For the function, f(x) =0, we have x = -6 and x =4
The given function is f(x) = x² + 2x - 24
For f(x) = 0
f(x) = x² + 2x - 24 =0
x² + 2x - 24 =0
Middle Term Splitting is a method to solve quadratic equations of the form ax² + bx +c, In middle-term splitting, we split the middle term into the factors of the constant terms, then we take common multiples from the terms and then convert the equation into factors, we equate each factor to zero and we get the desired results.
By splitting the middle term, we have:
x² + 6x -4x - 24 =0
x( x + 6) -4(x+6) =0
( x + 6 ) ( x - 4 ) = 0
( x + 6 ) = 0 and ( x - 4 ) = 0
x = -6 and x = 4
Hence, for f(x) =0, we have x = -6 and x =4
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factoring out: 25m + 10
Answer:
5(5m + 2)
Explanation:
To factor out the expression, we first need to find the greatest common factor between 25m and 10, so the factors if these terms are:
25m: 1, 5, m, 5m, 25m
10: 1, 2, 5, 10
Then, the common factors are 1 and 5. So, the greatest common factor is 5.
Now, we need to divide each term by the greatest common factor 5 as:
25m/5 = 5m
10/5 = 2
So, the factorization of the expression is:
25m + 10 = 5(5m + 2)
Find the domain of the function f(x)=√100x²
The domain of the function √(100x²) will be (-∞,∞) as the definition of domain states that the set of inputs that a function will accept is known as the domain of the function in mathematics.
What is domain?The range of values that we are permitted to enter into our function is known as the domain of a function. The x values for a function like f make up this set (x). A function's range is the set of values it can take as input.
What is function?A function in mathematics from a set X to a set Y assigns exactly one element of Y to each element of X. The sets X and Y are collectively referred to as the function's domain and codomain, respectively.
Here,
The function is √(100x²).
The domain would be (-∞,∞).
The set of inputs that a function will accept is known as the domain of the function in mathematics, and the domain of the function √(100x²) will be (-∞,∞), according to the definition of domain.
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A) 14x + 7y > 21 B) 14x + 7y < 21 C) 14x + 7y 5 21 D) 14x + 7y 221match with graph
As all the options are the same equation
so, we need to know the type of the sign of the inequality
As shown in the graph
The line is shaded so, the sign is < or >
The shaded area which is the solution of the inequaity is below the line
So, the sign is <
So, the answer is option B) 14x + 7y < 21
What is the solution set of x over 4 less than or equal to 9 over x?
we have
[tex]\frac{x}{4}\leq\text{ }\frac{9}{x}[/tex]Multiply in cross
[tex]x^2\leq36[/tex]square root both sides
[tex](\pm)x\leq6[/tex]see the attached figure to better understand the problem
the solution is the interval {-6,6}
the solution in the number line is the shaded area at right of x=-6 (close circle) and the shaded area at left of x=6 (close circle)
Write an equation of variation to represent the situation and solve for the missing information The time needed to travel a certain distance varies inversely with the rate of speed. If ittakes 8 hours to travel a certain distance at 36 miles per hour, how long will it take to travelthe same distance at 60 miles per hour?
The time needed to travel a certain distance varies inversely with the rate of speed, so:
[tex]\begin{gathered} let\colon \\ t=\text{time} \\ v=\text{rate of speed} \\ t\propto\frac{1}{v} \end{gathered}[/tex]8hours----------------------------->36mi/h
xhours----------------------------->60mi/h
[tex]\begin{gathered} \frac{8}{x}=\frac{36}{60} \\ \text{ Since the it varies inversely:} \\ \frac{8}{x}=(\frac{36}{60})^{-1} \\ \frac{8}{x}=\frac{5}{3} \\ \text{solve for x:} \\ x=\frac{3\cdot8}{5} \\ x=4.8h \end{gathered}[/tex]4.8 hours or 4 hours and 48 minutes
А ВC D0 2 4 68 10 12Which point best represents V15?-0,1)A)point AB)point Bpoint CD)point D
We have to select a point that is the best representative of the square root of 15.
We can calculate the square root of 15 with a calculator, but we can aproximate with the following reasoning.
We know that 15 is the product of 3 and 5. If we average them, we have 4.
If we multiply 4 by 4, we get 16, that is a little higher than 15.
If we go to the previous number (3) and calculate 3 by 3 we get 9, that is far from 15 than 16.
So we can conclude that the square root of 15 is a number a little less than 4.
In the graph, the point B is the one that satisfy our conclusion, as it is a point in the scale that is between 3 and 4, and closer to 4.
The answer is Point B
I'm attempting to solve and linear equation out of ordered pairs in slopes attached
We know that the equation of a line is given by
y = mx + b,
where m and b are numbers: m is its slope (shows its inclination) and b is its y-intercept.
In order to find the equation we must find m and b.
In all cases, m is given, so we must find b.
We use the equation to find b:
y = mx + b,
↓ taking mx to the left side
y - mx = b
We use this equation to find b.
1We have that the line passes through
(x, y) = (-10, 8)
and m = -1/2
Using this information we replace in the equation we found:
y - mx = b
↓ replacing x = -10, y = 8 and m = -1/2
[tex]\begin{gathered} 8-(-\frac{1}{2})\mleft(-10\mright)=b \\ \downarrow(-\frac{1}{2})(-10)=5 \\ 8-5=b \\ 3=b \end{gathered}[/tex]Then, the equation of this line is:
y = mx + b,
↓
y = -1/2x + 3
Equation 1: y = -1/2x + 3
2Similarly as before, we have that the line passes through
(x, y) = (-1, -10)
and m = 0
we replace in the equation for b,
y - mx = b
↓ replacing x = -1, y = -10 and m = 0
-10 - 0 · (-1) = b
↓ 0 · (-1) = 0
-10 - 0 = b
-10 = b
Then, the equation of this line is:
y = mx + b,
↓
y = 0x - 10
y = -10
Equation 2: y = -10
3Similarly as before, we have that the line passes through
(x, y) = (-6, -9)
and m = 7/6
we replace in the equation for b,
y - mx = b
↓ replacing x = -6, y = -9 and m = 7/6
[tex]\begin{gathered} -9-\frac{7}{6}(-6)=b \\ \downarrow\frac{7}{6}(-6)=-7 \\ -9-(-7)=b \\ -9+7=b \\ -2=b \end{gathered}[/tex]Then, the equation of this line is:
y = mx + b,
↓
y = 7/6x - 2
Equation 3: y = 7/6x - 2
4The line passes through
(x, y) = (6, -4)
and m = does not exist
When m does not exist it means that the line is vertical, and the equation looks like:
x = c
In this case
(x, y) = (6, -4)
then x = 6
Then
Equation 4: x = 6
5The line passes through
(x, y) = (6, -6)
and m = 1/6
we replace in the equation for b,
y - mx = b
↓ replacing x = 6, y = -6 and m = 1/6
[tex]\begin{gathered} -6-\frac{1}{6}(6)=b \\ \downarrow\frac{1}{6}(6)=1 \\ -6-(1)=b \\ -7=b \end{gathered}[/tex]Then, the equation of this line is:
y = mx + b,
↓
y = 1/6x - 7
Equation 5: y = 1/6x - 7
solve for x. then find the missing piece(s) of the parallelogram for #7
Let us find the angles of the parallelogram below
[tex]\begin{gathered} 2x+30 \\ x=40 \\ 2(40)+30 \\ 80+30 \\ 110^0 \end{gathered}[/tex][tex]\begin{gathered} 2x-10 \\ 2(40)-10 \\ 80-10 \\ 70^0 \end{gathered}[/tex]Theorem+: opposite angles of a parallelogram are the same
Hence the angles of the parallelogram are 110, 70, 110, and 70
Point P is in the interior of
∵ m< OZQ = m[tex]\because m\angle OZP=62[/tex]Substitute the measures of the given angles in the equation above
[tex]\therefore125=62+m\angle PZQ[/tex]Subtract 62 from both sides
[tex]\begin{gathered} \therefore125-62=62-62+m\angle PZQ \\ \therefore63=m\angle PZQ \end{gathered}[/tex]The measure of angle PZQ is 63 degrees
1. flight 1007 will hold 300 passengers the airline has booked 84% of the plane already. How many seats are open for the last-minute Travelers?2.Carly interviewed students to ask their favorite kind of television programs. 12 students claimed that they preferred comedies, 18 like drama 13 enjoy documentaries, and 7 voted for news programs what percentage of the students selected comedies?
The total passengers in the flight is P=300.
Determine the passenger who booked the seat in the flight.
[tex]\begin{gathered} Q=\frac{84}{100}\cdot300 \\ =252 \end{gathered}[/tex]The number of seats booked by passenger is 252.
Determine the seats available for last-minutes travelers.
[tex]\begin{gathered} S=300-252 \\ =48 \end{gathered}[/tex]So 48 seats available for the last minute travelers.
Add the rational expression as indicated be sure to express your answer in simplest form. By inspection, the least common denominator of the given factor is
Notice that the least common denominator is 9*2=18, therefore:
[tex]\begin{gathered} \frac{x-3}{9}+\frac{x+7}{2}=\frac{2(x-3)}{9\cdot2}+\frac{9(x+7)}{9\cdot2}, \\ \frac{x-3}{9}+\frac{x+7}{2}=\frac{2x-6}{18}+\frac{9x+63}{18}, \\ \frac{x-3}{9}+\frac{x+7}{2}=\frac{2x-6+9x+63}{18}, \\ \frac{x-3}{9}+\frac{x+7}{2}=\frac{11x+57}{18}\text{.} \end{gathered}[/tex]Answer:
[tex]\frac{x-3}{9}+\frac{x+7}{2}=\frac{11x+57}{18}\text{.}[/tex]The sum of 19 and twice a number
Answer:
2x+19
Step-by-step explanation:
Let x be the number
2x+19
What is the length of the dotted line in the diagram below? Round to the
nearest tenth.
Answer:
12.1 units
Step-by-step explanation:
Use the distance formula to find the distance between the points given.(-9,3), (7, -6)
Given the points:
[tex](-9,3),(7,-6)[/tex]You need to use the formula for calculating the distance between two points:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1^})^2[/tex]Where the points are:
[tex]\begin{gathered} (x_1,y_1) \\ (x_2,y_2) \end{gathered}[/tex]In this case, you can set up that:
[tex]\begin{gathered} x_2=7 \\ x_1=-9 \\ y_2=-6 \\ y_1=3 \end{gathered}[/tex]Then, you can substitute values into the formula and evaluate:
[tex]d=\sqrt{(7-(-9))^2+(-6-3)^2}[/tex][tex]d=\sqrt{(7+9)^2+(-9)^2}[/tex][tex]d=\sqrt{(16)^2+(-9)^2}[/tex][tex]d=\sqrt{256+81}[/tex][tex]d=\sqrt{337}[/tex][tex]d\approx18.36[/tex]Hence, the answer is:
[tex]d\approx18.36[/tex]Identify the vertex of the function below.f(x) - 4= (x + 1)2-onSelect one:O a. (-4,1)O b.(1,-4)O c. (-1,-4)O d.(-1,4)
The standard equation of a vertex is given by:
[tex]f(x)=a(x-h)^2+k[/tex]where (h,k) is the vertex.
Comparing with the given equation after re-arranging:
[tex]f(x)=(x+1)^2+4[/tex]The vertex of the function is (-1, 4)
12 + 24 =__(__+__)
Find the GCF. The first distributing number should be your GCF
A group of numbers' greatest common factor (GCF) is the biggest factor that all the numbers have in common. For instance, the numbers 12, 20, and 24 share the components 2 and 4.
Therefore, 12 and 24 have the most things in common. Figure 2: LCM = 24 and GCF = 12 for two numbers.
Find the other number if one is 12, then. What does 12 and 24's GCF stand for?
Example of an image for 12 + 24 = ( + ) Locate the GCF. You should distribute your GCF as the first number.
12 is the GCF of 12 and 24. We must factor each number individually in order to determine the highest common factor of 12 and 24 (factors of 12 = 1, 2, 3, 4, 6, 12; factors of 24 = 1, 2, 3,.
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4 Use the sequence below to complete each task. -6, 1, 8, 15, ... a. Identify the common difference (a). b. Write an equation to represent the sequence. C. Find the 12th term (a) our Wilson (All Things Algebral. 2011 Enter your answer(s) here
we have
-6, 1, 8, 15, ...
so
a1=-6
a2=1
a3=8
a4=15
a2-a1=1-(-6)=7
a3-a2=8-1=7
a4-a3=15-8=7
so
the common difference is
d=7
Part 2
write an equation
we have that
The equation of a general aritmetic sequence is equal to
an=a1+(n-1)d
we have
d=7
a1=-6
substitute
an=-6+(n-1)7
an=-6+7n-7
an=7n-13
Part 3
Find 12th term
we have
n=12
a12=-7(12)-13
a12=71
a group orders three large veggie pizzas each slice represents eighth of an entire Pizza the group eats 3/4 of a piece of how many slices of pizza are left
1 pizza contain 8 slices, so we can state a rule of three as:
[tex]\begin{gathered} 1\text{ Pizza ------ 8 slices} \\ \frac{3}{4}\text{ pizza ------ x} \end{gathered}[/tex]then, x is given by
[tex]x=\frac{(\frac{3}{4})(8)}{1}\text{ slices}[/tex]which gives
[tex]\begin{gathered} x=\frac{3}{4}\times8 \\ x=\frac{3\times8}{4} \\ x=3\times2 \\ x=6\text{ slices} \end{gathered}[/tex]that is, 3/4 of pizza is equivalent to 6 slices. So, there are 8 - 6 = 2 slices left of one pizza.
However, they bought 3 large pizzas and ate almost one of them. So, there are 2x8 = 16slices plus 2 slices, that is, 18 slices are left.
Percents build on one another in strange ways. It would seem that if you increased a number by 5% and thenincreased its result by 5% more, the overall increase would be 10%.7. Let's do exactly this with the easiest number to handle in percents.(a) Increase 100 by 5%(b) Increase your result form (a) by 5%.(C) What was the overall percent increase of the number 100? Why is it not 10%?
Answer:
a) 105
b) 110.25
c) Increase of 10.25%. It is not 100% because the second increase of 5% is over the first increased value, not over the initial value.
Step-by-step explanation:
Increase and multipliers:
Suppose we have a value of a, and want a increse of x%. The multiplier of a increase of x% is given by 1 + (x/100). So the increased value is (1 + (x/100))a.
(a) Increase 100 by 5%
The multiplier is 1 + (5/100) = 1 + 0.05 = 1.05
1.05*100 = 105
(b) Increase your result form (a) by 5%.
1.05*105 = 110.25
(C) What was the overall percent increase of the number 100? Why is it not 10%?
110.25/100 = 1.1025
1.1025 - 1 = 0.1025
Increase of 10.25%. It is not 100% because the second increase of 5% is over the first increased value, not over the initial value.
1 4/5 + (2 3/20 + 3/5) use mental math and properties to solve write your answer in simpleist form
Given data:
The given expression is 1 4/5 + (2 3/20 + 3/5).
The given expression can be written as,
[tex]\begin{gathered} 1\frac{4}{5}+(2\frac{3}{20}+\frac{3}{5}_{})=\frac{9}{5}+(\frac{43}{20}+\frac{3}{5}) \\ =\frac{9}{5}+\frac{43+12}{20} \\ =\frac{9}{5}+\frac{55}{20} \\ =\frac{36+55}{20} \\ =\frac{91}{20} \end{gathered}[/tex]Thus, the value of the given expression is 91/20.
The half-life of radium is 1690 years. If 70 grams are present now, how much will be present in 570 years?
Solution
Given that
Half life is 1690 years.
Let A(t) = amount remaining in t years
[tex]\begin{gathered} A(t)=A_0e^{kt} \\ \\ \text{ where }A_{0\text{ }}\text{ is the initial amount} \\ \\ k\text{ is a constant to be determined.} \\ \end{gathered}[/tex]SInce A(1690) = (1/2)A0 and A0 = 70
[tex]\begin{gathered} \Rightarrow35=70e^{1690k} \\ \\ \Rightarrow\frac{1}{2}=e^{1690k} \\ \\ \Rightarrow\ln(\frac{1}{2})=1690k \\ \\ \Rightarrow k=\frac{\ln(\frac{1}{2})}{1690} \\ \\ \Rightarrow k=-0.0004 \end{gathered}[/tex]So,
[tex]A(t)=70e^{-0.0004t}[/tex][tex]\Rightarrow A(570)=70e^{-0.0004(570)}\approx55.407\text{ g}[/tex]Therefore, the answer is 55.407 g
The beginning mean weekly wage in a certain industry is $789.35. If the mean weekly wage grows by 5.125%, what is the new mean annual wage? (1 point)O $829.80O $1,659.60O $41,046.20$43,149.82
Given:
The initial mean weekly wage is $ 789.35.
The growth rate is 5.125 %.
Aim:
We need to find a new annual wage.
Explanation:
Consider the equation
[tex]A=PT(1+R)[/tex]Let A be the new annual wage.
Here R is the growth rate and P is the initial mean weekly wage and T is the number of weeks in a year.
The number of weeks in a year = 52 weeks.
Substitute P=789.35 , R =5.125 % =0.05125 and T =52 in the equation.
[tex]A=789.35\times52(1+0.05125)[/tex][tex]A=43149.817[/tex][tex]A=43149.82[/tex]The new mean annual wage is $ 43,149.82.
Final answer:
The new mean annual wage is $ 43,149.82.
(9 •10^9)•(2•10)^-3)
First, let's distribute the exponent -3 for 2 and ten, like this:
[tex]\begin{gathered} 9\times10^9\times(2\times10)^{-3}^{} \\ 9\times10^9\times2^{-3}\times10^{-3} \end{gathered}[/tex]Now, we can apply the next property when we have a number raised to a negative power:
[tex]a^{-b}=\frac{1}{a^b}[/tex]Then:
[tex]\begin{gathered} 9\times10^9\times2^{-3}\times\frac{1}{10^3} \\ 9\times2^{-3}\times\frac{10^9}{10^3} \end{gathered}[/tex]And when we have a division of the same number raised to different powers we can apply:
[tex]\frac{a^b}{a^c}=a^{b-c}[/tex]then:
[tex]\begin{gathered} 9\times2^{-3}\times\frac{10^9}{10^3} \\ 9\times2^{-3}\times10^{9-3} \\ 9\times2^{-3}\times10^6 \\ 9\times\frac{1}{2^3}^{}\times10^6 \end{gathered}[/tex]Now, as we know, having 10 raised to 6 means that we are multiplying ten by ten 6 times, when we do this we get:
[tex]10\times10\times10\times10\times10\times10=1000000[/tex]And with 2 raised to three we get:
[tex]2\times2\times2=8[/tex]Then we have:
[tex]\begin{gathered} 9\times\frac{1}{8^{}}^{}\times1000000 \\ \frac{9\times1000000}{8^{}}^{} \\ \frac{9000000}{8^{}}^{} \\ \frac{4500000}{4}^{}=11250000 \end{gathered}[/tex]If b is a positive real number and m and n are positive integers, then.A.TrueB.False
we have that
[tex](\sqrt[n]{b})^m=(b^{\frac{1}{n}})^m=b^{\frac{m}{n}}[/tex]therefore
If b is a positive real number
then
The answer is true2. Luis hizo una excursión de 20 km 75 hm 75 dam 250 m en tres etapas. En la primera recorrió 5 km 5 hm, y en la segunda 1 km 50 dam más que en la anterior. ¿Cuánto recorrió en la tercera etapa? Expresa el resultado de forma compleja
Which function has the greatest average rate of change on the interval [1,5]
Answer:
Explanation:
Given: interval [1,5]
Based on the given functions, we start by computing the function values at each endpoint of the interval.
For:
[tex]\begin{gathered} y=4x^2 \\ f(1)=4(1)^2 \\ =4 \\ f(5)=4(5)^2 \\ =100 \\ \end{gathered}[/tex]Now we compute the average rate of change.
[tex]\begin{gathered} \text{Average rate of change = }\frac{f(5)-f(1)}{5-1} \\ =\frac{100-4}{5-1} \\ \text{Calculate} \\ =24 \end{gathered}[/tex]For:
[tex]\begin{gathered} y=4x^3 \\ f(1)=4(1)^3 \\ =4 \\ f(5)=4(5)^3 \\ =500 \end{gathered}[/tex][tex]\begin{gathered} \text{Average rate of change = }\frac{f(5)-f(1)}{5-1} \\ =\frac{500-4}{5-1} \\ =124 \end{gathered}[/tex]For:
[tex]\begin{gathered} y=4^x \\ f(1)=4^1 \\ =4 \\ f(5)=4^5 \\ =1024 \end{gathered}[/tex][tex]\begin{gathered} \text{Average rate of change = }\frac{f(5)-f(1)}{5-1} \\ =\frac{1024-4}{5-1} \\ =255 \end{gathered}[/tex]For:
[tex]\begin{gathered} y=4\sqrt[]{x} \\ f(1)=4\sqrt[]{1} \\ =4 \\ f(5)\text{ = 4}\sqrt[]{5} \\ \end{gathered}[/tex][tex]\begin{gathered} \text{Average rate of change = }\frac{f(5)-f(1)}{5-1} \\ =\frac{(4\sqrt[]{5\text{ }})\text{ -4}}{5-1}\text{ } \\ =1.24 \end{gathered}[/tex]Therefore, the function that has the greatest average rate is
[tex]y=4^x[/tex]What is the solution to the following equation?x^2+3x−7=0
Answer:
Explanation:
Given the equation:
[tex]x^2+3x-7=0[/tex]On observation, the equation cannot be factorized, so we make use of the quadratic formula.
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]Comparing with the form ax²+bx+c=0: a=1, b=3, c=-7
Substitute these values into the formula.
[tex]x=\dfrac{-3\pm\sqrt[]{3^2-4(1)(-7)}}{2\times1}[/tex]We then simplify and solve for x.
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A line passes through the point (-2,-7) and has a slope of 4
Answer:
y= 4x +1
Step-by-step explanation:
The equation of a line, in slope-intercept form, is given by y= mx +c, where m is the slope and c is the y-intercept.
Given that the slope is 4, m= 4.
Substitute m= 4 into y= mx +c:
y= 4x +c
To find the value of c, substitute a pair of coordinates the line passes through.
When x= -2, y= -7,
-7= 4(-2) +c
-7= -8 +c
c= -7 +8
c= 1
Substitute the value of c into the equation:
Thus, the equation of the line is y= 4x +1.
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what is 12 + 0.2 + 0.006 as a decimal and word form
twelve and two hundred six thousandths
The number of algae in a tub in a labratory increases by 10% each hour. The initial population, i.e. the population at t = 0, is 500 algae.(a) Determine a function f(t), which describes the number of algae at a given time t, t in hours.(b) What is the population at t = 2 hours?(c) What is the population at t = 4 hours?
a) Let's say initial population is po and p = p(t) is the function that describes that population at time t. If it increases 10% each hour then we can write:
t = 0
p = po
t = 1
p = po + 0.1 . po
p = (1.1)¹ . po
t = 2
p = 1.1 . (1.1 . po)
p = (1.1)² . po
t = 3
p = (1.1)³ . po
and so on
So it has an exponential growth and we can write the function as follows:
p(t) = po . (1.1)^t
p(t) = 500 . (1.1)^t
Answer: p(t) = 500 . (1.1)^t
b)
We want the population for t = 2 hours, then:
p(t) = 500 . (1.1)^t
p(2) = 500 . (1.1)^2
p(2) = 500 . (1.21)
p(2) = 605
Answer: the population at t = 2 hours is 605 algae.
c)
Let's plug t = 4 in our function again:
p(t) = 500 . (1.1)^t
p(4) = 500 . (1.1)^4
p(4) = 500 . (1.1)² . (1.1)²
p(4) = 500 . (1.21) . (1.21)
p(4) = 500 . (1.21)²
p(4) = 732.05
Answer: the population at t = 4 hours is 732 algae.