The value of x for each item is given as follows:
28. x = 5.
29. x = 3.44.
How to obtain the value of x in each item?For item 28, we apply the crossing chord theorem, which states that the products of the parts of the chords are equal, hence the value of x is obtained as follows:
16x = 10 x 8
16x = 80
x = 5.
For item 29, we apply the two secant theorem, hence the value of x is obtained as follows:
10(x + 10) = 12(12 + 25)
10x + 100 = 444
10x = 344
x = 3.44.
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Given the functions a(x) = 3x - 12 and b(x) = x-9, solve a[b(x)].
Oa[b(x)] = 3x²-21
O a[b(x)] = 3x² - 39
Oa[b(x)] = 3x - 21
Oa[b(x)] = 3x - 39
Answer:
Step-by-step explanation:
hello :
a(x) = 3x - 12 and b(x) = x-9, so
a[b(x)]=a(x-9) =3(x-9)-12
a[b(x)]=3x-9-12
a[b(x)]=3x+21
in triangle ABC, AC=13, BC=84, and AB=85. find the measure of angle C.
in triangle ABC, AC=13, BC=84, and AB=85. the measure of angle is C.5.1 degrees,
How can the angle be found?Using the The Law of Cosines in any triangle which can be expressed as;
[tex]c^2 = a^2 + b^2 - 2ab * cos(C)[/tex]
Then if we expressed the given sides in the cosine formula we have
[tex]13^2 = 85^2 + 84^2 - 2 * 85 * 84 * cos(C)[/tex]
Then we have [tex]169 = 7225 + 7056 - 14280 * cos(C)[/tex]
[tex]14280 * cosC = (7225 + 7056 - 169)[/tex]
[tex]cosC = \frac{14212}{ 14280}[/tex]
cos 0.9947
[tex]C = cos^-1(0.9947)[/tex]
C = 5.1 degrees
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suppose that fanf g are continuous functions,
12_(^28 f(x)dx=19, and 12 (^28 g(x)dx =35.
Find
12_ f^28 [kg(x)-2f(x)] dx, where the coefficient
k=7
those are intergral symbols with numbers on
top and bottom. please show work. thanks
The required result based on given continuous functions is: 207. See the explanation for same below.
A continuous function in mathematics is one in which a continuous variation (that is, a change without a jump) of the argument causes a continuous variation of the function's value.
Since G and F are continuous functions,
[tex]\int_{12}^{28}) \, f(x) dx[/tex] = 19
[tex]\int_{12}^{28}) \, f(x) dx[/tex] = 35
Therefore,
[tex]\int_{12}^{28}) \, [K g(x) - 2 f(x)] dx[/tex]
= K [tex]\int_{12}^{28}) \, g(x) dx -2 \int_{12}^{28}) \, f(x) dx[/tex]
So, given that K = 7, we have
7 x 35 - 2 x 19
= 245 - 38
= 207
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Complete the equations to solve 1{,}860 \div61,860÷61, comma, 860, divided by, 6.
\phantom{=}\greenD{1{,}860}\div{\blueD6}=1,860÷6empty space, start color #1fab54, 1, comma, 860, end color #1fab54, divided by, start color #11accd, 6, end color #11accd
=(\greenD{1{,}800}\div\,=(1,800÷equals, left parenthesis, start color #1fab54, 1, comma, 800, end color #1fab54, divided by
) \, + \,(\greenD{60}\div\,)+(60÷right parenthesis, plus, left parenthesis, start color #1fab54, 60, end color #1fab54, divided by
))right parenthesis
= 300 +=300+equals, 300, plus
==equals
The division of the figure based on the information is 14.09.
How to illustrate the information?It should be noted that the question is simply to divide 860 by 61.
The division based on the information will be illustrated thus:
= 860 ÷ 61
= 14.09
In conclusion, the correct option is 14.09.
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6) Ron bought a car 8 years ago for $15,694. If the rate of depreciation is 4.75% per year, what is the value of
Ron's car now? Round answer to nearest cent.
O a.) $10,642.94
Ob.) $10,332.94
Oc.) $10,632.94
Od.) $10,662.94
Solve [tex]2 cosx=4cosx sin^2x[/tex]
There are four solutions for the trigonometric equation 2 · cos x = 4 · cos x · sin² x are x₁ = π/4 ± 2π · i, x₂ = 3π/4 ± 2π · i, x₃ = 5π/4 ± 2π · i and x₄ = 7π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex].
How to solve a trigonometric equation
In this problem we must simplify the trigonometric equation by both algebraic and trigonometric means and clear the variable x:
2 · cos x = 4 · cos x · sin² x
2 · sin² x = 1
sin² x = 1/2
sin x = ± √2 /2
There are several solutions:
x₁ = π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]
x₂ = 3π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]
x₃ = 5π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]
x₄ = 7π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]
There are four solutions for the trigonometric equation 2 · cos x = 4 · cos x · sin² x are x₁ = π/4 ± 2π · i, x₂ = 3π/4 ± 2π · i, x₃ = 5π/4 ± 2π · i and x₄ = 7π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex].
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a 300-acre farm produced 20,000 bushels of corn last year. what is the minimum rate of production, in bushels per acre, that is needed this year, so the farm's two-year production total will be at least 49,100 bushels of corn?
The minimum production rate of bushels per acre of corn needed this year is 97 for two year's production to reach the goal of 49,100.
How to calculate the minimum production rate?To calculate the minimum production rate for this year, we must subtract the production rate of the previous year, with what is expected to be obtained this year.
49,100 - 20,000 = 29,100Then we must divide the minimum that must be obtained by the total area, by the 300 acres that we have, to know how much corresponds to each acre.
29,100 ÷ 300 = 97According to the above, each acre must produce 97 bushels of corn to reach the goal of 49,100 for both years.
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Find the coordinates of the vertices of the triangle if it’s rotated 270° counterclockwise.
If these coordinates are rotated 270° counterclockwise, the resulting coordinates will be;
X' = (3, -1)
Y' = (2, -2)
Z' = (-1, -3)
Rotation of coordinatesIf the coordinates (x, y) is rotated 270° counterclockwise, the resulting coordinate point will be;
(x, y) - > (y, -x)
From the given triangles, the coordinate points are expressed as
X = (1, 3)
Y = (2, 2)
Z = (3, -1)
If these coordinates are rotated 270° counterclockwise, the resulting coordinates will be;
X' = (3, -1)
Y' = (2, -2)
Z' = (-1, -3)
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Alan is building a garden shaped like a rectangle with a semicircle attached to one short side. If he has 70 feet of fencing to go around it, what dimensions will give him the maximum area in the garden? Round the answers to the nearest tenth.
The dimension that would give the maximum area is 20.8569
How to solve for the maximum area
Let the shorter side be = x
Perimeter of the semi-circle is πx
Twice the Length of the longer side
[tex][70-(\pi )x -x][/tex]
Length = [tex][70-(1+\pi )x]/2[/tex]
Total area =
area of rectangle + area of the semi-circle.
Total area =
[tex]x[[70-(1+\pi )x]/2] + [(\pi )(x/2)^2]/2[/tex]
When we square it we would have
[tex]70x +[(\pi /4)-(1+\pi)]x^2[/tex]
This gives
[tex]70x - [3.3562]x^2[/tex]
From here we divide by 2
[tex]35x - 1.6781x^2[/tex]
The maximum side would be at
[tex]x = 35/2*1.6781[/tex]
This gives us 20.8569
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Which expression is equivalent to
xay
8
700
8√√√x
y
8√√y
Mark this and return
128x56
√ 2x75
? Assume x > 0 and y> 0.
Save and Exit
The equivalent expression of [tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex] is [tex]\frac{8\sqrt y}{x}[/tex]
How to determine the equivalent expression?The expression is given as:
[tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex]
Divide 128 by 2
[tex]\sqrt{\frac{64x^5y^6}{x^7y^5}}[/tex]
Apply the law of indices to the variables
[tex]\sqrt{\frac{64y^{6-5}}{x^{7-5}}}[/tex]
Evaluate the differences
[tex]\sqrt{\frac{64y}{x^2}}[/tex]
Take the square root of 64
[tex]8\sqrt{\frac{y}{x^2}}[/tex]
Take the square root of x^2
[tex]\frac{8\sqrt y}{x}[/tex]
Hence, the equivalent expression of [tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex] is [tex]\frac{8\sqrt y}{x}[/tex]
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What is the volume of the cylinder?
The volume of a cylinder with a diameter of 16 feet and height of 10 feet is 2010.62 ft³
What is an equation?An equation is an expression that shows the relationship between two or more numbers and variables.
The volume of a cylinder with radius (r) and height (h) is:
Volume = πr²h
Given that h = 10 ft, r = 16/2 = 8 ft
The volume = π * 8² * 10 = 2010.62 ft³
The volume of a cylinder with a diameter of 16 feet and height of 10 feet is 2010.62 ft³
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Please look at the attachment and answer my question :(
Answer:
AB / BC = 2 / 3
A equals 9 and C = 13
AC = 13 - 9 = 4
AB + BC = 13
A) AB + BC = 4
B) AB / BC = 2/3 Therefore BC = AB / (2/3)
B) BC = 1.5 AB
A) BC = 4 - A/B
Multiplying equation A) by -1
A) -BC = -4 + AB then we add equation B)
B) BC = 1.5 AB then adding both equations
0 = 2.5 AB -4
2.5 AB = 4
AB = 1.6
Since A = 9 then the number at B is 9 + 1.6
equals 10.6
Step-by-step explanation:
Answer:
The number at B is 10.6
Step-by-step explanation:
Let the number at B be x.
We know that distance between two points on the number line is the absolute value of the difference of numbers at those points.
Then distances representing the lengths of segments AB and BC are:
AB = x - 9BC = 13 - xWe are given the ratio of segments:
AB/BC = 2/3Substitute and solve for x:
(x - 9)/(13 - x) = 2/33(x - 9) = 2(13 - x)3x - 27 = 26 - 2x3x + 2x = 26 + 275x = 53x = 53/5x = 10.6Write these decimals in order from smallest to largest: 0.507, 0.75, 0.5, 0.078
A random variable X has a gamma density function with parameters α= 8 and β = 2.
Without making any assumptions, derive the moment generating function of X and use to
determine the mean and variance of X.
I know you said "without making any assumptions," but this one is pretty important. Assuming you mean [tex]\alpha,\beta[/tex] are shape/rate parameters (as opposed to shape/scale), the PDF of [tex]X[/tex] is
[tex]f_X(x) = \dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} = \dfrac{2^8}{\Gamma(8)} x^7 e^{-2x}[/tex]
if [tex]x>0[/tex], and 0 otherwise.
The MGF of [tex]X[/tex] is given by
[tex]\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx[/tex]
Note that the integral converges only when [tex]t<2[/tex].
Define
[tex]I_n = \displaystyle \int_0^\infty x^n e^{(t-2)x} \, dx[/tex]
Integrate by parts, with
[tex]u = x^n \implies du = nx^{n-1} \, dx[/tex]
[tex]dv = e^{(t-2)x} \, dx \implies v = \dfrac1{t-2} e^{(t-2)x}[/tex]
so that
[tex]\displaystyle I_n = uv\bigg|_{x=0}^{x\to\infty} - \int_0^\infty v\,du = -\frac n{t-2} \int_0^\infty x^{n-1} e^{(t-2)x} \, dx = -\frac n{t-2} I_{n-1}[/tex]
Note that
[tex]I_0 = \displaystyle \int_0^\infty e^{(t-2)}x \, dx = \frac1{t-2} e^{(t-2)x} \bigg|_{x=0}^{x\to\infty} = -\frac1{t-2}[/tex]
By substitution, we have
[tex]I_n = -\dfrac n{t-2} I_{n-1} = (-1)^2 \dfrac{n(n-1)}{(t-2)^2} I_{n-2} = (-1)^3 \dfrac{n(n-1)(n-2)}{(t-2)^3} I_{n-3}[/tex]
and so on, down to
[tex]I_n = (-1)^n \dfrac{n!}{(t-2)^n} I_0 = (-1)^{n+1} \dfrac{n!}{(t-2)^{n+1}}[/tex]
The integral of interest then evaluates to
[tex]\displaystyle I_7 = \int_0^\infty x^7 e^{(t-2) x} \, dx = (-1)^8 \frac{7!}{(t-2)^8} = \dfrac{\Gamma(8)}{(t-2)^8}[/tex]
so the MGF is
[tex]\displaystyle M_X(t) = \frac{2^8}{\Gamma(8)} I_7 = \dfrac{2^8}{(t-2)^8} = \left(\dfrac2{t-2}\right)^8 = \boxed{\dfrac1{\left(1-\frac t2\right)^8}}[/tex]
The first moment/expectation is given by the first derivative of [tex]M_X(t)[/tex] at [tex]t=0[/tex].
[tex]\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}[/tex]
Variance is defined by
[tex]\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2[/tex]
The second moment is given by the second derivative of the MGF at [tex]t=0[/tex].
[tex]\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18[/tex]
Then the variance is
[tex]\Bbb V[X] = 18 - 4^2 = \boxed{2}[/tex]
Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is
[tex]M_X(t) = \displaystyle \sum_{k=0}^\infty \dfrac{M_X^{(k)}(0)}{k!} t^k[/tex]
where [tex]M_X^{(k)}(0)[/tex] is the [tex]k[/tex]-derivative of the MGF evaluated at [tex]t=0[/tex]. This is also the [tex]k[/tex]-th moment of [tex]X[/tex].
Recall that for [tex]|t|<1[/tex],
[tex]\displaystyle \frac1{1-t} = \sum_{k=0}^\infty t^k[/tex]
By differentiating both sides 7 times, we get
[tex]\displaystyle \frac{7!}{(1-t)^8} = \sum_{k=0}^\infty (k+1)(k+2)\cdots(k+7) t^k \implies \displaystyle \frac1{\left(1-\frac t2\right)^8} = \sum_{k=0}^\infty \frac{(k+7)!}{k!\,7!\,2^k} t^k[/tex]
Then the [tex]k[/tex]-th moment of [tex]X[/tex] is
[tex]M_X^{(k)}(0) = \dfrac{(k+7)!}{7!\,2^k}[/tex]
and we obtain the same results as before,
[tex]\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4[/tex]
[tex]\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18[/tex]
and the same variance follows.
claire and martin are trying to determine the length of a slug. Claire has a mesuring stick whose markings are 1 cm apart, and Martin has a measuring stick whose markings are 0.1cm apart.
Claire's measuring stick would make a more precise measurement
How to determine the precise measurement?The given parameters are:
Claire = 1 cm
Martin = 0.1 cm
The 0.1 cm marking would make a more accurate measurement.
This is so because the length of a slug would most likely be in decimal centimeters
However, this doesn't translate to a precise measurement.
Hence, Claire's measuring stick would make a more precise measurement
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will give brainly .Which statements are true regarding undefinable terms in geometry? Check all that apply.
A point has no length or width.
A point indicates a location in a coordinate plane.
A plane has one dimension, length.
A line has a definite beginning and end.
A plane consists of an infinite set of lines.
A line consists of an infinite set of points.
The true statements about terms in geometry that are undefinable are;
A point has no length or widthA point indicates a location on the coordinate planeA line consists of an infinite set of points.A plane consists of an infinite set of points.How can the true statements be found?Three undefinable terms in geometry are;
1) Point
2) Line
3) Plane
The above terms do not have a formal definition but they can be described based on their properties.
The other geometric terms are defined based on the above undefinable terms.
A point can be described as a location in space that is dimensionless and can be specified on the coordinate plane as an ordered pairs (x, y).
True statements about a point are therefore;
A point has no length or widthA point indicates a location on the coordinate planeA line is infinitely long, that has no beginning or end. It has one dimension, with no thickness or height.
Given that a point is dimensionless, a line can be considered a set of points.
A true statement is therefore;
A line consists of an infinite set of points.A plane is a two dimensional geometric figure that have infinite length and width.
An infinite set of lines that forms a two dimensional figure can be used to describe a plane.
The true statement with regards to a plane is therefore;
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What is the solution to the following system of equations?
x-3y=6
2x + 2y = 4
(-1,3)
(3,-1)
(1, -3)
(-3,1)
Answer:
(b) (3,-1)
Step-by-step explanation:
One can determine the correct answer by checking to see if it satisfies the equations.
CheckingThe first equation can be rewritten to ...
x = 6 +3y . . . . . . . add 3y to both sides
This makes it easier to check answer choices:
a) -1 = 6 +3(3) . . . . no
b) 3 = 6 +3(-1) . . . . yes . . . . . (3, -1) is the solution
c) 1 = 6 +3(-3) . . . . no
d) -3 = 6 +3(1) . . . . no
__
Additional comment
We can further convince ourselves this is the correct choice by seeing if it satisfies the second equation:
2x +2y = 4 . . . for (x, y) = (3, -1)
2(3) +2(-1) = 4 . . . . yes
Suppose f(x)=2^x. What is the graph of g(x)=1/3f(x)?
Please see the blue curve of the image attached below to know the graph of the function g(x) = (1/3) · 2ˣ.
How to graph a transformed function
Herein we have an original function f(x). The transformed function g(x) is the result of compressing f(x) by 1/3. Then, we find that g(x) = (1/3) · 2ˣ. Lastly, we graph both function on a Cartesian plane with the help of a graphing tool.
The result is attached below. Please notice that the original function f(x) is represented by the red curve, while the transformed function g(x) is represented by the blue curve.
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Suppose f(x) and g(x) are differentiable functions satisfying f(1) = f′(1) = 2. Let
h(x) = g(f(x)) − g(2x). Determine whether h′(1) > 1
Answer:
1777777777888876332222233
inverse function of f(x)=x-7/x+4
Final Answer: The inverse of f (x)=7x-4 is f^-1 (x)= (x+4)/7
i don’t understand!!!
In other words, 12 goes in the top box and 13 goes in the bottom. The fraction slash sign is not part of either box.
===========================================================
Explanation:
Cosine is the ratio of adjacent over hypotenuse.
cos(angle) = adjacent/hypotenuse
For the reference angle X, the adjacent leg is 36 units long. It's the leg closest or touching angle X.
The hypotenuse is always the longest side. It is always opposite the 90 degree angle. The hypotenuse in this case is 39 units.
Therefore,
cos(X) = 36/39 = (12*3)/(13*3) = 12/13
look at the pictures
The key feature that the function of f(x) and g(x) has in common is the domain.
What is the domain and range of a function?The domain of a function is the set of input or argument values for which the function is valid and well defined. The range is the set of the dependent variable for which a function is defined.
From the given information, we are to find the domain, range, x-intercept, and, y-intercept of the given equation:
[tex]\mathbf{f(x) = -4^x+5}[/tex]
[tex]\mathbf{g(x) = x^3 +x^2 -4x+5}[/tex]
For [tex]\mathbf{f(x) = -4^x+5}[/tex];
The domain of a function [tex]\mathbf{-4^x+5}[/tex] has no undefined points or constraints. Thus, the domain is -∞ < x < ∞.
The range f(x) < 5. The x-intercepts, when y is zero = [tex]\mathbf{(\dfrac{In(5)}{2In(2)},0)}[/tex]The y-intercepts; when x is zero = (0,4)For [tex]\mathbf{g(x) = x^3 +x^2 -4x+5}[/tex]
The domain of a function [tex]\mathbf{g(x) = x^3 +x^2 -4x+5}[/tex] has no undefined points or constraints. Thus, the domain is -∞ < x < ∞.
The range -∞ < f(x) < ∞. The x-intercepts, when y is zero = (-2.939, 0)The y-intercepts; when x is zero = (0,5)Learn more about the domain and range of a function here;
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Hi I would like to know how I can solve this problem.
The [tex]n[/tex]-th term is
[tex]U_n = \dfrac14 n^2 (n+1)^2[/tex]
so the 39th term is
[tex]U_{39} = \dfrac14 39^2 40^2 = \boxed{608,400}[/tex]
Observe that
[tex]2^3 + 4^3 + 6^3 = 2^3 + 2^3\times2^3 + 2^3\times3^3 = 8\left(1^3+2^3+3^3)[/tex]
which suggests that
[tex]V_n = 8U_n = \boxed{2n^2(n+1)^2}[/tex]
What point is 2/3 of the distance from point A(3, 1) to point B(3, 19)?
What point is 2/3 of the distance from point A(3, 1) to point B(3, 19)?
(3, 13)
(9/3, 13)
(2, 12)
(3, 19)
Answer: (3, 13)
Step-by-step explanation:
If we let the point be P, then AP:BP=2:1.
[tex]P=\left(\frac{(2)(3)+(1)(3)}{2+1}, \frac{(2)(19)+(1)(1)}{2+1} \right)=(3, 13)[/tex]
A solid right square prism is cut into 5 equal pieces parallel to its bases. The volume of each piece is V = 1/5s2h.Solve the formula for s. help asap
The solution of the formula for s is:
[tex]s = \sqrt{\frac{5V}{h}}[/tex]
What is the symbolic equation?The equation is:
[tex]V = \frac{1}{5}s^2h[/tex]
To solve for s, we isolate it, hence:
[tex]s^2 = \frac{5V}{h}[/tex]
[tex]s = \sqrt{\frac{5V}{h}}[/tex]
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To the nearest percent how much greater is 8 than 6
The percent greater of 8 than 6 to the nearest percent is 25%
PercentagePercent greater = difference / higher chance value × 100
= 2/8 × 100
= 0.25 × 100
Percent greater = 25%
Therefore, the percent greater of 8 than 6 to the nearest percent is 25%
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Suppose you have $1500 in your savings account at the end of a certain period of time. You invested $1000 at a 2.71% simple annual interest rate. How long, in years, was your money invested? State your result to the nearest hundredth of a year.
The time taken in years for money invested to accumulate to $1500 is 18.45 years
Simple interestPrincipal = $1000Interest rate = 2.71% = 0.0271t = ?Simple interest = Total savings - principal
= 1500 - 1000
= $500
Simple interest = principal × rate × time
500 = 1000 × 0.0271 × t
500 = 27.1 × t
500 = 27.1t
t = 18.4501845018450
Approximately,
t = 18.45 years
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Find the exact solutions of x2 − 3x − 5 = 0 using the quadratic formula. Show all work! 75 points please help!!!!!
Answer:
[tex]x=\dfrac{3+ \sqrt{29}}{2}, \quad \dfrac{3- \sqrt{29}}{2}[/tex]
Step-by-step explanation:
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Given quadratic equation:
[tex]x^2-3x-5=0[/tex]
Define the variables:
[tex]\implies a=1, \quad b=-3, \quad c=-5[/tex]
Substitute the defined variables into the quadratic formula and solve for x:
[tex]\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(1)(-5)}}{2(1)}[/tex]
[tex]\implies x=\dfrac{3 \pm \sqrt{9+20}}{2}[/tex]
[tex]\implies x=\dfrac{3 \pm \sqrt{29}}{2}[/tex]
Therefore, the exact solutions to the given quadratic equation are:
[tex]x=\dfrac{3+ \sqrt{29}}{2}, \quad \dfrac{3- \sqrt{29}}{2}[/tex]
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Answer:
[tex]\boxed {\frac{3+\sqrt{29}}{2}} \boxed{\frac{3-\sqrt{29}}{2}}[/tex]
Step-by-step explanation:
Quadratic Formula :
[tex]\boxed {\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}}[/tex]
We are given the equation x² - 3x - 5 = 0.
Here,
a = 1b = -3c = -5Solving :
3 ± √3² - 4(1)(-5) / 2(1)3 ± √29 / 2Hence, the solutions are :
[tex]\boxed {\frac{3+\sqrt{29}}{2}} \boxed{\frac{3-\sqrt{29}}{2}}[/tex]
paulette spent 45.50 on concert tickets before fees. there was a service fee of 6% added on to her purchase.
The original price of concert tickets before the addition of service fee is 42.93
PercentageAmount spent on tickets including service fee = 45.50Percentage if service fee = 6%Original cost before service fee = x45.50 = x + (6% of x)
45.50 = x + (0.06 × x)
45.50 = x + 0.06x
45.50 = 1.06x
x = 45.50/1.06
x = 42.9245283018867
Approximately,
x = 42.93
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The lengths of the sides of the right triangle above are a, 3, and c. What is a in terms of c?
The expression for a in terms of c is [tex]a^{2}= \sqrt{c^{2} -9}[/tex]. The correct option is the third option [tex]a^{2}= \sqrt{c^{2} -9}[/tex]
Pythagorean theoremFrom the question, we are to determine the expression for a in terms of c
In the given right triangle, we can write that
[tex]c^{2} = a^{2} +3^{2}[/tex] (Pythagorean theorem)
Thus,
[tex]c^{2} = a^{2} +9[/tex]
[tex]a^{2}= c^{2} -9[/tex]
[tex]a^{2}= \sqrt{c^{2} -9}[/tex]
Hence, the expression for a in terms of c is [tex]a^{2}= \sqrt{c^{2} -9}[/tex]. The correct option is the third option [tex]a^{2}= \sqrt{c^{2} -9}[/tex]
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