A child grew 314 inches one year and 434 inches the next year. How many inches did the child grow over the two years?

inches: ??

PLZ HELP ASAP

On solving the equation Ý = 140 + (-0.0667)X, the value for blood pressure is obtained as 129.328 mm Hg.

What is an equation?

A mathematical definition of an equation is a claim that two expressions are equal when they are joined by the equals sign ("=").

The equation is given as Ý = 140 + (-0.0667)X.

An amount of medication of 160 mg is found to result in a blood pressure of 127.74 mm Hg.

Here the value of X is X = 160.

The predicted blood pressure will be -

Ý = 140 + (-0.0667)160

Ý = 140 - 10.672

Ý = 129.328

The predicted blood pressure value is 129.328 mm Hg.

Here the predicted blood pressure value 129.328 mm Hg is greater than the actual blood pressure value 127.74 mm Hg.

Therefore, the observed value will be below the regression line.

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The angle measures for this problem are given as follows:

The sum of the measures of the internal angles of a triangle is of 180º.

The triangle in this problem is ABC, hence the measure of a is obtained as follows:

a + 68 + 50 = 180

a = 180 - (68 + 50)

a = 62º.

c and d are corresponding angles to angle a, as they are on the same position relative to parallel lines, hence their measures are given as follows:

Angle b is a corresponding interior angle with angle a, hence they are supplementary and it's measure is given as follows:

a + b = 180

62 + b = 180

b = 118º.

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The area of the triangle with vertices A(B, -9, 10), B(0, 1, 2), and C(-1, 2, 0) is approximately 107.81 square units.

To verify if the points A(1, 1, 3), B(-3, -8, 4), C(-1, -5, -3), and D(3, 4, -4) form the vertices of a parallelogram, we can check if the opposite sides are parallel.

The vector AB can be calculated by subtracting the coordinates of point A from point B:

AB = B - A = (-3 - 1, -8 - 1, 4 - 3) = (-4, -9, 1).

The vector DC can be calculated by subtracting the coordinates of point C from point D:

DC = D - C = (3 - (-1), 4 - (-5), -4 - (-3)) = (4, 9, -1).

If AB is parallel to DC, then the two vectors will be scalar multiples of each other. We can check this by calculating the scalar multiples of the components:

(-4/4) = (-9/9) = (1/-1).

Since the scalar multiples are equal, we can conclude that AB is parallel to DC. Similarly, we can verify that BC is parallel to AD and AC is parallel to BD.

Therefore, the points A(1, 1, 3), B(-3, -8, 4), C(-1, -5, -3), and D(3, 4, -4) form the vertices of a parallelogram.

To find the area of the parallelogram, we can calculate the magnitude of the cross product of vectors AB and AD (or BC and BD), and then divide it by 2.

The cross product AB x AD can be calculated as follows:

AB x AD = |i j k|

|-4 -9 1|

|4 9 -1|.

Expanding the determinant calculation:

= (9 * (-1) - 1 * 9) * i - ((-4) * (-1) - 1 * 4) * j + ((-4) * 9 - (-4) * 9) * k

= (-18)i - (0)j + (0)k

= (-18)i.

The magnitude of the cross product AB x AD is |(-18)i| = 18.

So, the area of the parallelogram is 18 square units.

Moving on to the second part of the question, to find the area of the triangle with vertices A(B, -9, 10), B(0, 1, 2), and C(-1, 2, 0), we can again use the cross product.

The vector AB can be calculated by subtracting the coordinates of point A from point B:

AB = B - A = (0 - (-9), 1 - (-9), 2 - 10) = (9, 10, -8).

The vector AC can be calculated by subtracting the coordinates of point A from point C:

AC = C - A = (-1 - (-9), 2 - (-9), 0 - 10) = (8, 11, -10).

The cross product AB x AC can be calculated as follows:

AB x AC = |i j k|

|9 10 -8|

|8 11 -10|.

Expanding the determinant calculation:

= (10 * (-10) - (-8) * 11) * i - (9 * (-10) - (-8) * 8) * j + (9 * 11 - 10 * 8) * k

= (-108)i - (-32)j + (1)k

= (-108)i + 32j + k.

The magnitude of the cross product AB x AC is |(-108)i + 32j + k| = sqrt((-108)^2 + 32^2 + 1^2) = sqrt(11617) ≈ 107.81.

Therefore, the area of the triangle with vertices A(B, -9, 10), B(0, 1, 2), and C(-1, 2, 0) is approximately 107.81 square units.

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sorry wrong answer dont use this answer

Step-by-step explanation:

a). A = {x  ∈ R  I  5x-8 < 7}

5x - 8 < 7 <=> 5x < 8+7  <=> 5x < 15  =>

x < 3  => A = (-∞ ; 3)

A ∩ N = {0 ; 1 ; 2}

A - N* = (-∞ ; 3) - {1 ; 2}

b).  A = { x ∈ R  I  7x+2 ≤ 9}

7x+2 ≤ 9 <=> 7x ≤ 7  => x ≤ 1 => x ∈ (-∞ ; 1]

A ∩ N = {0 ; 1}

A-N* = (-∞ ; 1)

c). A = { x ∈ R  I  I 2x-1 I < 5}

I 2x-1 I < 5  <=> -5 ≤ 2x-1 ≤ 5  <=>

-4 ≤ 2x ≤ 6  <=> -2 ≤ x ≤ 3  => x ∈ [-2 ; 3]

A ∩ N = {0 ; 1 ; 2 ; 3}

A - N* = [-2 ; 3) - {1 ; 2}

d). A = {x ∈ R I I 6-3x I ≤ 9}

I 6-3x I ≤ 9  <=> -9 ≤ 6-3x ≤ 9  <=>

-15 ≤ -3x ≤ 3 <=> -5 ≤ -x ≤ 3  =>

-3 ≤ x ≤ 5  => x ∈ [-3 ; 5]

A ∩ N = {0 ; 1 ; 2 ; 3 ; 4 ; 5}

A - N* = [-3 ; 5) - {1 ; 2 ; 3 ; 4}

3rd choice is answer

Step-by-step explanation:

Answer:

1.125

Step-by-step explanation:

Answer 1.125 ounces

Step-by-step explanation:

divide 4.5 by 4

Answer:

x=−27/26

and y=−192/13

Step-by-step explanation:

6x+y=−21;8x−3y=36

Step: Solve6x+y=−21for y:

6x+y=−21

6x+y+−6x=−21+−6x(Add -6x to both sides)

y=−6x−21

Step: Substitute−6x−21foryin8x−3y=36:

8x−3y=36

8x−3(−6x−21)=36

26x+63=36(Simplify both sides of the equation)

26x+63+−63=36+−63(Add -63 to both sides)

26x=−27

26x

26

=

−27

26

(Divide both sides by 26)

Answer:

see below

Step-by-step explanation:

y = x^2 +2x - 8

Factor

y = ( x+4)(x-2)

The zeros are

x+4 = 0  x-2 = 0    

x= -4    x= 2

(-4,0)  (2,0)

The axis of symmetry is 1/2 way between the zeros

(-4+2)/2 = -2/2 = -1

x = -1

The x coordinate of the vertex is at the axis of symmetry

The y value is found by substituting x=-1

y = (-1+4) (-1-2)= 3*-3 = -9

Vertex ( -1,-9)

The domain is all real numbers

The range is from -9 up

y≥ -9  since it opens upward

Answer:

240 :)

Step-by-step explanation:

easy all u have to do is just add all the sides together but the empty sides with no number u have to try to figure out how much meters that is.

they're both 36 because they're the same size as the meters on the other side.

so its going to be

60+36+36+36+36+36

hope this helps

good luck!

Answer:

50 degrees

Step-by-step explanation:

10 + 40 = 50

I hope this helps!!

Answer:

50 degrees.

Step-by-step explanation:

40 + 10 = 50

Answer:

dehydration

Step-by-step explanation:

Diurectics causes an increase in amount of urine. I.e it makes you pee more.

If you urinate too much, you run the risk of dehydration because you lose too much water.

Answer:

B. y = -5/4x + 3

Step-by-step explanation:

The slope is negative because the line is going from up to down.

Options A and C are wrong.

To find the slope divide by difference of two points.

(0, 3) and (4, -2)

(-2-3)/(4-0)

-5/4

The slope is -5/4.

The y-intercept is (0, 3)

y = mx + b, m is the slope and b is the y-intercept.

y = -5/4x + 3

The length of the third side of the triangle is 2x+13.

According to the question,

We have the following information:

A triangle has a perimeter of 7x + 8. The first side of the triangle has a length of 3x, and the second side has a length of 2x − 5.

We know that perimeter of triangle is the sum of all three sides of triangle.

Let's take its third side to be y.

So, we have the following expression:

3x+2x-5+y = 7x+8

5x-5+y = 7x+8

Adding 5 on both the sides:

5x+y = 7x+8+5

5x+y = 7x+13

Subtracting 5x from both the sides:

y = 7x-5x+13

y = 2x+13

Hence, the correct option is B.

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Answer:

B. 2x+13

Step-by-step explanation:

I got it right on the test

So the two possible values are 2.5 + (3√7/2)i and 2.5 - (3√7/2)i.

To simplify the expression (20 ± i√1008)/8, we can start by simplifying the square root of 1008.

√1008 can be simplified as follows:

√1008 = √(16 × 63)

= √16 × √63

= 4√63

Now we can substitute this value back into the expression:

(20 ± i√1008)/8 = (20 ± i(4√63))/8

Next, we can simplify the expression by dividing both the numerator and the denominator by 4:

(20 ± i(4√63))/8 = 20/8 ± i(4√63)/8

Simplifying further:

20/8 = 2.5

(4√63)/8 = √63/2

= (√9 × √7)/2

= 3√7/2

The simplified expression is:

(20 ± i√1008)/8 = 2.5 ± (3√7/2)i

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The measure of the central angle indicated is 270 degrees

The term "diameter" refers to a straight line segment that passes through the center of a circle or a sphere, connecting two points on the circumference or surface of the circle or sphere. In other words, it is the longest distance that can be measured between two points on the edge of the circle or sphere, passing through its center.

To find the measure of the central angle indicated, we need to first identify the endpoints of the diameter that contains points W, T, and X. Let's assume that this diameter is WX. Then, we can find the measure of the central angle WTX by finding the measure of the arc WT and dividing it by 2.

We know that the diameter WX passes through the midpoint of segment VT, which we can find by averaging the coordinates of V and T. Using the coordinates given in the diagram, we have:

V: (9x-2, 15x+10)

T: (15x+10, 9x-2)

Midpoint of VT: ((9x-2 + 15x+10)/2, (15x+10 + 9x-2)/2)

= (12x + 4, 12x + 4)

Since this midpoint lies on the diameter WX, we can find the coordinates of point X by reflecting the midpoint across the y-axis:

X: (-12x - 4, 12x + 4)

Now we can find the measure of the arc WT by finding the difference between the angles formed by radii WT and WX. Let's call the center of the circle O:

m∠WOT = 90 degrees (since WT is a diameter)

m∠WOX = 180 degrees (since WX is a diameter)

m∠TOX = m∠WOT - m∠WOX = -90 degrees

To convert this angle to a positive measure, we can add 360 degrees:

m(arc WT) = m∠WOT - m∠TOX + 360 degrees = 90 degrees - (-90 degrees) + 360 degrees = 540 degrees

Finally, we can find the measure of the central angle WTX by dividing the measure of arc WT by 2:

m∠WTX = m(arc WT)/2 = 540 degrees/2 = 270 degrees

Therefore, the measure of the central angle indicated is 270 degrees

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To prove that 3x ≤ y, assume the opposite, that is, 3x > y, rearrange the inequality substitute x < 2y and simplify, contradict the given condition that x < 2y, therefore, concluding that 3x ≤ y.

Start by assuming the opposite, that is, 3x > y.

From the given inequality,\(7xy \leq 3x^2 + 2y^2,\), we can rearrange to get:
\(7xy - 3x^2 \leq 2y^2\)

We can substitute \(x < 2y\) into this inequality:
\(7(2y)x - 3(2y)^2 \leq 2y^2\)

Simplifying, we get:
\(y(14x - 12y) \leq 0\)

Since y is a real number, this means that either y ≤ 0 or 14x - 12y ≤ 0.

If y ≤ 0, then 3x ≤ y is trivially true.

If 14x - 12y ≤ 0, then we can rearrange to get:
3x ≤ (12/14)y
3x ≤ (6/7)y
3x < y (since we assumed 3x > y)

But this contradicts the given condition that x < 2y, so our assumption that 3x > y must be false.

Therefore, we can conclude that 3x ≤ y.

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As per the question, All four students A, B, C, and D are loss-averse over money and have the same value function as below:v(x dollars)={√x     x ≥ 0   {-2√-x  x < 0They are also loss averse over mugs and have the same value function.

v(y mugs)={3y y ≥ 0
        {4y y < 0
Now, we have to find the values of a, b, c and d as below:
- Student A owns a mug and is willing to sell it for a price of a dollars or more. i.e v(a) = v(0) + v(a-A), where A is the initial endowment of A. According to the given function, v(0) = 0, v(a-A) = 3, and v(A) = 4.
So, a ≥ A+3/2
- Student B does not own a mug and is willing to pay up to b dollars for buying it. i.e v(B-b) = v(B) - v(0), where B is the initial endowment of B. According to the given function, v(0) = 0, v(B-b) = -4, and v(B) = -3.
So, b ≤ B+1/2
- Student C does not own a mug and is indifferent between getting a mug and getting c dollars. i.e v(c) = v(0) + v(c), where C is the initial endowment of C. According to the given function, v(0) = 0, v(c) = 3.
So, c = C/2
- Student D is indifferent between losing a mug and losing d dollars. i.e v(D-d) = v(D) - v(0), where D is the initial endowment of D. According to the given function, v(0) = 0, v(D-d) = -3.
So, d = D/2
2) In this case, value function for money changes to:v(x dollars)={√x     x ≥ 0
            {-√-x   x < 0
However, the value function for mugs remains the same:v(y mugs)={3y y ≥ 0
         {4y y < 0
Therefore, values for a, b, c, and d will remain the same as calculated in part (1).
3) In this case, students are not loss-averse. Value function for money:v(x dollars)={√x     x ≥ 0
            {-√-x   x < 0
Value function for mugs:v(y mugs)={3y y ≥ 0
The reference point is the status quo, i.e initial endowment. So,
- Student A owns a mug and is willing to sell it for a price of a dollars or more. The value of mug for A is 3 initially and he would sell it for 3 or more.
So, a ≥ A+3/2
- Student B does not own a mug and is willing to pay up to b dollars for buying it. The value of mug for B is 3 initially and he would buy it for 3 or less.
So, b ≤ B+3/2
- Student C does not own a mug and is indifferent between getting a mug and getting c dollars. The value of the mug for C is 3 initially.
So, c = 3
- Student D is indifferent between losing a mug and losing d dollars. The value of the mug for D is 3 initially.
So, d = 3
4) In this case, value function for money:v(x dollars)={√x     x ≥ 0
            {-√-x   x < 0
Value function for mugs: Mug will have a value of 4 for someone who owns it and 3 for someone who does not own it.
The reference point is the status quo, i.e initial endowment. So,
- Student A owns a mug and is willing to sell it for a price of a dollars or more. The value of mug for A is 4 initially and he would sell it for 4 or more.
So, a ≥ A+2
- Student B does not own a mug and is willing to pay up to b dollars for buying it. The value of mug for B is 3 initially and he would buy it for 3 or less.
So, b ≤ B+3/2
- Student C does not own a mug and is indifferent between getting a mug and getting c dollars. The value of the mug for C is 3 initially and he would like to buy it for 3.
So, c = 3
- Student D is indifferent between losing a mug and losing d dollars. The value of the mug for D is 3 initially.
So, d = 3.

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Answer:

it is A

Step-by-step explanation:

i had the question on ed

Answer:

y greater - than - or - equal - to StartFraction 2 over 15 EndFraction

Step-by-step explanation:

Answer:

Answer: Option 330 bars of soap Step-by-step explanation: If the amount is $ 1,200 and the simple interest for a year is 10% then the amount that Mark must pay is: If each soap bar costs $ 4 then the amount of soap bars you must sell in 1 year is: The answer is the first option  

Step-by-step explanation:

Answer:

330

Step-by-step explanation:

$1,200.00 + 10% = $1,320.00 or $1,200.00 + $120.00 = $1,320.00

$1,320.00 / $4.00 = 330

330 * $4.00 = $1,320.00

Answer:

56 i think

Step-by-step explanation:

Answer:

the ANSWER IS EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Assigning the truth values p = true, q = false, and r = true to the propositional variables results in a valid argument.

To show that the argument is valid, we need to assign truth values to the propositional variables p, q, and r and evaluate the given argument.

Let's assign truth values to p, q, and r as follows:

p = true
q = false
r = true

Now, let's substitute these truth values into the argument:
p ∧ ¬q (p → q) → r ∴ ¬r

Substituting the truth values:
true ∧ ¬false (true → false) → true ∴ ¬true

Simplifying further:
true ∧ true (false) → true ∴ false

true ∧ true is true, and false → true is true (since false implies anything).

So, the argument can be rewritten as:
true → true → true ∴ false

Since the antecedent of the implication is true, the implication itself is true. Thus, the argument is valid.

In summary, assigning the truth values p = true, q = false, and r = true to the propositional variables results in a valid argument.

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When extrapolating the grading curve into the clay zone, the errors that might occur are: inaccurate estimation of particle size distribution, assumption of uniformity, over-reliance on empirical relationships, neglecting soil fabric and structure, and limitations of laboratory testing.

1. Inaccurate estimation of particle size distribution: The grading curve represents the distribution of particle sizes in a soil sample. When extrapolating into the clay zone, it can be challenging to accurately estimate the particle sizes due to the fine nature of clay particles. The extrapolated curve may not reflect the true distribution, leading to errors in analysis and design.

2. Assumption of uniformity: Extrapolating the grading curve assumes that the particle size distribution remains consistent throughout the clay zone. However, clay soils can exhibit significant variations in particle size distribution within short distances. Ignoring this non-uniformity can result in incorrect interpretations and predictions.

3. Over-reliance on empirical relationships: Grading curves are often used in conjunction with empirical relationships to estimate various soil properties, such as permeability or shear strength. However, these relationships are typically developed for specific soil types and may not be applicable to clay soils. Relying solely on empirical relationships without considering the unique behavior of clay can lead to significant errors in analysis and design.

4. Neglecting soil fabric and structure: Clay soils often exhibit complex fabric and structure due to their small particle size. Extrapolating the grading curve without considering the fabric and structure can overlook important characteristics such as particle orientation, interparticle forces, and fabric anisotropy. These factors can significantly influence the behavior of clay soils and should be accounted for to avoid errors.

5. Limitations of laboratory testing: Extrapolating the grading curve into the clay zone relies on laboratory testing to determine the particle size distribution. However, laboratory testing may not accurately represent the in-situ conditions or account for the changes in soil behavior due to sampling disturbance or reactivity. These limitations can introduce errors in the extrapolation process.

To mitigate these errors, it is essential to consider alternative methods of characterizing clay soils, such as direct sampling techniques or specialized laboratory tests. Additionally, using site-specific data and considering the unique properties of clay soils can help improve the accuracy of the extrapolated grading curve. Consulting with geotechnical engineers or soil scientists can provide further insights and guidance in addressing these errors.

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The original slope's reciprocal will be the opposite of the perpendicular slope. To determine the intercept, b, enter the supplied point and the new slope into the slope-intercept form (y = mx + b). Rewrite the following equation in standard form: ax + by = c.

The values of the slope and y-intercept provide details on the relationship between the two variables, x and y. The slope shows how quickly y changes for every unit change in x. When the x-value is 0, the y-intercept shows the y-value.

Y = mx + b, where m denotes the slope and b the y-intercept, is how the equation of the line is expressed in the slope-intercept form. We can see that the slope of the line in our equation, y = 6x + 2, is 6.

The slope is m and the y-intercept is b in the equation y=mx+b in slope-intercept form. Some equations can also be rewritten so that they resemble slope-intercept form. For instance, y=x can be written as y=1x+0, resulting in a slope and y-intercept of 1 and 0, respectively.

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Answer:

23.3

Step-by-step explanation:

Answer:

Step-by-step explanation:

Coefficient: If a term has a variable, the numerical factor before the variable

Like Terms: A term that only has numerical factors and variables

Constant: Terms that have the exact same amount of variable factors

Equation: Two algebraic expressions that are equal

Solution: A value that makes an algebraic equation true

The correct match is given below:-

Coefficient:- If a term has a variable, the numerical factor is before the variable.

Like Terms:- A term that only has numerical factors and variables.

Constant:- Terms that have the exact same amount of variable factors.

Equation:- Two algebraic expressions that are equal.

Solution:- A value that makes an algebraic equation true.

The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.

Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.

The appropriate pairing is shown below: -

Coefficient: The numerical factor comes before the variable if a term has one.

Similar Terms: A term with only numerical variables and factors.

Terms with the same number of variables are said to be constants.

Equation: The equality of two algebraic expressions.

A value that makes an algebraic equation true is the answer.

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Answer:

50/100

Step-by-step explanation:

Answer:

GH

Step-by-step explanation:

A secant is a straight line which cuts the circle at 2 points

GH and FE both do this but FE is a special secant, that is the diameter

Then GH is a secant