{gathered} C(8)\text{ lies in the range 0}\leq x\leq12 \\ \text{Hence we use 4x where x = 8} \\ \text{The cost of 8 hot cocoa bombs = 4(8) = \$32} \end{gathered}[/tex]Step 2
To find the cost of 18 hot cocoa bombs
[tex]\begin{gathered} C(18)\text{ lies in the range 12}Step 3To find the C(30)
[tex]\begin{gathered} C(30)\text{ lies in the range x}\ge18 \\ \text{Hence we use 3x +15, where x = 30} \\ C(30)\text{ = 3(30) + 15 = 90 + 15 = \$105} \\ \end{gathered}[/tex]Step 4
What C(30) represents.
C(30) represents the cost of ordering 30 hot cocoa bombs which is $105
Which equation has at least one solution? Mark all that app A. 2x-1= 2 B. 3 y + 1) = 3y 1 C. 5p - (3 + p) = 6p + 1 D. 4/5m=1-1/5m E. 10 +0.5w =1/2w - 10 F. 4a + 3(a - 2) = 8a - (6 + a) Answer Choices:
Let's check the options
A.
2x - 1 = 2
2x= 3
x= 3/2=1.5
option A has atleast one solution
B
3y+ 1 = 3y
option B has no solution
C.
5p - (3 + p) = 6p + 1
5p - 3 - p = 6p + 1
4p - 6p = 1 + 3
-2p = 4
p =-2
option C has atleast one solution
D.
4/5 m = 1- 1/5 m
4/5 m + 1/5m = 1
1m = 1
m = 1
Option D has atleast one solution
E.
10 + 0.5w = 1/2w - 10
0.5 w - 1/2 w = -10 - 10
option E has no solution
F.
4a + 3(a-2) = 8a - (6+a)
4a +3a - 6 = 8a -6 - a
7a -6 = 7a - 6
option F has many solution. Hence it also has atleast one solution
Therefore;
option A, C, D and F has atleast one solution
There is $1.90 in a jar filled with quarters, dimes, and nickels. There are 2 more quarters than dimes and there are 2 more nickels than quarters. How many of each coin are there? quarters dimes [ ) nickels Enter the number that belongs in the green box.
5 quarters, 3 dimes, 7 nickels
Explanations:Let the number of quarters in the jar = q
Let the number of dimes in the jar = d
Let the number of nickels in the jar = n
1 quarter = $0.25
1 dime = $0.1
1 nickel = $0.05
The jar is filled with quarters, dimes, and nickels, totaling $1.90
This can be represented mathematically as:
0.25q + 0.1d + 0.05n = 1.90.........(1)
There are two more quarters than dime:
q = d + 2..............(2)
There are two more nickels than quarters
n = q + 2..............(3)
make d the subject of the formula in equation (2)
d = q - 2............(4)
Substitute equations (3) and (4) into equation (1)
0.25q + 0.1(q - 2) + 0.05(q + 2) = 1.90
0.25q + 0.1q + 0.05q - 0.2 + 0.1 = 1.90
0.4q - 0.1 = 1.90
0.4q = 1.90 + 0.1
0.4q = 2.0
q = 2.0/0.4
q = 5
n = q + 2
n = 5 + 2
n = 7
d = q - 2
d = 5 - 2
d = 3
There are 5 quarters, 3 dimes, 7 nickels
find the solution to the following system by substitution x + y = 20 y = 3x 8
Based on the substitution method, the solution of the system of the equation is x = 3 and y = 17.
Substitution method:
Substitution method is the way of finding the value of any one of the variables from one equation in terms of the other variable.
Given,
Here we have the system of equations
x + y = 20
y = 3x + 8
Now we need to find the solutions for these equation using the substitution method.
From the given details we know that the value of y is defined as 3x + 8.
So, we have to apply these value on the other equation in order to find the value of x,
x + (3x + 8) = 20
4x + 8 = 20
4x = 20 - 8
4x = 12
x = 3
Now apply the value of x into the other equation in order to find the value of y,
y = 3(3) + 8
y = 9 + 8
y = 17
Therefore, the solution of the equation is x = 3 and y = 17.
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4. Which inequality is represented by the graph?8642S-6428X4-6laO4x - 2y > 12O4x - 2y < 12O4x + 2y > 12O4x + 2y < 12
Hello there. To solve this question, we'll have to remember some properties about inequalities and its graphs.
First, we have to determine the equation of the line. For this, we have to find, by inspection, two points contained in that line:
We can easily find the points (0, -6) and (2, -2).
With this, we can find the equation of the line using the point-slope formula:
[tex]y-y_0=m\cdot(x-x_0)[/tex]Where (x0, y0) is a point of the line, as well as (x1, y1) and the slope m is given by:
[tex]m=\frac{y_1-y_0}{x_1-x_0}[/tex]Plugging the coordinates of the points, we get:
[tex]m=\frac{-2-(-6)}{2-0}=\frac{-2+6}{2}=\frac{4}{2}=2[/tex]Such that:
[tex]\begin{gathered} y-(-6)=2\cdot(x-0) \\ y+6=2x \end{gathered}[/tex]Rearranging it in the ax + by = c form,
[tex]2x-y=6[/tex]Multiply both sides of the equation by a factor of 2
[tex]4x-2y=12[/tex]Finally, notice that the values of y in the shaded region are greater than the values in the line, which means that the inequality we're looking for is:
[tex]4x-2y>12[/tex]All the point (x, y) satisfying this inequality are contained in the shaded region.
This not a test btw ! But can you please help me with this !
Using elimination:
[tex]\begin{gathered} (A)-3(B)\colon \\ 6x+12x-3y+3y=-4-15 \\ 18x=-19 \end{gathered}[/tex]Therefore, the answer is:
B) Multiply A by 1 and B by -3
A boat is heading towards a lighthouse, whose beacon-light is 140 feet above the water. The boat’s crew measures the angle of elevation to the beacon, 10∘∘ . What is the ship’s horizontal distance from the lighthouse (and the shore)? Round your answer to the nearest tenth of a foot if necessary.
see the figure below to better understand the problem
we have that
tan(10∘)=140/x -----> by TOA
solve for x
x=140/tan(10∘)
x=794 ft
therefore
The answer is 794 feetAnswer:
Step-by-step explanation:
tan 10=140/x
x=140 / tan 10
x=794
9. If L 1 equals 120 then what is the measure of its supplement <2=
Supplementary angles are angles whose addition sums up to 180 degrees.
Therefore, if angle 1 measures 120, then its supplement which is angle 2, must mean both add up to 180.
Hence, you have
Angle 1 + Angle 2 = 180
120 + Angle 2 = 180
Subtract 120 from both sides of the equation
Angle 2 = 180 - 120
Angle 2 = 60 degrees
By definiton, two angles are complimentary angles if they both add up to 90 degrees. Hence if angle L5 equals 50 degrees, then its compliment would be derived as 90 - 50 which equals 40. The compliment of angle L5 which is 50 degrees, equals 40 degrees.
I need some help with this! I know about the trig identitys and stuff like that, but I just get a little confused on how to apply sometimes.
we have that
Let
x ------> the distance in miles from a point on the ground (the red line)
In the right triangle of the figure
sin(6.5)=7,000/x
solve for x
x=7,000/sin(6.5)
using a calculator
x=61,835.70 ft
Convert to miles
Remember that
1 mile=5,280 ft
so
61,835.7 ft=61,835.7/5,280=11.71 miles
therefore
the answer is 11.71 milesAlyssa will correctly label the numbers 48.4, 48482, 48.09, and 48on the number line below.
The numbers under consideration are:
[tex]48.4,\text{ 48}\frac{1}{2},\text{ 48.09, 48}\frac{3}{5}[/tex]Converting all the numbers to decimal:
[tex]\begin{gathered} 48\frac{1}{2}=\text{ 48+0.5 = 48.5} \\ 48\frac{3}{5}=\text{ 48 + }0.6\text{ = 48.6} \end{gathered}[/tex]Therefore, the numbers can be written as:
48.4, 48.5, 48.09, and 48.6
Out of these numbers, only 48.6 is closest to 49
[tex]48\frac{3}{5}\text{ is closest to 49}[/tex]Given four numbers P, Q, R and S. The first three numbers form an arithmetic sequence while the last three form a geometric sequence. If the sum of the first and the fourth number is 16 and the sum of the second and the third number is 12, find these four numbers.
The most appropriate choice for arithmetic and geometric series will be given by-
P = 0, Q = 4 , R = 8, S = 16 or P = 15, Q = 9, R = 3, S = 1 are the required numbers
What is arithmetic and geometric series?
Arithmetic series are those series whose difference between two consecutive terms are same.
Geometric series are those series whose ratio between two consecutive terms are same.
Here,
P, Q and R forms an Arithmetic sequence
Let P = a - d , Q = a, R = a + d, Where a is the first term of the Arithmetic sequence and d is the common difference of the sequence.
Let S = b
Q, R and S forms a Geometric sequence
[tex]\frac{a + d}{a} = \frac{b}{a +d}[/tex]
[tex](a + d)^2 = ab\\a^2 + d^2 + 2ad = ab\\[/tex] ............... (1)
Now the sum of first and fourth number is 16
a - d + b = 16
b = 16 - a + d
Putting the value of b in (1),
[tex]a^2 + d^2 + 2ad = a(16 - a +d)\\a^2 + d^2 + 2ad = 16a -a^2+ad\\a^2+a^2 + d^2+2ad - ad - 16a = 0\\2a^2 + ad+d^2 -16a = 0[/tex]............ (2)
Sum of second and third number is 12
[tex]a + a + d = 12\\2a +d = 12\\d = 12-2a[/tex]
Putting the value of d in (2)
[tex]2a^2 + a(12 - 2a)+(12 - 2a)^2-16a = 0\\2a^2 + 12a - 2a^2+144-48a+4a^2 - 16a = 0\\4a^2-52a+144=0\\4(a^2-13a+36)=0\\a^2 -13a+36=0\\a^2-9a-4a+36=0\\a(a - 9)-4(a-9)=0\\(a-4)(a-9) = 0\\[/tex]
[tex]a - 4 = 0[/tex] or [tex]a - 9 = 0[/tex]
[tex]a = 4[/tex] or [tex]a = 9[/tex]
When a = 4,
[tex]d = 12 - 2\times 4\\d = 12 - 8\\d = 4[/tex]
[tex]b = 16 - 4+4\\b = 16[/tex]
P = 4 - 4 = 0
Q = 4
R = 4 + 4 = 8
S = 16
When a = 9,
[tex]d = 12 - 2\times 9\\d = 12 - 16\\d = -6[/tex]
[tex]b = 16 - 9-6\\b = 1[/tex]
P = 9 - (-6) = 15
Q = 9
R = 9 + (-6) = 3
S = 1
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An envelope is 15 centimeters wide, and it measures 17 centimeters along the diagonal. The envelope is __ centimeters tall.
An envelope is rectangular in shape.
Given the width = 15cm, and diagonal = 17cm
Let h represent the tall length of the envelope
Applying Pythagoras theorem, we have
[tex]\begin{gathered} 17^2=15^2+h^2 \\ 289=225+h^2 \\ h^2=289-225 \\ h^2=64 \\ h=\sqrt[]{64} \\ h=8\operatorname{cm} \end{gathered}[/tex]The envelope is 8cm tall
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
Rob earns $4200 per month at his new job. He pays the following taxes: 6.2% for social security 1.45% for Medicare 16% for federal income tax • 5.5% for state income tax Calculate his annual net income.
Answer:
$35,708.4
Explanation:
His net income will be the total that he earns less the taxes.
So, we need to calculate how much money does Rob pays for each tax.
Therefore, 6.2% of 4200 is equal to:
[tex]4200\times6.2\text{ \% =4200 }\times\frac{6.2}{100}=260.4[/tex]It means that Rob pays $260.4 each month for social security,
In the same way, 1.45% of 4200 is equal to:
[tex]4200\times1.45\text{ \% = 4200}\times\frac{1.45}{100}=60.9[/tex]16% of 4200 is equal to:
[tex]4200\times16\text{ \% = 4200}\times\frac{16}{100}=672[/tex]5.5% of 4200 is equal to:
[tex]4200\times5.5\text{ \% = 4200}\times\frac{5.5}{100}=231[/tex]Now, we can calculate the net income per month as:
$4200 - $260.4 - $60.9 - $672 - $231 = $2975.7
Finally, his annually net income will be the net income per month multiplied by 12 months:
$2975.7 x 12 = $35,708.4
So, the answer is $35,708.4
2. Given: ZMOP is a right angle RP I OP Prove: MO || RP
Given that;
[tex]\begin{gathered} \measuredangle MOP\text{ is a right angle.} \\ \measuredangle MOP=90^0 \end{gathered}[/tex]And;
[tex]\vec{RP}\perp\vec{OP}[/tex]Since line RP is perpendicular to line OP, Angle RPO must be a right angle.
[tex]\measuredangle RPO=90^0[/tex]Recall that for two parallel lines intersected by a straight line, Same side interior angles are supplementary.
[tex]A+B=180^0[/tex]So, for line MO to be parallel to line RP, the sum of angle MOP and angle RPO must be equal to 180 degree.
[tex]\measuredangle MOP+\measuredangle RPO=90+90=180^0[/tex]Since the sum of angle MOP and angle RPO is equal to 180 degree, then line MO is parallel to line RP.
[tex]\begin{gathered} \text{ Since} \\ \measuredangle MOP+\measuredangle RPO=180^0 \\ \text{Then;} \\ MO\Vert RP \end{gathered}[/tex]Proved
-10 is no less than 2 times a number plus 14
Let the number be x.
Then according to the question,
[tex]\begin{gathered} -10\ge2x+14 \\ -10-14\ge2x \\ -24\ge2x \\ x\ge-12 \end{gathered}[/tex]Thus, the number should be greater than or equal to -12.
ok my question is math algebra. consider the linear equation y-1=0 and grapthe two points
To find:
We need to find two points on the linear equation y-1=0 and to plot those points on graph.
Step by step solution:
We know that:
General coordinate of any two points on line y = 1:
= (x, 1)
So let us assume any two random points on the line:
= (1,1) and (2,1)
We will now mark them on the graph:
