Answers
A percentage of 100 · [p' - exp(- 4000 · k)] / p' of carbon-14 has decayed and lost in last 4000 years.
How to determine the remaining percentage of carbon-14 in ancient pollen
In this question we have the case of simple decay of a radioactive isotope (carbon-14) in time, which is represented by the following exponential formula:
p(t) = p' · exp(- k · t)
Where:
p' - Initial amount of pollen.p(t) - Current amount of pollen.k - Decay constant.t - Time, in years.If we know that t = 4000 years, then percentage of lost carbon-14 is:
p = 100 - 100 · p(t) / p'
p = 100 - [100 · exp(- 4000 · k)] / p'
p = 100 · {1 - [exp(- 4000 · k)] / p'}
p = 100 · [p' - exp(- 4000 · k)] / p'
An amount of 100 · [p' - exp(- 4000 · k)] / p' per cent has been decayed and lost.
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Related Questions
For each diagram, write and solve an inequality for x.
7.
2x - 5
X
What is the answer
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For the given diagram the inequality for x is x ≥ 6.
What is inequality?If two values differ, it can be determined whether one is larger, smaller, or simply not equal to the other.
A and B claim that they are not equal.
A must be less than B if A = B.
An is greater than b when a > b. These two constitute strict inequality.
A must be less than or equal to b if a and b are divided.
An is greater than b if a and b are equal.
We have given a right-angled triangles with hypotenuse 2x - 5 and base x
Using the Pythagoras theorem
c² = a² + b²
We know hypotenuse is always greater than base with that in mind lets assume x to be 1
then 2x - 5
= 2(1)-5
= -3 which is smaller than 1
It means x at least need to be equal to 6 or greater than 6 to Pythagoras theorem to be possible
When x = 6
= 2(6)-5
= 7 hypotenuse
Gives the inequality
x ≥ 6
Thus, For the given diagram the inequality for x is x ≥ 6.
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Pls help this is geometry and its a really important test
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Answer:#4:B m of angle 2=35° m of angle 3=55°
Step-by-step explanation:if line C is 180° degrees because a flat line in 180° then you see that adding the angle measure of 4 which is 35° +the right angle which is 90 degree=125° So the 180°-125°=55°=m of angle 3 and to find angle measure of 2 angle 1 is a right angle and the inside of a triangle opens to 180 degrees so you grab measure of angle 1=90° and measure of angle 3=55° add them 90°+55°=145° so 180°-145°=35°
simplify both questions and give full explanation please
Answers
Answer:
First question: 1/x^-7
Second question:
x^7
Step-by-step explanation:
To divide these problems (same variable base and exponents) subtract the exponents.
To get rid of a negative on the exponent, "push" the term across the fraction bar. Passing over the fraction bar changes the sign of the exponent. There are math reasons for this, its not random. But thats how it works. (Has to do with an exponent of -1 which will give you the reciprocal of your base).
Also, to multiply terms with the same base, add the exponents.
see image.
Hello! I just want to confirm my answer is correct and if there’s anything I should add? Thanks for your help!
Answers
Given:
given functions are
[tex]f(x)=x^4-4x^3-2x^2-12x+9,g(x)=\sqrt{x^2-2x-3},h(x)=\frac{-x^2+1}{x^2-2x-3}[/tex]Find:
(A) we have to compare the Domain and range of the function f(x) and g(x).
(B) We have to find the relationship between the break of h(x) and zeros of f(x).
Explanation:
The domain and Range of f(x) is
[tex]\begin{gathered} Domain(f)=(-\infty,\infty) \\ Range(f)=[0,\infty) \end{gathered}[/tex]Domain and Range of g(x) is
[tex]\begin{gathered} Domain(g)=(-\infty,-1]\cup[3,\infty) \\ Range(g)=[0,\infty) \end{gathered}[/tex]Domain of h(x) is
[tex]Domain(h)=(-\infty.-1)\cup(-1,3)\cup(3,\infty)[/tex]Now zeros of the function f(x) are
[tex]\begin{gathered} x^4-4x^3-2x^2+12x+9=0 \\ (x+1)^2(x-3)^2=0 \\ x=-1,-1,3,3 \end{gathered}[/tex]Therefore, zeors of the function f(x) are -1,-1,3,3.
Now,
(A)The difference between Domain of f(x) and g(x) is of the interval (-1,3). The domain of f(x) is all Real Numbers and Domain of g(x) is all the real number except the interval (-1,3).
The Range of both f(x) and g(x) is same.
(B) The breaks in the Domain of h(x) are equal to the zeros -1, 3 of f(x).
please help me out quickly
Answers
An identity matrix is a square matrix with 0s everywhere else and 1s on the major diagonal.
1. The given statement exists true.
2. The given statement exists false.
What is an inverse matrix transform?A rigid body transformation is the inverse matrix. It not only conforms to the original matrix's shape, but if you translate and rotate an object, you may return it to its original location by undoing the translations and rotations.
In general, AB = BA, even if A and B exist as both squares. If AB = BA, then we express that A and B commute. For a general matrix A, we cannot express that AB = AC yields B = C. (However, if we comprehend that A exists invertible, then we can multiply both sides of the equation AB = AC to the left by A-1 and get B = C.)
2. The given statement exists false.
An identity matrix is a square matrix with 0s everywhere else and 1s on the major diagonal.
Identity matrices of orders 1, 2, and 3 are those in which all the members of a square matrix are zero and none of the diagonal elements are zero. Keep in mind that when k = 1, a scalar matrix is an identity matrix. But it is obvious that any identity matrix is a scalar matrix.
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The first statement is true but the second given statement is false .
Inverse of a matrix: What does that mean?The idea of the inverse of a matrix is a multidimensional generalization of the idea of the reciprocal of a number. Just as the product of a number and its inverse is equal to 1, that of a square matrix and its inverse is equal to the identity matrix.
Even though A and B are squares, AB Equals BA in general. We can say that A and B commute if AB = BA. We cannot state that AB = AC produces B = C for a general matrix A. (However, if we are aware that A is invertible, we can multiply both sides of the expression AB = AC to the left by A-1 to obtain B = C.)
2. The aforementioned claim is untrue.
A square matrix with 1s on the principal diagonal and 0s everywhere else is known as an identity matrix.
When all of a square matrix's members are zero and none of the diagonal elements are zero, the matrix is said to be an identity matrix of order 1, 2, or 3. Do not forget that k = 1 .
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Which characteristic of the line that passes through the points (6,10) and (12,2)..
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SOLUTION:
Let us find the slope of the line:
[tex]m\text{ = }\frac{y_2-y_1}{x_2-x_1}[/tex]Where x1 = 6, x2 = 12, y1 = 10 and y2 = 2
[tex]\begin{gathered} m\text{ = }\frac{2-10}{12-6} \\ m\text{ = }\frac{-8}{6}\text{ = -}\frac{4}{3} \\ \\ \end{gathered}[/tex]The slope of the given line is -4/3
if you do this I will name my future kid after you
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We have to determine wheter the line segments MN and RS are parallel, perpendicular or neither. To do this we will find the slope of each of this segments; the slope is given by
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]For the line segment MN we have M(-2,2) and N(1,-3), then
[tex]\begin{gathered} m_1=\frac{-3-2}{1-(-2)} \\ =\frac{-5}{3} \end{gathered}[/tex]For the line segment RS we have R(-2,1) and (3,4), then
[tex]\begin{gathered} m_2=\frac{4-1}{3-(-2)} \\ =\frac{3}{5} \end{gathered}[/tex]Now that we have the slopes of each lines segments we have to remember two theorems.
T1. Two lines are parallel if and only if
[tex]m_1=m_2[/tex]T2. Two lines are perpendicular if and only if
[tex]m_1m_2=-1[/tex]Once we know this theorems we can answer the question.
First, we notice that the slopes of this segments are not equal so we can conlcude that they are not parallel.
Let's see if they are perpendicular, to do this we multiply the slopes
[tex]\begin{gathered} m_1m_2=(-\frac{5}{3})(\frac{3}{5}) \\ =-\frac{15}{15} \\ =-1 \end{gathered}[/tex]Since the result of their multiplication is -1 we conclude that this lines segments are perpendicular.
Which of the following could be the ratio between the lengths of the two legsof a 30-60-90 triangle?Check all that apply.A. 2:13OB. 5:15C. 2:15OD. 1 : 15E. 1 : 2UO F. 5:3
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Let a, b, c be the length of sides of the triangle lying opposite to angles measuring 30, 60 and 90 degrees, respectively.
Applying the sine law,
[tex]\frac{a}{\sin30}=\frac{b}{\sin60}=\frac{c}{\sin 90}[/tex]27. The figure is formed from rectangles. Find the total area. The diagram is not to scale.3 ft5 ft3 ft8 ftA. 26 ft?B. 40 ftC. 73 ft?D. 34 ft?
Answers
To find the total area, we can divide the total figure into two figures as follows:
Then, we have that the total area is the sum of both areas of the figures:
[tex]A_{\text{total}}=5ft\cdot5ft+3ft\cdot3ft=25ft^2+9ft^2=34ft^2[/tex]Therefore, the total area is equal to 34 sq. feet (option D).
Need Help on this problem
Answers
Answer:
Marginal Average Cost Function C'(x) = 5.7
Step-by-step explanation:
The Marginal Average Cost is just the first differential of the Cost Function
[tex]\dfrac{d}{dx} C(x) = \dfrac{d}{dx}(161) + \dfrac{d}{dx}(5.7x)\\\\\\The first differential of a constant is 0 and the first differential of ax is a \\\\So ,\\\\\dfrac{d}{dx} C(x) = 0 + 5.7 = 5.7[/tex]
This means that as x increases by 1 unit, cost increases by 5.7 units. x presumably refers to the number of units produced
The Marginal Revenue can be found the same way
[tex]\dfrac{d}{dx} R(x) = \dfrac{d}{dx}6x - \dfrac{d}{dx}(0.08x^2)\\\\\dfrac{d}{dx}6x = 6\\\\\dfrac{d}{dx}(0.08x^2) = (2)(0.08)x[/tex]
[tex]= 0.16x[/tex] since [tex]\dfrac{d}{dx} (x^2) = 2x[/tex]
The cost of creating a software program is $5000. Every extra feature added to the software costs $100. Thetotal charge of the software with x extra features is given by the function f(x) = 100x + 5000. How will the graphof this function change if the basic cost is raised to $5200 and the cost of each extra feature is increased to$120?
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The original function is:
[tex]f(x)=100x+5000[/tex]Which means that it has a slope of 100, and a starting point of 5000.
As we can see on the image above, the first point is (0,5000), which is the starting point of the function. The second point is (2, 5200), which means that for a 2 unit increase on the features, we got 200 increase on the price, leading to a slope of 100.
If we change the basic cost to 5200, the starting point of the function will increase by 200 units, if we change the slope to 120, then for every feature we add, the cost will increase faster. In summary, the cost of the program will be higher to start, and it will also increase faster. This is illustrated in the image below:
[tex]g(x)=120x+5200[/tex]In green we have the new function, and in red we have the old one. We can see that the new one increases faster, and is always above the old one.
Please help I don't understand how to do this. I am stuck
For this exercise, you will demonstrate the tests to show that y=x^19-8x^7+7x^4 is neither an even function nor an odd function.
A) When you apply the test for evenness and simplify the resulting equation, you get
y=
B) When you apply the test for oddness and simplify the resulting equation, you get
y=
Answers
The given equation is proved that y=x¹⁹-8x⁷+7x⁴ is neither an even nor an odd function.
Given the equation is y=x¹⁹-8x⁷+7x⁴
Replace x and -x and check to see if the resultant equation matches the
original equation to see whether the function is even
Write the original equation into a function first by swapping y for f(x).
f(x) = x¹⁹-8x⁷+7x⁴
now, replace x with -x
f(-x) = (-x)¹⁹-8(-x)⁷+7(-x)⁴
f(-x) = -x¹⁹+8x⁷+7x⁴
now check whether f(x) = f(-x)
since f(x)≠f(-x), the function is not even.
If you want to determine whether a function is odd, you should check to see if f(-x) = -f (x). The function is odd if the two equations are the same.
-f(x) = -(x¹⁹-8x⁷+7x⁴)
-f(x) = -x¹⁹+8x⁷-7x⁴
now check whether f(-x) = -f(x)
since f(-x)≠-f(x), the function is not odd.
hence the equation is neither an even nor an odd function.
Therefore, the equation is proved that it is neither even nor odd.
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It takes 6 slices of bread, 9 oz of cheese, and 2 oz of butter tomake three grilled-cheese sandwiches. What is the cost per sandwichif bread (18 slices) costs $1.90, 1 lb of cheese costs $2.49, and1/2 lb of butter costs $1.29?a. $.78b. $4.87c. $2.12d. $1.62
Answers
Step 1: Write the equivalence of each unit
[tex]\begin{gathered} 1lb\Rightarrow16\text{ oz} \\ 1\text{ lb of cheese }\Rightarrow\text{ 16 oz of che}ese \\ \frac{1}{2}\text{ lb of butter }\Rightarrow\text{ 8 oz of butter} \end{gathered}[/tex]Step 2: Calculate the cost of producing 3 grilled-cheese sandwiches
[tex]\begin{gathered} 18\text{ slices of bread cost \$1.90} \\ 1\text{ slice will cost x} \\ \Rightarrow x=\frac{1.90}{18} \\ 6\text{ slices of bread will cost=}\frac{1.90}{18}\times6=\text{ \$0.63} \end{gathered}[/tex][tex]\begin{gathered} 1\text{ lb of cheese cost \$2.49 } \\ \text{ Since 1 lb is equal to 16 oz} \\ 16\text{ oz of cheese cost \$2.49} \\ 1\text{ oz of cheese will cost \$x} \\ \Rightarrow\text{ \$x=}\frac{2.49}{16}\text{ } \\ 9\text{ oz will cost }\Rightarrow\text{ }\frac{2.49}{16}\times9=\text{ \$1.40} \end{gathered}[/tex][tex]\begin{gathered} \frac{1}{2}\text{ lb of butter cost \$1.29} \\ \text{ This implies } \\ 8\text{ oz of butter will cost \$1.29} \\ 1\text{ oz of butter will cost \$x} \\ \Rightarrow\text{ \$x =}\frac{1.29}{8} \\ 2\text{ oz will then cost =}\frac{1.29}{8}\times2=\text{ \$0.32} \\ \end{gathered}[/tex]From the above calculations we can calculate the cost of producing three grilled-cheese sandwiches as
[tex]0.63+1.40+0.32=\text{ \$2.35}[/tex]Thus, the cost per sandwich is given as
[tex]\frac{2.35}{3}=\text{ \$0.78}[/tex]Hence, the cost per sandwich is $0.78
Option A is the right answer
Suppose you’re given the following table of values for the function f(x), and you’re told that the function is off:
Answers
Odd function f(x):
f(-x) = -f(x)
Let's see what is happening for x = 0:
f(-0) = -f(0)
But -0 = 0 so: f(0) = -f(0) => 2f(0) = 0 => f(0) = 0
But, from the table: f(0) = 2
So the function can't be an odd function, D is the correct answer
Anyone know how to work this?
Answers
The income from the sales of potatoes in 2021 is equal to €510.
How to calculate the income from potatoes?From the information provided, we can logically deduce that the total income (revenue) that was generated from the sales of the four (4) vegetables in 2021 is equal to 2040 Euros (€).
Note: Each interval of the bar chart on the y-coordinate represents €40.
Assuming the pie chart was drawn to scale, we can reasonably infer and logically deduce that each of its sector represent 25%. This ultimately implies that, the income for the sales of potatoes in 2021 is 0.25 of the total income (revenue) that was generated from the sales of the four (4) vegetables in 2021.
Mathematically, the income for the sales of potatoes in 2021 can be calculated as follows:
Income from potatoes = 25/100 × 2040
Income from potatoes = 0.25 × 2040
Income from potatoes = €510.
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The volume of this triangular prism is 20,580 cubic millimeters. What is the value of p?
Answers
We have that the formula of the volume of the triangular prism is:
[tex]V=A\cdot h[/tex]Where A is the area of the base and h is the height. Then we can write the volume like this:
[tex]\begin{gathered} A=\frac{p\cdot b}{2} \\ \Rightarrow V=(\frac{p\cdot b}{2})\cdot h=\frac{p\cdot b\cdot h}{2} \end{gathered}[/tex]now, if b=20, h=42 and V=20580, then we substitute and solve for p:
[tex]\begin{gathered} V=\frac{p\cdot b\cdot h}{2} \\ \Rightarrow20580=\frac{p\cdot20\cdot42}{2}=420\cdot p \\ \Rightarrow p=\frac{20580}{420}=49 \\ p=49 \end{gathered}[/tex]therefore, the value of p is 49 milimeters
formulas need help will send picutre
Answers
In the formula
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_2}[/tex]putting in the given values of R_1, R_2, and R_3 gives
[tex]\frac{1}{R}=\frac{1}{10.0}+\frac{1}{14.0}+\frac{1}{35.0}[/tex]We add the fractions on the right-hand- side to get:
[tex]\frac{1}{R}=\frac{1\cdot7}{10.0\cdot7}+\frac{1\cdot5}{14.0\cdot5}+\frac{1\cdot2}{35.0\cdot2}[/tex][tex]\frac{1}{R}=\frac{7}{70}+\frac{5}{70}+\frac{2}{70}[/tex][tex]\frac{1}{R}=\frac{14}{70}=\frac{1}{5}[/tex]Taking the reciprocal both sides gives
[tex]R=5.0[/tex]Which is our answer!
Find the distance between (5, -7) and (-2,-4)
Answers
Answer: 7.62
Step-by-step explanation:
I am not understanding this question can you help me solve this? Please I desperately need your help
Answers
Given:
[tex]\[g\left(x\right)=\left\{\begin{matrix}\sqrt[3]{x+5} \\ -{{x}^2}+11\end{matrix}\text{ }\begin{matrix}x\le-4 \\ x>-4\end{matrix}\right.\][/tex]To find:
The value of:
[tex]g(-4)[/tex]Explanation:
For x = -4, we can consider the below function,
[tex]g(x)=\sqrt[3]{x+5}[/tex]Substituting the value x = -4,
[tex]\begin{gathered} g(-4)=\sqrt[3]{-4+5} \\ =\sqrt[3]{1} \\ =1 \end{gathered}[/tex]Final answer:
Hence, the required solution is:
[tex](A)\text{ 1}[/tex]Two candles start burning at the same time. one candle is 15 cm tall and burn and burns at a rate of 5 cm every 6 hours. The other candle is 25 cm tall at a rate of 2 1/2 cm every hour
How tall will the candles be when they first burn down to the same height?
Answers
The height of the candles when they would be the same height is 10cm.
What is the height when the two candles would be the same?The first step is to determine how much of each candle is left after one hour.
Candle left after c hours = height of the candle - rate at which the candle burns per hour x c
Rate at which the candle burns per hour = 5/6
15 - 5/6c
Candle left after c hours = 25 - 2 1/2c
When the two candles are of the same height, the two above equations would be equal
15 - 5/6c = 25 - 2 1/2c
Combine similar terms: 25 - 15 = 2 1/2 - 5/6c
Add similar terms: 10 = 1 2/3 c
Divide both sides by 3/5: 10 = 5/3c
c = 10 x 3 / 5
c = 6 hours
Height in 3 hours = 15 - 5/6(6) = 15 - 5 = 10cm
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B / 7 equals 3 what is b
Answers
Answer:
B = 21
Step-by-step explanation:
B/7 =3
cross multiply
B = 21
Range for function y=-3x+12
Answers
The range of y=-3x+12 is (-∞,∞) or (y | y ∈ R)
What is Domain and Range ?
The components of a function are its domain and range. The domain of a function is the set of all possible input values, whereas the range of a function is its potential output. Range, Domain, and Function
Since no domain is specified to determine the precise range, As a result, the range of the supplied equation, y=-3x+12, only includes real integers.
Interval Notation : (−∞,∞)
The range is the set of all valid y values. You can also use the graph to find the range. The range is ,
Interval range : (−∞,∞)
Hence, The Range is (-∞,∞) or (y | y ∈ R)
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Help I will gib brain
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Answer:
x-axis: (18, 0)
y-axis: (0, 12)
quadrant I: (4.5, 10.6)
quadrant II: (-15, 21)
quadrant III: [tex](-\frac{3}{4}, -\frac{4}{9})[/tex]
quadrant IV: (3, -7)
Step-by-step explanation: (view attached screenshot)
(18, 0) is 18 units left, 0 units up. This make the point line up exactly on the x-axis
(0, 12) is 0 units left, and 12 units up. This will cause it to be in the line of y-axis.
(4.5, 10.6) both x and y values in the point are positive, this means that it's in the first quadrant
(-15, 21) x value is negative, y value is positive. This means that the point is in the second quadrant
[tex](-\frac{3}{4}, -\frac{4}{9})[/tex] both x and y values are negative. This means that the point is in the third quadrant.
(3, -7) x value is positive, y value is negative. This means that it's in the fourth quadrant.
Graph the exponential function.G(x)=(1/3)^xPlot five point on the graph of the function.
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Graph the function:
[tex]G(x)=(\frac{1}{3})^x[/tex]We'll use the following values of x: {-2, -1, 0, 1, 2}.
Substituting:
[tex]G(-2)=(\frac{1}{3})^{-2}=3^2=9[/tex][tex]G(-1)=(\frac{1}{3})^{-1}=3^1=3[/tex][tex]G(0)=(\frac{1}{3})^0=1[/tex][tex]G(1)=(\frac{1}{3})^1=0.333[/tex][tex]G(2)=(\frac{1}{3})^2=0.111[/tex]The graph of the function is shown below:
Order twenty-six eighths, the cube root of 32, negative pi, and negative four and two thirds from greatest to least.
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The ordering of the numbers from greatest go least will be twenty-six eighths, the cube root of 32, negative pi, and negative four and two thirds
How to order true numbers?Based on the information, we want to order twenty-six eighths, the cube root of 32, negative pi, and negative four and two thirds.
It should be noted that 26/8 = 3.25
Cube root of 32 = 3.19
Negative pi = -3.14
Negative four and two thirds = -4.67
The arrangement is given above.
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help help help pleaseeeee!!!!!!
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a) The linear equation that models the price-sales relationship for toy is C(x) = -500x +5500
b) The forecast calls for 2250 sales at a $6.50 pricing.
Define slope.The ratio of the increase in elevation between two points to the run in elevation between those same two points is referred to as the slope.
A line's equation is represented by:
y = mx +b
, where
The slope, or m, represents the rate of change.
The value of y at x = 0 is represented by the y-intercept or b.
Item a:
In this issue:
Two points are (6, 2500) and (8, 1500).
The slope is calculated by dividing the change in y by the change in x, so:
m = [tex]\frac{1500-2500}{8-6}[/tex]
m = [tex]\frac{-1000}{2}[/tex]
m = -500
Thus,
y = -500x +b
Point (6,2500) indicates that, which we utilize to find b, is true when.
y = -500x +b
2500 = -500(6) + b
b = 5500
Thus
y = -500x + 5500
Item b:
When x = 6.5, sales are y, so:
y = -500(6.5) + 5500
y = 2250
The forecast calls for 2250 sales at a $6.50 pricing.
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The sum of the even numbers between 77 and 535 is how much less than the sum of the odd numbers between78 and 536?
Answers
The sum of the even numbers between 77 and 535 is:
[tex]78+80+82+\ldots+530+532+534=70074[/tex]Now, the sum of the odd numbers between 78 and 536 is:
[tex]77+79+81+\cdots+531+533+535=70380[/tex]Thus, the sum of the even numbers between 77 and 535 is 306 less than the sum of the odd numbers between 78 and 536 because
[tex]70380-70074=306[/tex]write an expression in simplest form for the perimeter of a right triangle with leg lengths of 12a^4 and 16a^4an expression is____a^4
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Given the lengths of the right triangle:
[tex]\begin{gathered} 12a^4^{} \\ \text{and } \\ 16a^4 \end{gathered}[/tex]To find the perimeter, use the formula:
[tex]P=a+b+\sqrt[]{a^2+b^2}[/tex]Thus, we have:
[tex]P=12a^4+16a^4+\sqrt[]{(12a^4)^2+(16a^4)^2}[/tex][tex]\begin{gathered} P=28a^4+\sqrt[]{144a^4+256a^4} \\ P=28a^4+\sqrt[]{400a^4} \\ P=28a^4+20a^4 \\ P=48a^4 \end{gathered}[/tex]Therefore, the expression is:
[tex]48a^4[/tex]Melissa is planning a rectangular vegetable garden with a square patch for tomatoes. She wants the length of the garden to exceed three times the length of the tomato patch by two feet. She also wants the garden’s width to exceed the width of the tomato patch by five feet.Part AMelissa wants to know how the width and length of the garden relate to the length of the square tomato patch. If each side of the tomato patch is x feet, write the functions WG(x) and LG(x) to represent the garden’s width and length, respectively.Part BWrite the function AG(x) representing the area of the garden in terms of x.Part CIf the sides of the square tomato patch are seven feet, find the area of the garden.Part DMelissa decides to reserve a patch in her vegetable garden for growing bell peppers. She wants its width to be half the width of the tomato patch, and its length must exceed the length of the tomato patch by two feet. Write the functions WB(x) and LB(x) representing the width and length, respectively, of the bell pepper patch.Part EWrite the function AB(x) representing the area of the bell pepper patch in terms of x.Part FWrite the function ATB(x) representing the combined area of the tomato patch and the bell pepper patch.Part GYou’ve written functions to represent the area of the tomato patch and the area of the bell pepper patch. Now write the function AR(x) for the remaining planting area in the garden.Part HIf Melissa wants the area of the bell pepper patch to be 31.5 square feet, find the area of the remaining space in the garden after planting tomatoes and bell peppers.
Answers
Part A
The functions would be:
[tex]\begin{gathered} WG(x)=x+5 \\ LG(x)=3x+2 \end{gathered}[/tex]Part B
[tex]\begin{gathered} AG(x)=WG(x)\cdot LG(x) \\ \rightarrow AG(x)=(x+5)(3x+2) \\ \rightarrow AG(x)=3x^2+2x+15x+10 \\ \\ \Rightarrow AG(x)=3x^2+17x+10 \end{gathered}[/tex]Part C
Let's evaluate x = 7 in AG(x)
[tex]\begin{gathered} AG(7)=3(7^2)+17(7)+10 \\ \rightarrow AG(7)=276 \end{gathered}[/tex]Thereby, the area of the garden would be 276 square feet
Part D
The functions would be:
[tex]\begin{gathered} WB(x)=\frac{x}{2} \\ LB(x)=x+2 \end{gathered}[/tex]Part E
[tex]\begin{gathered} AB(x)=WB(x)\cdot LB(x) \\ \rightarrow AB(x)=(\frac{x}{2})(x+2) \\ \\ \Rightarrow AB(x)=\frac{x^2}{2}+x \end{gathered}[/tex]Part F
[tex]\begin{gathered} ATB(x)=x^2+AB(x) \\ \rightarrow ATB(x)=x^2+\frac{x^2}{2}+x \\ \\ \Rightarrow ATB(x)=\frac{3}{2}x^2+x \end{gathered}[/tex]Part G
[tex]\begin{gathered} AR(x)=AG(x)-ATB(x) \\ \rightarrow AR(x)=3x^2+17x+10-\frac{3}{2}x^2-x \\ \\ \Rightarrow AR(x)=\frac{3}{2}x^2+16x+10 \end{gathered}[/tex]Part H
We have a function for the area of the bell pepper patch in terms of x, the measurement of the lenght and width of the tomato patch. This is:
[tex]AB(x)=\frac{x^2}{2}+x[/tex]We know the value of this area. This way, we can solve the equation for x,
[tex]31.5=\frac{x^2}{2}+x\rightarrow63=x^2+2x\rightarrow x^2+2x-63=0[/tex]Using the cuadratic formula, and ignoring non-positive results, we'll get that
[tex]x=7[/tex]Now, plugging in this value in AR(x),
[tex]\begin{gathered} AR(7)=\frac{3}{2}(7^2)+16(7)+10 \\ \Rightarrow AR=195.5 \end{gathered}[/tex]This way, we can conclude that the remaining space in the garden after planting tomatoes and bell peppers is 195.5 square feet
Solve the following for 'x'x+1+3 2= xO A. -1O B. 3O c. 3OD. 1
Answers
Step 1: "braking" fractions
[tex]\begin{gathered} \frac{x}{3}+\frac{x+1}{2}=x \\ \frac{x+1}{2}=\frac{x}{2}+\frac{1}{2} \end{gathered}[/tex]We replace the second in the original equation:
[tex]\frac{x}{3}+\frac{x}{2}+\frac{1}{2}=x[/tex]Step 2: rearraging the equation (the terms with x on one side, numbers on the other)
[tex]\begin{gathered} \frac{x}{3}+\frac{x}{2}+\frac{1}{2}=x \\ \frac{x}{3}+\frac{x}{2}=x-\frac{1}{2} \\ \frac{x}{3}+\frac{x}{2}-x=-\frac{1}{2} \end{gathered}[/tex]Step 3: adding fractions
Since
[tex]\begin{gathered} \frac{1}{3}+\frac{1}{2}-1=\frac{1}{3}+\frac{1}{2}-\frac{1}{1} \\ =(\frac{1}{3}+\frac{1}{2})-\frac{1}{1} \\ \end{gathered}[/tex]We know that
[tex]\begin{gathered} (\frac{1}{3}+\frac{1}{2})=\frac{1\cdot2+1\cdot3}{3\cdot2} \\ =\frac{2+3}{6} \\ =\frac{5}{6} \end{gathered}[/tex]Replacing it:
[tex]\begin{gathered} (\frac{1}{3}+\frac{1}{2})-\frac{1}{1}=\frac{5}{6}-\frac{1}{1} \\ =\frac{5\cdot1-6\cdot1}{6\cdot1} \\ =\frac{5-6}{6} \\ =-\frac{1}{6} \end{gathered}[/tex]Then
[tex]\begin{gathered} \frac{x}{3}+\frac{x}{2}-x=-\frac{1}{2} \\ -\frac{1}{6}x=-\frac{1}{2} \end{gathered}[/tex]Step 4: finding x
[tex]\begin{gathered} -\frac{1}{6}x=-\frac{1}{2} \\ \frac{1}{6}x=\frac{1}{2} \\ 6\cdot\frac{1}{6}x=6\cdot\frac{1}{2} \\ x=3 \end{gathered}[/tex]Answer: C.x=3