We have to prove that [tex]\sqrt{2}+\sqrt{5}[/tex] is irrational. We can prove this statement by contradiction.
Let us assume that [tex]\sqrt{2}+\sqrt{5}[/tex] is a rational number. Therefore, we can express:
[tex]a=\sqrt{2}+\sqrt{5}[/tex]
Let us represent this equation as:
[tex]a-\sqrt{2}=\sqrt{5}[/tex]
Upon squaring both the sides:
[tex](a-\sqrt{2})^{2}=(\sqrt{5})^{2}\\a^{2}+2-2\sqrt{2}a=5\\a^{2}-2\sqrt{2}a=3\\\sqrt{2}=\frac{a^{2}-3}{2a}[/tex]
Since a has been assumed to be rational, therefore, [tex]\frac{a^{2}-3}{2a}[/tex] must as well be rational.
But we know that [tex]\sqrt{2}[/tex] is irrational, therefore, from equation [tex]\sqrt{2}=\frac{a^{2}-3}{2a}[/tex] the expression [tex]\frac{a^{2}-3}{2a}[/tex] must be irrational, which contradicts with our claim.
Therefore, by contradiction, [tex]\sqrt{2}+\sqrt{5}[/tex] is irrational.
Step-by-step explanation:
Let us assume [tex]\bf \sqrt{2}+\sqrt{5} [/tex] as rational.
[tex]\therefore \: \:\bf \sqrt{2}+\sqrt{5} = \dfrac{a}{b}[/tex]
(where a and b are coprimes)[tex]\implies\:\:\bf \sqrt{5} = \dfrac{a}{b} - \sqrt{2} [/tex]
[tex]\implies\:\:\bf(\sqrt{5})^{2} = \bigg( \dfrac{a}{b} -\sqrt{2}\bigg)^{2} [/tex]
squaring on both sides[tex]\implies\:\:\bf 5 = \dfrac{a^{2} }{b^{2}} - 2 \bigg(\dfrac{a}{b}\bigg) (\sqrt{2}) + (\sqrt{2})^{2}[/tex]
[tex]\implies\:\:\bf 5 = \dfrac{a^{2} }{b^{2}} - 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg) + 2[/tex]
[tex]\implies\:\:\bf 5-2 = \dfrac{a^{2}}{b^{2}}- 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)[/tex]
[tex]\implies\:\:\bf 3 = \dfrac{a^{2}}{b^{2}}- 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)[/tex]
[tex]\implies\:\:\bf 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)= \dfrac{a^{2}}{b^{2}} -3[/tex]
[tex]\implies\:\:\bf 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)=\dfrac{a^{2}-3b^{2}}{b^{2}}[/tex]
[tex]\implies\:\:\bf 2 \sqrt{2} = \bigg( \dfrac{a^{2}-3b^{2}}{b^{2}} \bigg) \bigg( \dfrac{b}{a}\bigg) [/tex]
[tex]\implies\:\:\bf \sqrt{2} =\dfrac{a^{2}-3b^{2} }{2ab}[/tex]
[tex]\bf Irrational \cancel{=} Rational [/tex][tex]\textsf Hence, our assumption is wrong [/tex]
[tex] \therefore \:\:\sf {{\bf{\sqrt{2}+\sqrt{5}}} is irrational }[/tex]
[tex]\mathfrak\purple{{\purple{\bf{H}}}ence\: {\purple{\bf{P}}}roved}[/tex]
Jill had an AGI of $25,000. She had $2800 in medical expenses, paid $6000 in rent, and had to buy a new uniform for work, which was not reimbursed by her employer. Which expense(s) can she itemize on her tax return? A.Nonreimbursed work expenses, mortgage interest, and medical expenses B.Mortgage interest and medical expenses C.Mortgage interest only D.Medical expenses and nonreimbursed work expenses.
Answer:
medical expenses and non-reimbursed work expenses
Step-by-step explanation:
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x * 1 + x/1 = _______.
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For what value of x:
is the square of the binomial x+1 twenty greater than the square of the binomial x–3?
Final answer:
The value of x for which the square of the binomial (x+1) is twenty greater than the square of the binomial (x-3) is 3.5.
Explanation:
The student is asking for a value of x for which the square of the binomial (x+1) is twenty greater than the square of the binomial (x-3). To find this value, we set up an equation based on the given information:
(x + 1)² = (x - 3)² + 20
First, we expand both squares:
x² + 2x + 1 = x² - 6x + 9 + 20
Now, simplify and move all terms to one side to solve for x:
8x = 28
Divide both sides by 8 to find the value of x:
x = 28 / 8
x = 3.5
Therefore, the value of x for which the square of (x+1) is twenty greater than the square of (x-3) is 3.5.
Find dy/dx by implicit differentiation. 8 cos x sin y = 6
The sum of 5 consecutive numbers is 135
Answer:
Step-by-step explanation:
The answer is 25+26+27+28+29=135
Is the correct answer I basically divided 135 by 5 and got 27 then I just worked around that number to get the answer.
Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 3z = 9
The largest rectangular box volume in the first octant, with one vertex on [tex]\(x + 2y + 3z = 9\),[/tex] is [tex]\(\frac{486}{125}\).[/tex]
To find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex on the plane [tex]\(x + 2y + 3z = 9\),[/tex]we can set up the problem using optimization techniques.
Let the coordinates of the vertex of the box that lies on the plane [tex]\(x + 2y + 3z = 9\)[/tex] be[tex]\((x, y, z)\).[/tex] Since the other vertices are on the coordinate planes, the dimensions of the box are [tex]\(x\), \(y\), and \(z\).[/tex]
The volume [tex]\(V\)[/tex]of the rectangular box is given by:
[tex]\[V = x \cdot y \cdot z\][/tex]
Given that this vertex lies on the plane [tex]\(x + 2y + 3z = 9\),[/tex] we have the constraint:
[tex]\[x + 2y + 3z = 9\][/tex]
We need to maximize [tex]\(V\)[/tex] subject to this constraint. To do this, we can use the method of Lagrange multipliers. We introduce a Lagrange multiplier [tex]\(\lambda\)[/tex] and define the Lagrangian function:
[tex]\[\mathcal{L}(x, y, z, \lambda) = x y z + \lambda (9 - x - 2y - 3z)\][/tex]
To find the critical points, we take the partial derivatives of [tex]\(\mathcal{L}\)[/tex] with respect to [tex]\(x\), \(y\), \(z\), and \(\lambda\)[/tex] and set them to zero:
[tex]\[\frac{\partial \mathcal{L}}{\partial x} = yz - \lambda = 0\]\[\frac{\partial \mathcal{L}}{\partial y} = xz - 2\lambda = 0\]\[\frac{\partial \mathcal{L}}{\partial z} = xy - 3\lambda = 0\]\[\frac{\partial \mathcal{L}}{\partial \lambda} = 9 - x - 2y - 3z = 0\][/tex]
From the first three equations, we can express [tex]\(\lambda\)[/tex] as follows:
[tex]\[\lambda = yz\]\[\lambda = \frac{xz}{2}\]\[\lambda = \frac{xy}{3}\][/tex]
Equating these expressions for [tex]\(\lambda\):[/tex]
[tex]\[yz = \frac{xz}{2} \implies 2yz = xz \implies x = 2y \quad \text{(if \(z \neq 0\))}\]\[yz = \frac{xy}{3} \implies 3yz = xy \implies y = 3z \quad \text{(if \(x \neq 0\))}\][/tex]
Substituting [tex]\(y = 3z\) and \(x = 2y = 2(3z) = 6z\)[/tex] into the constraint [tex]\(x + 2y + 3z = 9\):[/tex]
Now, using [tex]\(z = \frac{3}{5}\):[/tex]
[tex]\[y = 3z = 3 \left(\frac{3}{5}\right) = \frac{9}{5}\]\[x = 6z = 6 \left(\frac{3}{5}\right) = \frac{18}{5}\][/tex]
The dimensions of the box are:
[tex]\[x = \frac{18}{5}, \quad y = \frac{9}{5}, \quad z = \frac{3}{5}\][/tex]
The volume[tex]\(V\)[/tex] is:
[tex]\[V = x \cdot y \cdot z = \left(\frac{18}{5}\right) \left(\frac{9}{5}\right) \left(\frac{3}{5}\right) = \frac{18 \cdot 9 \cdot 3}{5^3} = \frac{486}{125} = 3.888\][/tex]
Therefore, the volume of the largest rectangular box is:
[tex]\[\boxed{\frac{486}{125}}\][/tex]
The volume of the largest rectangular box with one vertex on the plane x + 2y + 3z = 9 is found using Lagrange multipliers. The maximum volume is 4.5 cubic units. The calculations involve the gradient method and substitution.
To find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 3z = 9, we need to maximize V = xyz subject to the constraint x + 2y + 3z = 9.
We can use the method of Lagrange multipliers for this problem:
Define the function we want to maximize, f(x, y, z) = xyz.Introduce the constraint as a new function, g(x, y, z) = x + 2y + 3z - 9 = 0.Set up the system of equations using the gradient of the function and the constraint: ∇f = λ∇g.This gives us the following system of equations:
yz = λxz = 2λxy = 3λx + 2y + 3z = 9From these equations, we can solve for x, y, z, and λ:
λ = yzλ = xz / 2λ = xy / 3Equating and solving, we obtain x = 1.5, y = 1.5, and z = 2.
Finally, substituting these values into V = xyz gives the volume V = (1.5) imes (1.5) imes 2 = 4.5.
What is the arc length of an angle of 2π 3 radians formed on the unit circle? A) π 3 B) 2π 3 C) 4π 3 D) 5π 3
The arc length of an angle of 2π/3 radians formed on the unit circle is 2π/3.
To find the arc length of an angle of 2π/3 radians on the unit circle, we can use the formula:
Arc Length = Radius * Angle
Since we are considering the unit circle, the radius is 1. Therefore, the arc length is equal to the measure of the angle.
In this case, the angle is 2π/3 radians. So, the arc length is 2π/3.
The correct option is A) π/3.
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find the volume of this prism
Larry used a pattern of colors to make a cube train he use Red Cube a blue cube a Red Cube and another Red Cube before he started the pattern again he use 15 cubes how many red cubes did Larry use
write a polynomial (x+6)(x-2)(x-1)
Which expression represents the sum of 2/3m - 1 1/6 and 5/6m - 1 1/3
1. A paper cup designed to hold popcorn is in the shape of a cone. The diameter of the cup is 12 centimeters and the height is 16 centimeters. What is the volume of popcorn the cup could hold? Use 3.14 for pi. Enter your answer, as a decimal, in the box.
Answer:
Volume of popcorn cup = 602.88 cm^3
Step-by-step explanation:
Volume of a cone = 1/3 πr^2 h
Given: π = 3.14, r = 12/2 = 6 cm, h = 16 cm
Now plug in these values in the above formula, we get
Volume of popcorn cup = 1/3 * 3.14*6^2*16
= 1/3*3.14 *36*16
= 3.14 *12*16
Volume of popcorn cup = 602.88 cm^3
Hope this will helpful.
Thank you.
Steven, a tailor, got an order to make a blazer. The customer specifically asked him to save 5/6 of a foot of the given cloth to make a pocket square. However, Steven accidentally saved 5/12 of a foot. What is the difference between the requested cloth and the saved cloth? A. 0.1466' B. 0.4166' C. 0.4265' D. 04066'
A coin is tossed 6 times what is the probability of getting all heads
On eight book shelves there are 44 books per shelf. How many books are there altogether? If 1/4 of these books are novels, how many novels would there be?
Suppose you have 15 days until your field trip and you need to raise $900 there are 10 students going on the field trip they will each help fundraise how much should each student have raised in 1 week?
In a survey of 1756 adults 37% responded yes to the Severy question how many adults answer yes?
Start with the number n = 54527. Divide n by 5 and round the result up to an integer. Keep repeating the division and rounding step until the resulting number is less than 5. How many divisions are performed? You can use a calculator for this problem, but you should not have to actually perform all of the divisions.
The question involves applying the concept of significant figures and rounding numbers during divisions to determine how many times 54527 must be divided by 5 before it becomes less than 5 without manually performing each division.
The question asks for the number of times the number n = 54527 needs to be divided by 5 and rounded up until it is less than 5. This is a problem that can be solved by understanding exponential decay and the concept of significant figures. It is also an exercise in rounding numbers appropriately. To determine the number of divisions without actually performing each division, one can use logarithms.
The concept of significant figures is important in this context, as each division reduces the number of significant figures by approximately one (since we're dividing by a number that has only one significant figure, 5). The rule is that when dividing, the number of significant figures in the result should be the same as the smallest number of significant figures in the input values.
Here's the process:
Apply logarithms to find the exponent x in 5ˣ = n.Recognize that each division by 5 reduces the exponent by 1.Calculate the number of times x must be reduced by 1 until the value is less than 1, which corresponds to the original number being less than 5.The original calculation without the repetition of divisions would use logarithms to solve for x in 5ˣ = 54527, or log5(54527). However, for the purposes of this example and to avoid calculator work, we can estimate that since 5⁴ = 625 and 5⁵ = 3125, it will take more than 4 but significantly fewer than 10 divisions (as 510 is much greater than 54527) to make the number less than 5.
Construct arguments-Janie served four same size pizzas. Explain how to find how many slices of pizza served if the angle for each slice turns through a right angle
A wet bicycle tire leaves a trace of water on the floor. The tire has a radius of 14 inches, and the bicycle wheel makes 3 full rotations before stopping. How long is the trace of water left on the floor? Give your answer in terms of pi.
A scientist finds that on one side of a mountain 35 cacti have purple flowers and 16 have white flowers. If he goes to the other side of the mountain, what is the experimental probability that the first cactus he comes across has white flowers?
Answer: The experimental probability that the first cactus he comes across has white flowers is 16/51.
Step-by-step explanation:
Since,
[tex]\text{Experimental probability} = \frac{\text{Number of event occurrence}}{\text{Number of trials}}[/tex]
Let W represents the event of occurrence of white flower and P represents the occurrence of purple flower,
Then, According to the question,
n(P) = 35 and n(W) = 16
Also, the total number of trials, n(S) = 35 + 16 = 51
Thus, the probability of occurring white flower is,
[tex]P(W)=\frac{n(W)}{n(S)}=\frac{16}{51}[/tex]
Find the general solution of the given second-order differential equation. y'' − y' − 30y = 0 webassign
The general solution of the given second-order differential equation
y'' - y' - 30y = 0 is,
⇒ y = C₁ e⁶ˣ + C₂ e⁻⁵ˣ
What is mean by Function?A relation between a set of inputs having one output each is called a function. and an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).
Given that;
The second-order differential equation is,
⇒ y'' − y' − 30y = 0
Now, We can simplify as;
⇒ y'' − y' − 30y = 0
This gives the general form as;
⇒ m² - m - 30y= 0
⇒ m² - 6m + 5m - 30 = 0
⇒ m (m - 6) + 5 (m - 6) = 0
⇒ (m + 5) (m - 6) = 0
⇒ m = - 5 or m = 6
Hence, The general solution of the given second-order differential equation y'' - y' - 30y = 0 is,
⇒ y = C₁ e⁶ˣ + C₂ e⁻⁵ˣ
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35 less than 7 times a number is 98. what is the number?
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.3 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.)
The angle's rate of change when the bottom of the ladder is 8ft from the wall is
[tex]\dfrac{-1.3}{6} rad/s[/tex]
It is given the Length of Ladder [tex](h)[/tex] is [tex]10ft.[/tex]. and the distance between the bottom of the ladder to the wall [tex](r)[/tex] is [tex]8ft.[/tex] as shown in the below figure.
By using the Pythagoras Theorem
[tex]b=\sqrt{10^{2}-8^{2} }\\=\sqrt{36} \\=6ft.[/tex]
and
[tex]cos(\theta)= r/h\\cos(\theta)=r/10[/tex]
Differentiating both sides with respect to [tex]'t'[/tex] by using the chain rule
[tex]-sin(\theta)\dfrac{\mathrm{d}\theta }{\mathrm{d} t}=\dfrac{1}{10} \dfrac{\mathrm{d}r }{\mathrm{d} t}\\\\\dfrac{\mathrm{d}\theta }{\mathrm{d} t}=\dfrac{1}{10} \dfrac{\mathrm{d}r }{\mathrm{d} t} \dfrac{1}{-sin(\theta)} } ......(eq. 1)[/tex]
given
[tex]\dfrac{\mathrm{d}r }{\mathrm{d} t}=1.3ft./s\\sin(\theta)=\frac{6}{10}[/tex]
putting this in eq.1, we get
[tex]\dfrac{\mathrm{d} \theta}{\mathrm{d} t} = \dfrac{1.3}{10} (\dfrac{1}{-\frac{6}{10} }) \\\dfrac{\mathrm{d} \theta}{\mathrm{d} t} =\dfrac{-1.3}{6} rad/s[/tex]
So the angle's rate of change when the bottom of the ladder is 8ft from the wall is[tex]\dfrac{-1.3}{6} rad/s[/tex].
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siplifier
cos(π/7)+cos(2π/7)+cos(3π/7)+cos(4π/7)
What is the solution to the system of equations graphed on the coordinate plane?
(3,4) would be the answer on e2020/edge
Is the number of total molecules on the left side of a balanced equation always equal to the number of total molecules on the right side of the equation? explain your answer.3. is the number of total molecules on the left side of a balanced equation always equal to the number of total molecules on the right side of the equation? explain your answer?
Since no reaction creates or destroys atoms, every balanced chemical equation must have equal numbers of atoms of each element on each side of the equation. However, the number of molecules does not necessarily have to be the same.
The answer is
No, the total number of molecules can be equal or not
Answer:
No, the number of total molecules on the left side of a balanced equation is not equal to the number of total molecules on the right side of the equation. A molecule is the smallest number of atoms bonded together for a chemical reaction. The total number of atoms must be the same, but not molecules. The reactants and products will bond together in different ways leading to different numbers of reactants and products
Step-by-step explanation:
this is for pennfoster
I don't understand this at all
I need help with this question
Population y grows according to the equation dy/dt=ky, where k is a constant and t is measured in years. if the population doubles every 10 years, then the value of k is
Given that the population doubles every ten years, k may be found using the population growth equation dy/dt=ky by computing ln(2)/10, or roughly 0.0693 annually.
We begin by thinking about the solution to the differential equation dy/dt = ky, where the population doubles every ten years, in order to determine the value of k. This differential equation can be solved generally as y(t) = y(0)e^kt, where y(0) is the beginning population.
Since there are ten years between population doubling, we can write: y(10) = 2y(0) = y(0)e^10k.
The result of dividing both sides by y(0) is 2 = e^10k.
We get: ln(2) = 10k by taking the natural logarithm on both sides.
After calculating k, we have k = ln(2)/10 ≈ 0.0693.
Thus, k's value is around 0.0693 per year.