Find the area of the triangle ABC with the given parts. Round to the nearest tenth when necessary

The period of the function is 1/60 minutes

As a result, blood pressure changes between each heartbeat as a consequence of these oscilllations happening every second.

Max value = 110 mmHg

the minimum = 50 mmHg.

This suggests that systolic pressure remains at 110 mmHg and diastolic pressure is still maintained at 50 mmHg.

blood pressure peaks occur at times t = (1/2 + 2 * k) / 120 seconds.

(a) The given function is:

p(t) = 80 + 30 * sin(120 * pi * t)

This is equivalent to sinusoidal function in the form of:

p(t) = A + B * sin(C * t)

Where:

A is the baseline value,

B is the amplitude, and

C determines the frequency of the function.

Information given in the problem

A = 80, B = 30, and C = 120 * pi.

The period of a sinusoidal function is given by:

Period = 2 * pi / C

Period = 2 * pi / (120 * pi) = 1/60 minutes

The period of the function is 1/60 minutes, which means that the blood pressure oscillates every 1/60 minutes or 1 second. As a result, blood pressure changes between each heartbeat as a consequence of these oscilllations happening every second.

(b) maximum and minimum values of a sinusoidal function

Max value = A + B

Min value = A - B

Substituting the values of A and B:

Max value = 80 + 30 = 110 mmHg

Min value = 80 - 30 = 50 mmHg

Max value = A + B giving values of 110 mmHg

the minimum is A - B delivering 50 mmHg.

This suggests that systolic pressure remains at 110 mmHg and diastolic pressure is still maintained at 50 mmHg.

(c)  the time at which the blood pressure is at its maximum, we solve for t when the sinusoidal function is at its peak.

sin(120 * pi * t) = 1

Taking the inverse sine of both sides:

120 * pi * t = pi/2 + 2 * pi * k (where k is an integer)

Solving for t:

t = (1/2 + 2 x k)/120 (for k = 0, 1, 2, ......)

implying that blood pressure peaks occur at times t = (1/2 + 2 * k) / 120 seconds.

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The expressions for [tex]x^6[/tex] can be varied, one such basic expression is [tex](x^n)^m[/tex], where n and m are positive integers equal to 2 and 3.

A grouping of numbers, variables, and operations like addition, subtraction, multiplication, division, exponentiation and more is known as an expression. A single number or variable can be an expression it can also include a combination of terms or functions.

By simplifying the [tex]x^6[/tex] we can get the expression-

[tex]x^6=(x^2)^3 \\\\x^6=(x^3)^2\\\\x^6=x^2 * x^4 \\\\x^6=(x^2 + 1 - 1)^6\\\\x^6=(x^3 - x^2 + x - 1 + 1)^2[/tex]

all these expressions represent [tex]x^6[/tex].

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Domain where x is not equal to 3 and a range where y is not equal to 2 for function  (x+5)/(x-3)

Domain = Set of all input values of a function.

range = set of all output values of a function.

Given: Domain:  x≠3 ; range = y≠2

We do not include a value for domain if it makes the expression indeterminant .

Since all the functions in options are fractions, here the denominator does not equal to 0.

But in option C and D, the denominator can be zero if x=3.

So , domain for then it x ∈ R-3

for option C if 2=2(x+5)/(x-3)

x-3=x+5

-3=5 which is not possible

Where as in option D, 2=(x+5)/(x-3)

2x-6=x+5

x=11

Hence, domain where x is not equal to 3 and a range where y is not equal to 2 for function  (x+5)/(x-3)

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Which function has a domain where x does not =3 and a range where y does not =2? A. f(x)=(x-5)/(x+3) B. f(x)=2(x+5)/(x+3) C. 2(x+5)/(x-3) D. (x+5)/(x-3)

The coordinates of the focus of the parabola y = -(1/8)x² - 2x - 4 is (-8, 2).

y = -(1/8)x² - 2x - 4

Given by the equation y = -(1/8)x² - 2x - 4,

To find the focus of the parabola we first need to put the equation in standard form:

y = a(x - h)² + k

Where (h, k) is the vertex of the parabola, and a is a constant that determines the shape and size of the parabola.

Completing the square on the x terms, we get:

y = -(1/8)(x² + 16x + 64) - 4 + 8

= -(1/8)(x + 8)² + 4

Comparing this to the standard form, we see that the vertex is (-8, 4) and a = -1/8.

Since the coefficient of x² is negative, the parabola opens downwards.

The focus of the parabola is located at a distance of 1/(4a) units from the vertex, on the axis of symmetry, which is a vertical line passing through the vertex.

Substituting the value of a, we get:

= 1/(4a)

= 1/(4(-1/8))

= 2

Therefore, the focus of the parabola is located 2 units below the vertex, on the line x = -8.

So the focus has coordinates (-8, 2).

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The volume of the image is given as follows:

10626 ft³.

The volume of a prism is calculated as the multiplication of the base area of the prism by the height of the prism.

The bottom of the prism in this problem square base of side length 23 ft, along with height of 18 ft, hence:

Bottom volume = 18 x 18 x 23

Bottom volume = 7452 ft³.

The top of the prism also has square base of side length 23 ft, along with triangular height of 12 ft, hence:

Top volume = 0.5 x 12 x 23 x 23 (multiplies by 0.5 as the top is a triangle).

Top volume = 3174 ft³.

Hence the total volume of the prism is given as follows:

7452 + 3174 = 10626 ft³.

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Answer:

Step-by-step explanation:The function y=log(x) is translated 1 unit right and 2 units down.

To calculate the three-sigma control limits, we first need to find the mean and standard deviation of the sample.

Here, we have,

The mean is:

μ = (4 + 12 + 16 + 8 + 9 + 6 + 5 + 12 + 15 + 7 + 6 + 4 + 2 + 11) / 14 = 8.071

The standard deviation is calculated using the standard deviation formula and is arrived at:

σ = 4.319

The three-sigma control limits are:

Upper control limit = μ + 3σ = 8.071 + (3 × 4.319) = 20.027

Lower control limit = μ - 3σ = 8.071 - (3 × 4.319) = -3.886

b. We can check if the process is in control by looking at whether any of the data points fall outside of the control limits.

From the given data, we can see that the maximum number of complaints is 16, which is well within the upper control limit of 20.027. The minimum number of complaints is 2, which is also well within the lower control limit of -3.886.

Therefore, based on the given data, we can conclude that the process is in control.

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Answer:

I think its 221

Step-by-step explanation:

Answer:

The common ratio is 5.

30 ÷ 5 = 6.

Let n = 1 be the first term of this sequence.

[tex]a(n) = 6( {5}^{n} )[/tex]

Answer: 3

Step-by-step explanation:

The scale factor is the proportional ratio from one figure to another.

Notice the corresponding lengths are CE≅JS, TC≅TJ, and TE≅TS.

Comparing CE and JS, the length for CE is 4, and the length for JS is 12.

[tex]4*3=12[/tex]

This means the scale factor is 3. To be sure, let's try the other side.

TC and TE have the same lengths. They are both 5 in length. TS is 15 in length.

[tex]5*3=15[/tex]

This confirms that the scale factor is 3.

The graph that correctly shows the boundary line for y ≤ 2x − 4, is B. Graph B.

The inequality y ≤ 2x − 4, shows that the y - intercept is - 4. What this means is that the boundary line must pass through ( 0, - 4 ). As Graphs C and D do not cross ( 0, - 4 ), neither of them can be the boundary line.

Also, when representing an inequality on a graph, the signs of ≤ and ≥ are presented with a solid line. > and < are the ones presented with a broken line. This means that Graph A cannot the boundary line for the inequality.

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Answer: Linear

Step-by-step explanation:

I can tell from the table because there is a difference of 4 between each y value.

I dont know what the other blanks will be because i cant see the options. But it shouldnt be too hard to fill out

Answer:

I am not sure if you wanted the answer in mm or m.  I gave the answer is m

v =  0.58875[tex]m^{3}[/tex]

Step-by-step explanation:

v = 1/3[tex]\pi r^{2} h[/tex]

v = 1/3 (3.14)([tex].075^{2}[/tex])(100)    The diameter 150mm is .0150m  Take have of that to find the radius

v =  0.58875[tex]m^{3}[/tex]

Answer:

The result for the difference in lung cancer worry between males and females was not statistically significant in Hahn's (2017) study (t(45)=0.69, p > 0.05). The results for the differences in worry between smokers and nonsmokers, and those who did and did not graduate high school, were statistically significant with p values less than 0.05.

Step-by-step explanation:

The determinant of each of the given matrices are:

1) -4269

2) -1988

3) -68.92

The determinant of a matrix is a number that is specially defined only for square matrices. Determinants are mathematical objects that are very useful in the analysis and solution of systems of linear equations.

Thus:

1) [tex]\left[\begin{array}{ccc}7&&31\\142&&19\\\end{array}\right][/tex]

The determinant of this 2 x 2 matrix is:

(7 * 19) - (142 * 31) = -4269

2) [tex]\left[\begin{array}{ccc}7&11&21\\55&8&2\\-16&1&-9\end{array}\right][/tex]

The determinant of this 3 x 3 matrix is:

7((8*-9) - (1*2)) + 11((55*-9) - (2*-6)) + 21((55*1) - (8*-16)) = -1988

3) [tex]\left[\begin{array}{ccc}5.07&6.02&2.01\\-30.7&2.5&3.5\\3.55&-1.1&2.35\end{array}\right][/tex]

The determinant of this 3 x 3 matrix is:

5.07[(2.5 * 2.35) - (-1.1 *  3.5)] + 6.02[(-30.7 * 2.35) - (3.5 * 3.55)] + 2.01[(-30.7 * -1.1) - (2.5 * 3.55)] = -68.92

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Answer: Given the following equations, determine the x value(s) that result in an equal output for both functions. f(x) = 3* g(x)=4x+1. C0 and 2.

Step-by-step explanation:

Given the following equations, determine the x value(s) that result in an equal output for both functions. f(x) = 3* g(x)=4x+1. C0 and 2.

Answer:16.484

Step-by-step explanation:

find the area of a regular polygon with n sides and radius r, we can use the formula:

Area = (n * r^2 * sin(2*pi/n)) / 2

where pi is the mathematical constant pi (approximately equal to 3.14159).

Plugging in n = 14 and r = 1, we get:

Area = (14 * 1^2 * sin(2*pi/14)) / 2

= (14 * sin(pi/7)) / 2

≈ 16.484

Therefore, the area of the polygon with 14 sides and a radius of 1 unit is approximately 16.484 square units.

Answer:

C is correct because parallel lines cut by a transversal form congruent alternate interior angles.

Answer:

One.

Step-by-step explanation:

This polynomial has only one variable, which is x*. We can treat z and 7 as constants. Therefore, the polynomial is linear and can be written as:

5x* + c

where c = 7 - 3z.

A linear polynomial has either one root (when the coefficient is non-zero) or zero roots (when the coefficient is zero). Therefore, the given polynomial has exactly one root, which is:

x* = -c/5 = -(7 - 3z)/5 = (3z - 7)/5

Answer:

If 2:3 = x:12, then:

2/3 = x/12

Multiplying both sides by 12, we get:

x = 2/3 x 12 = 8

Therefore, the value of x is 8.

The equation of the dilated line is 9x + 15y = 10.

In Geometry, a scale factor can be defined as the ratio of two corresponding side lengths or diameter in two similar geometric objects, which can be used to either vertically or horizontally enlarge (increase) or reduce (compress) a function representing their size.

Generally speaking, the transformation rule for the dilation of a geometric object or equation based on a specific scale factor of 3 is given by this mathematical expression:

(x, y)      →    (SFx, SFy)

Where:

Therefore, the transformation rule for this dilation is given by;

(x, y)      →    (3x, 3y)

3(3x + 5y) = 10

9x + 15y = 10.

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(1) F(s) = 5+1 / (s^2 + 28s + 2)

To perform partial fraction expansion, we first need to factor the denominator:

s^2 + 28s + 2 = (s + 14 - sqrt(194))(s + 14 + sqrt(194))

We can then write:

F(s) = A / (s + 14 - sqrt(194)) + B / (s + 14 + sqrt(194))

where A and B are constants to be determined.

Multiplying both sides by the denominator and simplifying, we get:

5+1 = A(s + 14 + sqrt(194)) + B(s + 14 - sqrt(194))

Setting s = -14 - sqrt(194), we get:

5+1 = B(2sqrt(194))

Solving for B, we get:

B = (5+1) / (2sqrt(194))

Setting s = -14 + sqrt(194), we get:

5+1 = A(2sqrt(194))

Solving for A, we get:

A = (5+1) / (2sqrt(194))

Thus, the partial fraction expansion of F(s) is:

F(s) = [(5+1) / (2sqrt(194))] / (s + 14 + sqrt(194)) + [(5+1) / (2sqrt(194))] / (s + 14 - sqrt(194))

To find the inverse Laplace transform, we can use the table of Laplace transforms or MATLAB. Using MATLAB, we get:

ilaplace(F(s)) = (5+1) / (2sqrt(194)) * (exp(-14t) / sqrt(194)) * (cosh(sqrt(194)t) + sinh(sqrt(194)t))

(2) F(s) = s^2 + 3s + 1 / (s + 2)(s^2 + 8s + 1)

To perform partial fraction expansion, we first need to factor the denominator:

s^2 + 8s + 1 = (s + 4 - sqrt(15))(s + 4 + sqrt(15))

We can then write:

F(s) = A / (s + 2) + B / (s + 4 - sqrt(15)) + C / (s + 4 + sqrt(15))

where A, B, and C are constants to be determined.

Multiplying both sides by the denominator and simplifying, we get:

s^2 + 3s + 1 = A(s + 4 - sqrt(15))(s + 4 + sqrt(15)) + B(s + 2)(s + 4 + sqrt(15)) + C(s + 2)(s + 4 - sqrt(15))

Setting s = -4 + sqrt(15), we get:

-4 + sqrt(15) = A(-4 + sqrt(15) + 4 + sqrt(15))

Solving for A, we get:

A = (-4 + sqrt(15)) / (2sqrt(15))

Setting s = -4 - sqrt(15), we get:

-4 - sqrt(15) = A(-4 - sqrt(15) + 4 + sqrt(15))

Solving for A, we get:

A = (-4 - sqrt(15)) / (-2sqrt(15))

Setting s = -2, we get:

-1 = B(-2)(-2 + 4 + sqrt(15))

Solving for B, we get:

B = (-1) / (2sqrt(15) + 4)

Setting s =

The solution is,  the domain is: x ∈ (-∞, ∞).

Here, we have,

When we have two functions, f(x) and g(x), the composite function:

(f°g)(x)

is just the first function evaluated in the second one, or:

f( g(x))

And the domain of a function is the set of inputs that we can use as the variable x, we usually start by thinking that the domain is the set of all real numbers, unless there is a given value of x that causes problems, like a zero in the denominator, for example:

f(x) = 1/(x + 1)

where for x = -1 we have a zero in the denominator, then the domain is the set of all real numbers except x = -1.

Now, we have:

f(x) = x^2

g(x) = x + 9

then:

(f ∘ g)(x) = (x + 9)^2

And there is no value of x that causes problems here, so the domain is the set of all real numbers, that, in interval notation, is written as:

x ∈ (-∞, ∞)

(g ∘ f)(x)

this is g(f(x)) = (x^2) + 9 = x^2 + 9

And again, here we do not have any problem with a given value of x, so the domain is again the set of all real numbers:

x ∈ (-∞, ∞)

(f ∘ f)(x) = f(f(x)) = (f(x))^2 = (x^2)^2 = x^4

And for the domain, again, there is no value of x that causes a given problem, then the domain is the same as in the previous cases:

x ∈ (-∞, ∞)

(g ∘ g)(x) = g( g(x) ) = (g(x) + 9) = (x + 9) +9 = x + 18

And again, there are no values of x that cause a problem here,

so the domain is:

x ∈ (-∞, ∞)

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complete question:

Consider the following functions. f(x) = x2, g(x) = x + 9 Find (f ∘ g)(x). Find the domain of (f ∘ g)(x). (Enter your answer using interval notation.) Find (g ∘ f)(x). Find the domain of (g ∘ f)(x). (Enter your answer using interval notation.) Find (f ∘ f)(x). Find the domain of (f ∘ f)(x). (Enter your answer using interval notation.) Find (g ∘ g)(x). Find the domain of (g ∘ g)(x). (Enter your answer using interval notat

6,000 mL is equal to 6 litres

A unit of measurement is a definite magnitude of a quantity, defined and adopted by convention or by law, that is used as a standard for measurement of the same kind of quantity.

We know that 6,000 mL  is equal to 6L

One litre is equal to 1000 ml

1 l = 1000 ml

6 l=6000 ml

Hence,  6,000 mL is equal to 6 litres

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Answer:

It is a function

Step-by-step explanation:

It's a function because it does not over lapp

Transformation is a function that takes points on the plane and maps them to other points on the plane.

Transformations can be applied one after the other in a sequence where you use the image of the first transformation as the pre image for the next transformation.

Find the image for each sequence of transformations.

 Using geometry software, draw a triangle and label the

vertices A, B, and C. Then draw a point outside the

triangle and label it P.

Rotate △ABC 30° around point P and label the image as

△A′B′C ′. Then rotate △A′B′C ′ 45° around point P and

label the image as △A″B″C ″. Sketch your result.

 Make a conjecture regarding a single rotation that will map △ABC to △A″B″C″.

Check your conjecture, and describe what you did.

 Using geometry software, draw a triangle and label the

vertices D, E, and F.

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The essay, "I, Pencil," illustrates that production of an ordinary wood pencil used for writing d. involves the cooperative efforts of millions of people, whose actions are directed through markets.

I, Pencil, an essay by Leonard Read portrays that the manufacture of a common wood pencil utilized for writing necessitates the collective efforts of millions of human beings whose activities are strictly regulated through markets.

The paper conveys that though an ordinary pencil looks like such a straightforward item, it is actually created from a complex system of people, procedures, and materials from far and wide. From timber harvesters who procure the wood, to the miners who take graphite out from the ground, to the employees in manufacturing plants who produce the finished article, the lead holder necessitates the orderly collaboration of multiple individuals and enterprises.

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