what does it mean if you have hypertension when you first sit down but it drops to normal in 3 minutes

Y-intercept - 4

Slope - 3

Slope intercept form - y=3x + 4

Hope it's correct and it helped:)

Answer:

The Median is 75

Step-by-step explanation:

median is the middle number in a set of given numbers

arranging in order will be

30,64,70,80,90,110

the middle numbers are 70 and 80

Median=80+70/2

=150/2

Median=75

The value of the given double integral ∫∫_S 3dS 3ds, where S is the surface parametrized by r(u, v) = < u², uv, ½v² >, 0 ≤u≤ 1,0 ≤v≤1, is 1.5.

To evaluate the double integral, we can use the surface area element formula in the parametric form: dS = ||∂r/∂u × ∂r/∂v|| dude. Here, ∂r/∂u and ∂r/∂v are the partial derivatives of r(u, v) with respect to u and v, respectively. Taking the cross product and magnitude, we obtain ||∂r/∂u × ∂r/∂v|| = u²v√(1 + v²).

Calculate the partial derivatives of the vector function r(u, v) with respect to u and v:

∂r/∂u = < 2u, v, 0. >

∂r/∂v = < 0, u, v >

Compute the cross product of the partial derivatives to obtain the surface normal vector:

N = ∂r/∂u × ∂r/∂v

= < 2u, v, 0. > × < 0, u, v >

= < -v², -2uv, 2u² >

Calculate the magnitude of the surface normal vector:

||N|| = √((-v²)² + (-2uv)² + (2u²)²)

= √(v⁴ + 4u²v² + 4u⁴)

Set up the integral over the given parameter domain:

∫∫_S 3dS = ∫∫_D ||N|| dA

Here, D represents the parameter domain, which is the square region in the uv-plane defined by 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1.

Convert the double integral from the uv-plane to the corresponding limits in u and v:

∫∫_S 3dS = ∫[0,1]∫[0,1] ||N|| dudv

Substitute the magnitude of the surface normal vector ||N|| into the integral:

∫∫_S 3dS = ∫[0,1]∫[0,1] √(v⁴ + 4u²v² + 4u⁴) dudv

Now, we have set up the integral in terms of u and v. To evaluate it numerically, you can either integrate it symbolically or use numerical methods.

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Answer: The correct answer would be B,D,E

Step-by-step explanation: I AM TAKING THE TEST

Answer: 3

Step-by-step explanation:

The set of possible values for "p" is ((4-√7)/9, (4+√7)/9).

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Answer:

2

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

Here is your answer with solutions!!

Answer:

maybe 3 seconds.

but I am not sure

The probability of selecting 8 female students from a sample of 8 students from a student body with 60% female students is 0.2187, or 21.87%.

The probability of selecting a sample of 8 students from a student body of 60% female students is calculated using binomial probability. The binomial probability formula is used to calculate the probability of a certain number of successes in a certain number of independent trials. In this case, the probability of selecting 8 students, with 60% being female students, can be calculated using the binomial probability formula.

The probability can be calculated using the following equation:
\(P(x=8) = (n!/((n-x)!x!)) * p^x * q^{(n-x)}\)

Where:

In this case, n = 8, x = 8, p = 0.6, and q = 0.4. Plugging these values into the equation gives us a probability of 0.2187. This means that there is a 21.87% chance of selecting 8 female students out of a sample of 8 students from a student body with 60% female students.

It is important to remember that binomial probability is only used when there are two possible outcomes in each trial (i.e. success or failure). Additionally, it is important to remember that the equation only applies when the trials are independent of each other.

In conclusion, the probability of selecting 8 female students from a sample of 8 students from a student body with 60% female students is 0.2187, or 21.87%.

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Answer:

its option B

Step-by-step explanation:

Answer:

the answer is d

Step-by-step explanation:

The parametric equation for the line passing through points a(3, 0, -2) and b(5, 1, -3) is x = 3 + t(2), y = t(1), z = -2 + t(-1).

To find the parametric equation for the line passing through points a(3, 0, -2) and b(5, 1, -3), we use the general form of a parametric equation, where x, y, and z are expressed in terms of a parameter t.

We can start by obtaining the directional vector of the line, which is the difference between the coordinates of point b and point a: (5 - 3, 1 - 0, -3 - (-2)) = (2, 1, -1).

Next, we express x, y, and z in terms of t using the directional vector. For x, we have x = 3 + t(2), where the coefficient 2 corresponds to the change in x for each unit change in t. Similarly, for y, we have y = t(1), and for z, we have z = -2 + t(-1), indicating the changes in y and z with respect to t.

Combining these expressions, we obtain the parametric equation for the line passing through points a and b as x = 3 + t(2), y = t(1), and z = -2 + t(-1).

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The time is needed between injections is 3.6 hours, i.e., the drug is to be injected again when the number of milligrams reaches 6 mg.

We have the exponential function of number of milligrams D (ht) of a certain drug that is in a patient's bloodstream h hours after the drug is injected is

\(D(h)=25 {e}^{ - 0.4 h}\)

We have to solve for h (the numbers of hours) that would have passed when the D(h) (the amount of medication in the patient's bloodstream) equals 6 mg in order to know when the patient needs to be injected again.

\(6 = 25 {e}^{ - 0.4h} \)

\( \frac{6}{25} = \frac{25}{25} {e}^{ - 0.4h} \)

\(0.24= {e}^{ - 0.4h} \)

Taking logarithm both sides of above equation , we get,

\( \ln(0.24) = \ln( {e}^{ - 0.4h)} \)

Using the properties of natural logarithm,

\( \ln(0.24) = - 0.4h\)

\( - 1.427116356 = - 0.4h\)

\(h = \frac{1.42711635}{0.4} = 3.56779089\)

=> h = 3. 6

So, after 3.6 hours, the patient needs to be injected again.

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Title: Quadruple Crossing Number of Knots and Links by C. Adams

In the article titled "Quadruple Crossing Number of Knots and Links" by C. Adams, published in the Mathematical Proceedings of the Cambridge Philosophical Society, volume 156(2), pages 241-253 in 2014, the author explores the concept of the quadruple crossing number of knots and links.

The quadruple crossing number of a knot or link is defined as the minimum number of times any diagram of the knot or link must cross itself in order to represent the knot or link. C. Adams delves into the study of this intriguing aspect of knot theory, aiming to determine the quadruple crossing number for various knots and links.

Through meticulous analysis and calculations, Adams examines different types of knots and links and investigates their crossing numbers. The author presents mathematical proofs, formula derivations, and knot diagrams to illustrate the concepts and results obtained.

By exploring the quadruple crossing number, Adams aims to deepen our understanding of the complexity and structure of knots and links, shedding light on their mathematical properties and connections to other branches of mathematics.

C. Adams' article on the quadruple crossing number of knots and links provides valuable insights into the field of knot theory. By determining the minimum number of crossings required to represent various knots and links, the author contributes to our understanding of the complexity and intricacy of these mathematical objects. The findings presented in this article serve as a foundation for further research and exploration in the field of knot theory, and they demonstrate the significance of the quadruple crossing number as a fundamental parameter in the study of knots and links.

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Answer:

1 plus 1 is 2

Step-by-step explanation:

If you can download this app and type this question, I am pretty sure you can figure it out

Answer:

1+1=2

1+1=2

1+1=2

1+1=2

1+1=2

1+1=2

1+1=2

Answer:

When x=-1, y=x^2+x-4 is -4. When x=1, y=x^2+x-4 is -2.

Step-by-step explanation:

First, insert the value of x into the equation.

y=-1^2+-1-4

Then, solve for y.

y=-1^2+-1-4

(Any negative squared is always a positive!)

y=1-1-4

y=0-4

y=-4

Repeat this to find the value of y when x=1.

y=1^2+1-4

(Remember, 1 to the power of anything is always 1.)

y=1+1-4

y=2-4

y=-2

The polynomial function of least degree possible using the given information.

Real roots: −1 (with multiplicity 2), 1 and (2,

f(2)) = (2, 7) is f(x) = x³ - 3x² + 3x - 1.

Since the polynomial function has real roots at -1 (with multiplicity 2) and 1, we know that the factors of the polynomial are (x + 1)² and (x - 1).

Let the polynomial be of the form f(x) = ax³ + bx² + cx + d. We know that f(2) = 7, so:

a(2³) + b(2²) + c(2) + d = 7

8a + 4b + 2c + d = 7

Now we need to use the fact that the roots are -1 (with multiplicity 2) and 1. Since the polynomial has factors of (x + 1)² and (x - 1), we can write the polynomial as:

f(x) = a(x + 1)²(x - 1)

Expanding this expression, we get:

f(x) = a(x² - 2x + 1)(x - 1)

f(x) = ax³ - 3ax² + 3ax - a

Now we can equate the coefficients of this expression with the coefficients of our assumed polynomial f(x) = ax³ + bx² + cx + d:

a = a

-3a = b

3a = c

-a = d

Therefore, our polynomial function is:

f(x) = x³ - 3x² + 3x - 1

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Answer:

Brandon is 7 years old

Step-by-step explanation:

(27 - 6) / 3 = 7

27 = Sam's age

6 = years greater compared to/than means subtraction

3 =  times greater compared to/than means divion.

27 - 6 is in parentheses because it must be done first.

Please give brainlyest

Have a nice day

;)

Answer: x=-5

Step-by-step explanation:

The variables should’ve been moved to the left side:

4x+6-2x=-4

Move the constant with it’s like term:

4x-2x=-4-6

Combine like terms:

2x=-10

Divide by both sides and you have x=-5

There are 792 poker hands consisting of all face cards.

To determine the number of poker hands consisting of all face cards, we need to consider the number of ways we can select 5 face cards from the 12 available face cards.

Since we are selecting a specific number of items from a larger set without considering the order, we can use combinations to calculate the number of poker hands.

The number of combinations of selecting k items from a set of n items is given by the formula:

C(n, k) = n! / (k!(n-k)!)

In this case, we want to select 5 face cards from the set of 12 face cards, so we can calculate:

C(12, 5) = 12! / (5!(12-5)!)

C(12, 5) = 12! / (5! * 7!)

Calculating the factorial terms:

12! = 12 * 11 * 10 * 9 * 8 * 7!

5! = 5 * 4 * 3 * 2 * 1

7! = 7 * 6 * 5 * 4 * 3 * 2 * 1

Plugging in the values:

C(12, 5) = (12 * 11 * 10 * 9 * 8 * 7!) / (5 * 4 * 3 * 2 * 1 * 7!)

Simplifying the expression:

C(12, 5) = (12 * 11 * 10 * 9 * 8) / (5 * 4 * 3 * 2 * 1)

C(12, 5) = 792

Therefore, there are 792 poker hands consisting of all face cards.

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The coordinate vector of p(t) = 1 + 4t + 7t² relative to B is (2, 6, -1).

What is the coordinate vector?

A coordinate vector is a numerical representation of a vector that explains the vector in terms of a specific ordered basis. A simple example would be a position in a 3-dimensional Cartesian coordinate system with the basis being the system's axes.

Here, we have

Given: The set b = (1 + t², t + t², 1 + 2t + t²) is a basic for P₂.

Now, let coordinate vector of P(t) = 1 + 4t + 7t² relative to B is (C₁, C₂, C₃).

Then,

1 + 4t + 7t² = C₁(1 + t²) + C₂(t + t²) + C₃(1 + 2t + t²)

(C₁+C₃) + ( C₂+2C₃)t + (C₁+C₂ +C₃)t² = 1 + 4t + 7t²

C₁+C₃ = 1

C₂+2C₃ = 4

C₁+C₂ +C₃ = 7

Now, to find C₁, C₂, C₃ we solve the system.

The augmented matrix of the given system is:

= \(\left[\begin{array}{ccc}1&0&1|1\\0&1&2|4\\1&1&1|7\end{array}\right]\)

Now, we apply row reduction and we get

R₃ = R₃ - R₁

= \(\left[\begin{array}{ccc}1&0&1|1\\0&1&2|4\\0&1&0|6\end{array}\right]\)

R ⇔ R

= \(\left[\begin{array}{ccc}1&0&1|1\\0&1&0|6\\0&1&2|4\end{array}\right]\)

R₃ = R₃ - R₁

= \(\left[\begin{array}{ccc}1&0&1|1\\0&1&0|6\\0&0&2|-2\end{array}\right]\)

R₃ = 1/2R₃

= \(\left[\begin{array}{ccc}1&0&1|1\\0&1&0|6\\0&0&1|-1\end{array}\right]\)

R₁ = R₁ - R₃

= \(\left[\begin{array}{ccc}1&0&0|2\\0&1&0|6\\0&0&1|-1\end{array}\right]\)

C₁ = 2, C₂ = 6, C₃ = -1

Hence, the coordinate vector of p(t) = 1 + 4t + 7t² relative to B is (2, 6, -1).

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For 23: \( \cosh (ix) \) The real part is the cosh function evaluated at the imaginary part of the argument: \( \cosh (0) = 1 \) The imaginary part is the sinh function evaluated at the imaginary part of the argument: \( \sinh (0) = 0 \)

The absolute value is the magnitude of the complex number: \( |23| = \sqrt{1^2 + 0^2} = 1 \)  24: \( \cos (ix) \) The real part is the cos function evaluated at the imaginary part of the argument: \( \cos (0) = 1 \) The imaginary part is the sin function evaluated at the imaginary part of the argument: \( \sin (0) = 0 \)  The absolute value is the magnitude of the complex number: \( |24| = \sqrt{1^2 + 0^2} = 1 \)

For 25: \( \sin (x-iy) \) The real part is the sin function evaluated at the real part of the argument: \( \sin (x) \) The imaginary part is the negative of the sin function evaluated at the imaginary part of the argument: \( -\sin (-y) \)  The absolute value is the magnitude of the complex number: \( |25| = \sqrt{(\sin(x))^2 + (-\sin(-y))^2} \)

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23. Real part: \(\( \cos(23) \), Imaginary part: 0, Absolute value: \( | \cos(23) | \)\)

24. Real part: \(\( \cosh(24) \), Imaginary part: 0, Absolute value: \( | \cosh(24) | \)\)

25. Real part:\(\( \sin(x) \cosh(y) \), Imaginary part: \( \cos(x) \sinh(y) \),\)

Absolute value: \(\( \sqrt{ (\sin(x) \cosh(y))^2 + (\cos(x) \sinh(y))^2 } \)\)

26. Real part: \(\( \cosh(2) \cos(3) \), Imaginary part: \( \sinh(2) \sin(3) \),\)

Absolute value:\(\( | \cosh(2) \cos(3) + \sinh(2) \sin(3)i | \)\)

27. Real part: \(\( \sin(4) \cosh(3) \), Imaginary part: \( \cos(4) \sinh(3) \),\)

Absolute value: \(\( \sqrt{ (\sin(4) \cosh(3))^2 + (\cos(4) \sinh(3))^2 } \)\)

28. Real part:\(\( \tanh(1) \cos(\pi) \), Imaginary part: \( \sinh(1) \sin(\pi) \),\)

Absolute value:\(\( | \tanh(1) \cos(\pi) + \sinh(1) \sin(\pi)i | \)\)

To find the real part, imaginary part, and absolute value of the given expressions, let's evaluate them one by one:

23. The expression \( \cosh (i x) \) represents the hyperbolic cosine function of the imaginary number \( i x \). Since \( \cosh (ix) = \cos(x) \) for any real value of \( x \), the real part is \( \cos (23) \), the imaginary part is 0, and the absolute value is \( | \cos (23) | \).

24. The expression \( \cos (i x) \) represents the cosine function of the imaginary number \( i x \). Since \( \cos (ix) = \cosh(x) \) for any real value of \( x \), the real part is \( \cosh (24) \), the imaginary part is 0, and the absolute value is \( | \cosh (24) | \).

25. The expression \( \sin (x-i y) \) represents the sine function of the complex number \( x-i y \). The real part is \( \sin(x) \cosh(y) \), the imaginary part is \( \cos(x) \sinh(y) \), and the absolute value is \( \sqrt{ (\sin(x) \cosh(y))^2 + (\cos(x) \sinh(y))^2 } \).

26. The expression \( \cosh (2-3 i) \) represents the hyperbolic cosine function of the complex number \( 2-3i \). The real part is \( \cosh(2) \cos(3) \), the imaginary part is \( \sinh(2) \sin(3) \), and the absolute value is \( | \cosh(2) \cos(3) + \sinh(2) \sin(3)i | \).

27. The expression \( \sin (4+3i) \) represents the sine function of the complex number \( 4+3i \). The real part is \( \sin(4) \cosh(3) \), the imaginary part is \( \cos(4) \sinh(3) \), and the absolute value is \( \sqrt{ (\sin(4) \cosh(3))^2 + (\cos(4) \sinh(3))^2 } \).

28. The expression \( \tanh(1-i\pi) \) represents the hyperbolic tangent function of the complex number \( 1-i\pi \). The real part is \( \tanh(1) \cos(\pi) \), the imaginary part is \( \sinh(1) \sin(\pi) \), and the absolute value is \( | \tanh(1) \cos(\pi) + \sinh(1) \sin(\pi)i | \).

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The vertices of the image of the figure LMNO following a 90° rotation about the origin are; L'(3, 5), M'(6, 6), N'(3, 4), O'(6, 4), indicating that the corresponding segment L'O' and M'N' are on the lines x = 6 and x = 3, the correct option is option B.

B. x = 6 and x = 3

A rotation is the circular motion about a specified center of rotation of the points on the figure of the preimage.

The coordinates of the vertices of the figure LMNO are; L(-5, 3), M(-6, 6), N(-4, 6), O(-4, 3)

The transformation of the figure LMNO = 90° clockwise rotation about the origin.

The coordinate of the image of the point (x, y) following a rotation of 90° about the origin is the point (y, -x).

Therefore, the coordinates of the image of the figure LMNO following a rotation of 90° about the origin are found as follows;

L(-5, 3) \(\underset{\longrightarrow}{R_{90^{\circ}\ clockwise}}\) L'(3, 5)

M(-6, 6) \(\underset{\longrightarrow}{R_{90^{\circ}\ clockwise}}\) M'(6, 6)

N(-4, 6) \(\underset{\longrightarrow}{R_{90^{\circ}\ clockwise}}\) N'(6, 4)

O(-4, 3) \(\underset{\longrightarrow}{R_{90^{\circ}\ clockwise}}\) O'(3, 4)

The points L' and O' are colinear with the line x = 3, and the points M' and N' are colinear with the line x = 6

Therefore, the corresponding segment to the segment LO, L'O' lie on the the line x = 3, and the corresponding segment to the segment MN, M'N' lie on the line x = 6

The correct option is B. x = 6, and x = 3

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The (instantaneous) rate of change of \($f(t)$\) when \($t=3$\) is 9. Given that \($f(t)=t^2+3t-4$\).

(a) What is the average rate of change of \($f(t)$\) over the interval 3 to 4?

The average rate of change of \($f(t)$\) over the interval [3,4] is given by:

\($$\begin{align*}\frac{f(4)-f(3)}{4-3}&=\frac{(4)^2+3(4)-4-[(3)^2+3(3)-4]}{1}\\ &=\frac{16+12-4-9-9+4}{1}\\ &=\frac{10}{1}\\ &=10 \\\end{align*}\)

Therefore, the average rate of change of \($f(t)$\)over the interval 3 to 4 is 10.

(b) What is the (instantaneous) rate of change of \($f(t)$\) when \($t=3$\)?

The instantaneous rate of change of \($f(t)$\) at \($t=3$\) is given by:

\($$\begin{align*}\lim_{h\to0}\frac{f(3+h)-f(3)}{h}&=\lim_{h\to0}\frac{(3+h)^2+3(3+h)-4-[(3)^2+3(3)-4]}{h}\\ &=\lim_{h\to0}\frac{9+6h+h^2+9+3h-4-9}{h}\\ &=\lim_{h\to0}\frac{h^2+9h}{h}\\ &=\lim_{h\to0}(h+9)\\ &=9\\\end{align*}\)

Therefore, the (instantaneous) rate of change of \($f(t)$\) when\($t=3$\) is 9.

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Answer:

126 units²

Step-by-step explanation:

Draw 5 vertical lines BR, CQ, EO, GM, HL and we will have 4 trapezoids and 2 rectangles.

Trapezoid ABRS ≅ trapezoid IHLK

Trapezoid DEOP ≅ trapezoid FEON

Rectangle BCQR ≅ rectangle GHLM

Area of trapezoid is \(\frac{b_{1} +b_{2} }{2}h\)

Area of rectangle is \(lw\)  

The total area of the figure is: ( 2 × \(A_{ABRS}\) + 2 × \(A_{BCQR}\) + 2 × \(A_{DEOP}\) )

In trapezoid ABRS: \(b_{1}\) = 6, \(b_{2}\) = 8 and h = 3 ⇒

2 × \(A_{ABRS}\) = 2 × [3(6 + 8) ÷ 2] = 42 units²

In rectangle BCQR: l = 6 and w = 3 ⇒

2 × \(A_{BCQR}\) = 2 × (6 × 3) = 36 units²

In trapezoid DEOP: \(b_{1}\) = 6, \(b_{2}\) = 10 and h = 3 ⇒

2 × \(A_{DEOP}\) = 2 × [3(6 + 10) ÷ 2] = 48 units²  

\(A_{total}\) = 42 + 36 + 48 = 126 units²  

Answer:

1. 234.57

2.150.79

3. too lazy... lol sorry

Step-by-step explanation:

The standard form of the following equation is x-4y=8.

An equation is a mathematical statement that is made up of two expressions connected by an equal sign. Equation, statement of equality between two expressions consisting of variables and/or numbers.

Given:

The given equation is

y+2=1/4x

According to given question we have

We know that the standard form of the equation of a line is

Ax + By = C

where ,A is a positive integer and B, C are integers.

Compare the given equation

Rearrange the  given equation as

y+2=1/4x

Multiply 4 both sides we get,

4y+8=x

Compare the given equation

x-4y=8

Therefore, the standard form of the following equation is x-4y=8.

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