EXPLANATION
2) 812: The place of the underlined digits corresponds to UNITS.
VALUE --> 812: eight hundred twelve
Circle describe and correct each error Graph y-5(x-2)Point =(5, 2)M=3
Solution;
[tex]\begin{gathered} y-5=3(x-2) \\ y-5=3x-6 \\ y=3x-1 \\ \end{gathered}[/tex][tex]\begin{gathered} Slope(m)=3 \\ \end{gathered}[/tex][tex]\begin{gathered} Intercept\text{ on x axis; ie y=o} \\ 3x=1 \\ x=\frac{1}{3} \end{gathered}[/tex][tex]\begin{gathered} Intercept\text{ on y axis; ie x=0} \\ y=-1 \end{gathered}[/tex]To the describe the error on the graph;
The graph is supposed to pass through point(1/3, -1).
Find the equation of a line that has the points (3,4) and (-6,5) . Write the answer in slope intercept form
For any line passing through the points (x1,y1) and (x2,y2), its slope is given by:
a = (y2 - y1)/(x2 - x1)
And its intercept is given by:
b = y1 - ax1 = y2 - ax2
And the equation of this line, in the slope-intercept form, is y = ax + b
Then, for a line passing through the points (3,4) and (-6,5), we have:
a = (5 - 4)/(-6 - 3) = -1/9
Then, the intercept is:
b = 4 + 3/9 = 13/3
Therefore, the equation of the line is given by:
y = -x/9 + 13/3
The cost of general admission and student tickets to a high school football game are $7 and $4 respectively. The ticket sales for one game totaled $11,200. The equation that represents this situation is 7x + 4y = 11,200, what is the y-intercept of the graph and what does it represent?A. The y-intercept is 1,600. This tells us that only student's attended the high school football game.B. The y-intercept is 1,600. This tells us that none of the student's attended the high school football game.C. The y-intercept is 2,800. This tells us that none of the student's attended the high school football game.D. The y-intercept is 2,800. This tells us that only student's attended the high school football game.
Given:
The equation that represents the given situation is
[tex]7x+4y=11200.[/tex]Required:
We need to find the y-intercept of the given function.
Explanation:
The y-intercept is the intersection point where the line crosses the y-axis.
Substitute x =0 in the equation to find the y-intercept.
[tex]7(0)+4y=11200.[/tex][tex]4y=11200.[/tex]Divide both sides by 4.
[tex]\frac{4y}{4}=\frac{11200}{4}[/tex][tex]y=2800.[/tex]The y-intercept is 2,800.
We know that x represents the cost of general admission and y represents the student tickets to a high school football.
We get y =2800.
This tells us that only student's attended the high school football game.
Final answer:
The y-intercept is 2,800. This tells us that only student's attended the high school football game.
Write the SLOPE INTERCEPT form equation given the following information * 3 8) through: (2, 2), slope = 1
The intercept on the equation of line is represented by c, where the equation of line is y = mx + c
Now, ( 2,2 ) means x=2, y=2 , the slope is m which is -3/2 ( from the question )
Hence, substitute all into the equation of line to find c
Recall, equation y = mx + c
[tex]\begin{gathered} 2\text{ = }\frac{-3}{2}\text{ ( 2 ) + c } \\ 2\text{ = -3 + c } \\ c\text{ = 2 + 3 ( adding 3 to both sides ) } \\ c\text{ = 5 } \end{gathered}[/tex]Hence, we say that the intercept of the slope C is 5
The equation form is y = -3/2 x + 5 ( since m = -3/2 and c = 5 )
Linear Inequalities (Level 1)Sep 30, 10:57:41 PMWatch help videoSolve the following inequality for p. Write your answer in simplest form.8p + 10 > -5p - 1P
Solve the following inequality for p. Write your answer in simplest form.
8p + 10 > -5p - 1
group terms
8p+5p > -1-10
13p > -11
p > -11/13
Kristi Yang borrowed $12,000. The term of the loan was 150 days, and the annual simple interest rate was 6.5%. Find the simple interest due on the loan. (Round your answer to the nearest cent.)
For an initial ammount borrowed I, an term of the loan t, and an annual interest rate r, the simple interest S due on the loan is given by:
[tex]S=I\cdot r\cdot\frac{t}{365}[/tex]For I = $12000, r = 0.065 and t = 150 days, we have:
[tex]\begin{gathered} S=12000\cdot0.065\cdot\frac{150}{365} \\ S=12000\cdot0.065\cdot0.41096 \\ S=\text{ \$320.55} \end{gathered}[/tex]Determine the equation of the line that passes through the point (-1, 2) and isperpendicular to the line y = -2.
1) In this question, let's find the equation, using the point-slope formula:
[tex](y-y_0)=m(x-x_0)[/tex]2) Notice that since we want a perpendicular line we can write a perpendicular line to y=2, as x=-1/2 for -1/2 is the opposite and reciprocal to 2 (the necessary condition to get a perpendicular line).
So, the slope of that perpendicular line is -1/2
3) Let's plug into that Point-Slope formula, the slope m= -1/2 and the point:
[tex]\begin{gathered} (y-2)=-\frac{1}{2}(x+1) \\ y-2=-\frac{1}{2}x-\frac{1}{2} \\ y=-\frac{1}{2}x-\frac{1}{2}+2 \\ y=-\frac{1}{2}x+\frac{3}{2} \end{gathered}[/tex]4) Thus, the answer is:
[tex]y=-\frac{1}{2}x+\frac{3}{2}[/tex]Determine the intercepts of the line y=51 - 13 y-intercept: ( x-intercept: ( )
Explanation
[tex]y=5x-13[/tex]the y-intercept is the point where the line intersects the y-axis, it is when x=0
and
the x-intercept is the point where the line intersects the x-axis,this is when y=0
so
Step 1
y-intercept
let x=0 and replace
[tex]\begin{gathered} y=5x-13 \\ y=5\cdot0-13 \\ y=0-13 \\ y=-13 \end{gathered}[/tex]so, the ordered pair is ( 0,-13)
Step 2
x-intercept
let y=0 and solve for x
[tex]\begin{gathered} y=5x-13 \\ 0=5x-13 \\ \text{add 13 in both sides} \\ 0+13=5x-13+13 \\ 13=5x \\ \text{divide both sides by 5} \\ \frac{13}{5}=\frac{5x}{5} \\ \frac{13}{5}=x \end{gathered}[/tex]hence, the ordered pair is (13/5, 0)
I hope this helps you
For the following two numbers, find two factors of the first number such that their product is the first number and their sum is the second number,40, 14
First we need to factorate the number 40:
[tex]40=2\cdot2\cdot2\cdot5[/tex]The possible numbers we can create using these factors are 2, 4, 5, 8, 10 and 20.
So If the product of the two factors (let's call them 'a' and 'b') is 40 and the sum is 14, we have:
[tex]\begin{gathered} a\cdot b=40 \\ a+b=14 \\ \\ \text{From the second equation:} \\ b=14-a \\ \\ \text{Using this value of b in the first equation:} \\ a(14-a)=40 \\ 14a-a^2=40 \\ a^2-14a+40=0 \end{gathered}[/tex]Using the quadratic formula to solve this equation, we have:
[tex]\begin{gathered} a_1=\frac{-b+\sqrt[]{b^2-4ac}}{2a}=\frac{14+\sqrt[]{196-160}}{2}=\frac{14+6}{2}=10 \\ a_2=\frac{-b-\sqrt[]{b^2-4ac}}{2a}=\frac{14-6}{2}=4 \\ \\ a=10\to b=14-10=4 \\ a=4\to b=14-4=10 \end{gathered}[/tex]So the factors which product is 40 and the sum is 14 are 4 and 10.
3) 51. Which graphic organizer correctly groups the following numbers? 3.4 -2 3 -1.2
Interger numbers: are whole numbers (numbers without fractional part). Can be negtive and possitive numbers.
Whole numbers: Positive whole numbers
Rational numbers: A number that can be made by dividing two intergers. A number that can be written as a fraction
As in the given numbers not all are whole numbers it can not be intergers or whole.
All the given numbers can be written as fractions:
[tex]\begin{gathered} 3.4=\frac{34}{10}=\frac{17}{5} \\ \\ -2=-\frac{2}{1} \\ \\ 3=\frac{3}{1} \\ \\ -1.2=-\frac{12}{10}=-\frac{6}{5} \end{gathered}[/tex]Then, all are rational numbers
I need help solving this practice problem If you can , answer (a) and (b) separately so I can tell which is which
Step 1:
Write the expression
[tex](3x^5\text{ - }\frac{1}{9}y^3)^4[/tex]Step 2:
a)
[tex]\begin{gathered} (3x^5\text{ - }\frac{1}{9}y^3)^4 \\ =^4C_0(3x^5)^4(-\frac{1}{9}y^3)^0+^4C_1(3x^5)^3(-\frac{1}{9}y^3)^1+^4C_2(3x^5)^2(-\frac{1}{9}y^3)^2+ \\ +^4C_1(3x^5)^1(-\frac{1}{9}y^3)^3+^4C_0(3x^5_{})^0(-\frac{1}{9}y^3)^4 \end{gathered}[/tex]Step 3:
b) simplified terms of the expression
[tex]\begin{gathered} Note\colon \\ ^4C_0\text{ = 1} \\ ^4C_1\text{ = 4} \\ ^4C_2\text{ = 6} \\ ^4C_3\text{ = 4} \\ ^4C_4\text{ = 1} \end{gathered}[/tex]Next, substitute in the expression
[tex]\begin{gathered} =\text{ 1}\times81x^{20}\times1\text{ - 4}\times27x^{15}\text{ }\times\text{ }\frac{y^3}{9}\text{ + 6 }\times9x^{10}\times\frac{y^6}{81}\text{ - 4}\times3x^5\text{ }\times\text{ }\frac{y^9}{729} \\ +\text{ 1 }\times\text{ 1 }\times\frac{y^{12}}{6561}\text{ } \end{gathered}[/tex][tex]=81x^{20}-12x^{15}y^3\text{ + }\frac{2}{3}x^{10}y^6\text{ - }\frac{4}{243}x^5y^9\text{ + }\frac{1}{6561}y^{12}[/tex]Find the indicated probability using the standard normal distribution.P(-2.18
we have
Z1=0
Z2=-2.18
so
using the z-scores table values
P(-2.18DAN have coordinates D(-6, -1) the altitude drawn to side DN
Explanation:
The slope of the altitude drawn to side DN is the reciprocal and opposite to the slope of side DN, because the altitude is perpendicular to the side.
First we have to find the slope of side DN. The formula for the slope of a line with points (x1, y1) and (x2, y2) is:
[tex]m=\frac{y_1-y_2}{x_1-x_2}[/tex]In this problem the points are D(-6, -1) and N(-3, 10). The slope of side DN is:
[tex]m_{DN}=\frac{-1-10}{-6-(-3)}=\frac{-11}{-6+3}=\frac{-11}{-3}=\frac{11}{3}[/tex]Therefore the slope of the altitude is:
[tex]m_{\text{altitude}}=-\frac{1}{m_{DN}}=-\frac{1}{\frac{11}{3}}=-\frac{3}{11}[/tex]Answer:
The slope of the altitude is -3/11
There are 4 options on the dessert menu at a restaurant. Bill and Laura like all of the choices equally, so theyeach choose a dessert at random from the menu. What is the probability that Bill will choose apple pie andLaura will choose strawberry cheesecake for dessert? Express your answer as a decimal. If necessary, roundyour answer to the nearest thousandth.O 0.083O 0.250 0.938O 0.063
So Bill and Laura like all of the choices equally. This means that the probability of each of them choosing one out of the four is the same for each options. So this probability is 1 out of 4: 1/4. So the probability of Bill choosing apple pie is 1/4 and that of Laura choosing strawberry cheesecake is also 1/4. The probability of both making those choices at the same time is given by the product of the mentioned probabilities. Then the probability we are looking for is:
[tex]\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}=0.063[/tex]Then the answer is the fourth option.
in 1996, funding for a program increased by 0.50 billion dollars from the fudning in 1995. in 1997 the increase was 0.56 billion dollars over the funding in 1996. for those three years the funding was 11.40 billion dollars. how much was funded in each of these three years
Using simple mathematical operations, the funding for 3 years is:
Funding in 1995: $3.28 billionFunding in 1996: $3.78 billionFunding in 1997: $4.34 billionWhat are mathematical operations?An operation is a function in mathematics that transforms zero or more input values into a clearly defined output value. The operation's arity is determined by the number of operands. The rules that specify the order in which we should solve an expression involving many operations are known as the order of operations. PEMDAS stands for Parentheses, Exponents, Multiplication, Division, and Addition Subtraction (from left to right).So, let the funding in 1995 be 'x'.
Now, the equation will be formed as:
x + (x + 0.50) + (x + 0.50 + 0.56) = 11.40x + x + 0.50 + x + 0.50 + 0.56 = 11.403x + 1.56 = 11.403x = 11.40 - 1.563x = 9.84x = 9.84/3x = 3.28Hence,
Funding in 1995: $3.28 billionFunding in 1996: $3.28 + $0.50 = $3.78 billionFunding in 1997: $3.78 + $0.56 = $4.34 billionTherefore, using simple mathematical operations, the funding for 3 years is:
Funding in 1995: $3.28 billionFunding in 1996: $3.78 billionFunding in 1997: $4.34 billionKnow more about mathematical operations here:
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Rounding the problem to the nearest tenth if necessary and find the missing length?
Step 1:
[tex]\text{Triangle PQR is similar to triangle GHP}[/tex]Step 2:
Write the corresponding sides of the similar triangle
[tex]\begin{gathered} \\ PQ\text{ }\cong\text{ PG} \\ RP\text{ }\cong\text{ PH} \\ \frac{PQ}{PG}\text{ = }\frac{RP}{PH} \\ \\ \frac{PQ}{91}=\frac{72}{56}\text{ } \end{gathered}[/tex]Next
Cross multiply
[tex]\begin{gathered} 56PQ\text{ = 72 }\times\text{ 91} \\ PQ\text{ = }\frac{6552}{56} \\ PQ\text{ = 117} \end{gathered}[/tex]Final answer
PQ ? = 117
Find the algebraic form g(x) of the function g whose graph is produced by the following transformations on thegraph of f(x) = ×: The graph of f is reflected vertically, expanded horizontally by a factor of 2, shifted right 6 units, and shifteddown 4 units. Include graphs off and g as a part of your answer.
Given the function f(x) = x
We will find the function g(x) whose graph is produced by the following transformations on the graph of f(x)
First, The graph of f is reflected vertically
So,
[tex]\begin{gathered} f(x)\rightarrow f(-x) \\ g(x)=-x \end{gathered}[/tex]Second, expanded horizontally by a factor of 2
So,
[tex]\begin{gathered} f(-x)\rightarrow f(-2x) \\ g(x)=-2x \end{gathered}[/tex]Finally, shifted right 6 units, and shifted down 4 units.
So,
[tex]\begin{gathered} f(-2x)\rightarrow f(-2(x-6))-4 \\ g(x)=-2(x-6)-4 \end{gathered}[/tex]Simplifying the function g(x):
[tex]g(x)=-2x+8[/tex]The graph of the function (f) and (g) will be as shown in the following figure:
Express sin 285 as a function of the reference angle.Question 19 options:sin75sin105-sin75sin-75
Given:
[tex]\sin 285^{\circ}[/tex]To express it as a function of reference angle:
Since the angle lies in the fourth quadrant.
So, the reference angle will be,
[tex]\begin{gathered} \theta_{ref}=360-\theta_4 \\ \theta_{ref}=360^{\circ}-285 \\ \theta_{ref}=75^{\circ} \end{gathered}[/tex]Therefore, the function of the reference angle is,
[tex]\sin 75^{\circ}[/tex]In order to accumulate enough money for a down payment on a house, a couple deposits $513 per month into an account paying 6% compounded monthly. Ifpayments are made at the end of each period, how much money will be in the account in 3 years?Type the amount in the account: $(Round to the nearest dollar)
Step 1- Write out the Future Value Ordinary Annuity formula:
[tex]FV=C\times\frac{(1+r)^n-1}{r}[/tex]Where,
[tex]\begin{gathered} FV=\text{ the future value} \\ C=\text{monthly payments} \\ r=\text{ the interest rate} \\ n=\text{ the number of payments} \end{gathered}[/tex]Step 2- Write out the given values and substitute them into the formula:
[tex]\begin{gathered} C=\$513,r=0.06, \\ n=3\times12=36 \end{gathered}[/tex]Substituting the given values into the formula, we have:
[tex]FV=513\times\frac{(1+0.06)^{36}-1}{0.06}[/tex]Hence,
[tex]FV=513\times\frac{(1.06)^{36}-1}{0.06}[/tex]Hence, the future value is approximately:
[tex]FV\approx\$61109.00[/tex]Hence, the amount in the account in 3 years is:
$61109.00
Doug travels 5 times as fast as Gloria. Traveling in opposite directions, they are 858 miles apart after 6.5 hours. Find their rates or travel.
Distance = rate x time
Gloria = X mph
Doug = 5X mph
Where X is the rate of speed
And we have that they are 858 miles apart after 6.5 hours
6.5 X + 5( 6.5 X) = 858 mph
6.5X + 32,4X= 858 mph
39X = 858mph
X = (858mph)/39 = 22
5X = 110
So Gloria is traveling at 22 mph and Doug is traveling at 110 mph. This meand that Gloria rate of speed is 22 and Doug is 110
I'll show you the picture of the question I'm struggling with
The proportion of the first row with the second row is
[tex]\frac{7}{12}=\frac{8}{16}=\frac{6}{12}=\frac{10}{20}=\frac{2}{4}=\frac{1}{2}[/tex]that is, we have the proportionality 1:2, which correspond to the pink square.
Find the area and perimeter with these points.(-11,-8)(-11,0)(0,0)(0,-8)
we have the coordinates
(-11,-8)
(-11,0)
(0,0)
(0,-8)
step 1
plot the give points
using a graphing tool
see the attached figure below
The figure is a rectangle
where
L=0-(-11)=11 units (subtract x-coordinates)
W=0-(-8)=8 units (subtract y-coordinates)
step 2
Find out the area
A=L*W
A=11*8=88 unit2
step 3
Find out the perimeter
P=2(L+W)
P=2(11+8)
P=2(19)=38 units
Let f(x)=x2-2 and g(x)=9-X. Perform the composition or operation indicated.(fg)(-7)
Given the functions below
[tex]\begin{gathered} f(x)=x^2-2 \\ g(x)=9-x \end{gathered}[/tex]We are to find (fg)(-x)
SOLUTION
First of all, we have to get (fg)(x)
[tex](fg)(x)=(x^2-2)(9-x)[/tex]Expand the function
[tex]\begin{gathered} (fg)(x)=x^2(9-x)-2(9-x) \\ (fg)(x)=9x^2-x^3-18+2x \\ \therefore(fg)(x)=-x^3+9x^2+2x-18 \end{gathered}[/tex]Let us now solve for (fg)(-7)
[tex]\begin{gathered} (fg)(-7)=-(-7)^3+9(-7)^2+2(-7)-18 \\ (fg)(-7)=343+441-14-18=752 \end{gathered}[/tex]Hence,
[tex](fg)(-7)=752[/tex]
14a) Kevin asked a random group of students what theirfavorite class was, and the results are below.If Kevin were to randomly select a boy to explain whichclass was his favorite, what is the probability that he willpick a boy who likes History?A. 19.6%BoysGirlsMath810B. 29.3%English1517C. 35.3%Science105History 189D. 52.9%
If Kevin were to randomly select a boy from this table, this means that we need to add up the total number of boys in this sample in order to find the denominator of the fraction we are dealing with.
Let's add the number of boys up:
[tex]8+15+10+18=51[/tex]Now, the probability that a randomly selected boy likes History is the number of boys who said they like History divided by the total number of boys.
We already have the total number of boys, so now we need to find the number of boys who said they like History, which is 18 based off of the table.
The probability that Kevin will pick a boy who likes History is C) 35.3%.
[tex]\frac{18}{51}=.353[/tex]154
X 97 can you solve this problem
Answer:
14,938
Step-by-step explanation:
i mean multiply it out
A scientist records 61 more
shooting stars in the fall than in the spring.
There are 15 shooting stars in the spring.
How many shooting stars are in the fall?
Answer:
I call this a simple solution because it does not involve probability distributions. It just requires simple knowledge of probability, that’s it!
The question says any 15-minute interval, so one hour can be thought of as four 15-minute intervals.
Now, the question asks the probability of seeing at least one shooting star in those four 15-minute intervals.
Such a probability is the same as the complement of the probability of not seeing any shooting star in those four 15-minute intervals.
All of those four 15-minute intervals are independent of each other. So, their combined probability is the product of the individual probabilities.
The individual probability of not seeing any shooting star in an interval of 15 minutes is 1 - 0.2 = 0.8.
So, the combined probability of not seeing any of the shooting stars in the four intervals is: 0.8 * 0.8 * 0.8 * 0.8 = 0.8⁴ = 0.4096
Now, the complement of its probability is 1 - 0.4096 = 0.5904
So, that is the answer or the probability of seeing at least one shooting star in an interval of an hour.
Step-by-step explanation:
Why?
Before diving into the Poisson process, let me explain why such a demonstration is helpful even if the problem can be solved simply.
The problems we encounter in the real-life can’t be solved by mere simple probability formulas. That is the reason more complex concepts are developed in mathematics. They help model real-life scenarios. When the scenarios are not exactly the same as the mathematical model, we do some assumptions and approximate the scenarios with mathematical models. Then, we do the modeling.
Mathematical modeling requires prior practice. To visualize the mathematical models and naturalize ourselves to such (abstract) models, good practice with the problems always helps. So, even if these problems can be solved easily using simpler methods, such problems provide us a good opportunity to practice mathematics and modeling.
Assumptions
At first, let’s see if the assumptions of the Poisson process hold here. The number of events can be counted. The occurrences of the events are independent of each other. The average rate at which the events occur can be calculated and let’s assume two events can not occur exactly at the same instant in time.
So, all the assumptions of the Poisson process hold here. That gives us the green light to move forward.
Mathematical Work
Let’s assume the rate of λ per minute. Then, the rate corresponds to 15λ per quarter.
Now, the formula for the probability distribution is:
P(X = x) = μ^x * e^-μ / x!
Before we go further, let’s understand the mathematical statement. The mathematical statement expresses the probability of seeing x events in a time period. μ is the average number of events seen in the time period. The time period can be a minute, an hour, or even a day. It can be any time period.
At first, let’s see the first statement — In any 15-minute interval, there is a 20% probability that you will see at least one shooting star.
Here, the interval is of 15 minutes. So, the average number of events in the interval is 15λ. The probability of seeing 0 events (i.e X = 0) is:
P(X = 0) = μ⁰ * e^-(15λ) / 0! = 1 / e^(15λ) — — (1)
(1) is equal to 80% as the probability of at least one event happening is 20%.
So, 1 / e^(15λ) = 0.8
or, e^(15λ) = 1.25
or, 15λ = ln(1.25) = 0.2231
=> λ = 0.0149 / min
So, the average number of events (shooting stars) is 0.0149 per minute.
Now, the average number of events in an hour is:
0.0149 * 60 = 0.8926 / hour
So, the Poisson distribution expression becomes:
P(X = x) = 0.8926^x * e^-0.8926 / x! — — (2)
Here, the time period is an hour.
Now, let’s find the probability of not having even a single event in an hour:
P(X = 0) = 0.8926⁰ * e^-0.8926 / 0! = 0.4096
So, the probability of having at least one event is the complement of the above probability.
So, P(X≥1) = 1 - P(X=0) = 1 - 0.4096 = 0.5904
Step-by-step explanation:
(7x10^1(4x10^-7
(5.55 x 10^4) - ( 3.41 x 10^4)
(9 x 10^7) divided (3 x 10^3)
Work needs to be shows !!!
Answer:
(5.55 * 10^4) - (3.41 * 10^4)
=21,400
(9 * 10^7) divided (3 * 10^3)
= 30,000
Step-by-step explanation:
(5.55 * 10^4) - (3.41 * 10^4)
= (5.55 * 10,000) - (3.41 * 10,000)
= 55,500 - 34,100
= 21,400
(9 * 10^7) divided (3 * 10^3)
= (9 * 10,000,000) ÷ (3 * 1,000)
= 90,000,000 ÷ 3,000
= 30,000
Sorry but i don't understand the "(7x10^1(4x10^7". Your question is invalid.
One year there was a total of 44 commercial and noncommercial orbital launches worldwide. In addition, the number number of commercial orbital launches. Determine the number of commercial and noncommercial orbital launches was two more thank twice the number of commercial orbital launches (HURRY I NEED ANSWER)
The number of commercial orbital is 14, and the number of noncommercial orbital is 30.
What is algebra?
When numbers and quantities are represented in formulas and equations by letters and other universal symbols.
Given that,
The total number of commercial and noncommercial orbital launches worldwide = 44
Also, the number of noncommercial orbital is two more than twice of commercial orbital
Let the number of commercial orbital =x
Then number of noncommercial orbital = 2x+2
Since, total number of commercial and noncommercial orbital = 44
x + 2x +2 = 44
3x = 42
x = 14
The number of commercial orbital = x = 14
The number of noncommercial orbital = 2x+2 = 2×14+2 = 30
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Evaluate the following quotient. Leave your answer in scientific notation.(7.2 x 103) = (5 x 10)AnswerХ
To do a quotient between two numbers in scientific notation we must do the usual division with the number, and subtract the exponents, for example:
[tex]9\times10^7\div3\times10^3=3\times10^4[/tex]Therefore, we must divide
[tex]7.2\times10^3\div5\times10^5[/tex]Do the division:
[tex]\frac{7.2}{5}=1.44[/tex]Now we do the division of the exponents
[tex]\frac{10^3}{10^5}=10^{3-5}=10^{-2}[/tex]Now we put it all together:
[tex]7.2\times10^3\div5\times10^5=1.44\times10^{-2}[/tex]The graph of the quadratic function with vertex (2,3) is shown in the figure belowFind the domain and range
Solution
Domain
[tex]Domain=(-\infty,\infty)[/tex]Range
[tex]Range=(-\infty,3][/tex]