Subtract. (29v)−(−13v+59)
2
9
v
-
-
1
3
v
+
5
9

59v+59
5
9
v
+
5
9

19v+59
1
9
v
+
5
9

59v−13
5
9
v
-
1
3

59v−59
5
9
v
-
5
9

The polynomial is  (x - 4)(x - 7)/(x + 4)(x + 5).

The question asks to add the two fractions (x^(2)-4x+9)/(x^(2)+9x+20) and (x-37)/(x^(2)+9x+20). To solve this problem, we can first simplify the denominator by factoring the polynomial:

Denominator: (x^(2)+9x+20) = (x + 4)(x + 5)


Now, we can rewrite the two fractions with this new denominator:

(x^(2)-4x+9)/(x + 4)(x + 5) + (x-37)/(x + 4)(x + 5)


Then, we can use the distributive property to expand the fractions and combine like terms:

(x^(2)-4x+9 + x-37)/(x + 4)(x + 5)

= (x^(2) - 3x - 28)/(x + 4)(x + 5)


Finally, we can simplify the numerator by combining like terms:

= (x^(2) - 3x - 28)/(x + 4)(x + 5)

= (x - 4)(x - 7)/(x + 4)(x + 5)

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For given functions, "The y-intercept of f(x) is equal to the y-intercept of g(x)" is the correct answer i.e. A.

What is the definition of a function?

In mathematics, a function is a relation between two sets of elements, called the domain and the range, such that each element in the domain corresponds to exactly one element in the range.

More specifically, a function is a rule that assigns each element of the domain (input) to a unique element in the range (output). The notation for a function f with domain D and range R is typically written as:

f: D → R

f is a function mapping elements from the domain D to elements in the range R.

Now,

To find the y-intercept of a function, we set x = 0 and evaluate the function.

For [tex]f(x) = -4(1.09)^x[/tex], when x = 0, we get:

[tex]f(0) = - 4(1.09)^0 = - 4[/tex]

So, the y-intercept of f(x) is -4.

For g(x), we are given a table of values, and we can see that when x = 0, g(x) = -4. Therefore, the y-intercept of g(x) is also -4.

Hence,

          The first option "The y-intercept of f(x) is equal to the y-intercept of g(x)" is the correct answer.

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[tex]x=\textit{kgs of solution at 30\%}\\\\ ~~~~~~ 30\%~of~x\implies \cfrac{30}{100}(x)\implies 0.3 (x) \\\\\\ y=\textit{kgs of solution at 15\%}\\\\ ~~~~~~ 15\%~of~y\implies \cfrac{15}{100}(y)\implies 0.15 (y) \\\\\\ \textit{60 kgs of solution at 20\%}\\\\ ~~~~~~ 20\%~of~60\implies \cfrac{20}{100}(60)\implies 0.2 (60)\implies 12 \\\\[-0.35em] ~\dotfill[/tex]

[tex]\begin{array}{lcccl} &\stackrel{kgs}{quantity}&\stackrel{\textit{\% of kgs that is}}{\textit{nitrogen only}}&\stackrel{\textit{kgs of}}{\textit{nitrogen only}}\\ \cline{2-4}&\\ \textit{1st Fert.}&x&0.3&0.3x\\ \textit{2nd Fert.}&y&0.15&0.15y\\ \cline{2-4}&\\ mixture&60&0.2&12 \end{array}~\hfill \begin{cases} x + y = 60\\\\ 0.3x+0.15y=12 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{x+y=60}\implies y=60-x \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{using the 2nd equation}}{0.3x+0.15y=12}\implies \stackrel{\textit{substituting from above}}{0.3x+0.15(60-x)=12} \\\\\\ 0.3x+9-0.15x=12\implies 0.15x=3\implies x=\cfrac{3}{0.15} \\\\\\ \boxed{x=20}\hspace{5em}\stackrel{ 60~~ - ~~20 }{\boxed{y=40}}[/tex]

Method of inverse of matrix is given:  

To obtain the inverse of the given matrix, M, without using the Matlab inv() command, you can use the following steps:

1. Compute the determinant of M, which is equal to 2.

2. Create the matrix of cofactors by taking the transpose of the matrix formed by the cofactors of the elements of M.

3. Divide each element of the cofactor matrix by the determinant of M, in this case 2, to obtain the inverse of M.

Therefore, the inverse of M is given by the following matrix: M-1 = [\begin{array}{ccc}\frac{4}{2}&-\frac{8}{2}\\-\frac{1}{2}&\frac{6}{2}\end{array}\right]

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Answer:

false

Step-by-step explanation:

the pothagirum therum say that theres many reason its false

Answer: True

Step-by-step explanation:

True because we do not know the letters value, so in this case you only add the numbers 14 and 2. You will end up with the answer 16x after adding the letter/valuable.

Answer:

area=66

perimeter=42

Step-by-step explanation:

area = (12 x 3) + (5 x 6)

=36 + 30

=66

perimeter = 12 +9 +5 +6 +7 +5

=42

1. 9 months 2) $163.06, because it is the remaining balance after paying off the principal and interest accrued for that month. 3) $348.16, because it is the balance remaining on the lower interest card after paying off the higher interest card and making the monthly payment for that month. 4) 9 months 5) $168.79, because it is the difference between the total amount paid to each card when paying off the higher interest card first versus paying off the lower interest card first.

Interest is the fee paid for the use of borrowed money, usually expressed as a percentage of the borrowed amount.

Therefore, 1. 9 months 2) $163.06, because it is the remaining balance after paying off the principal and interest accrued for that month. 3) $348.16, because it is the balance remaining on the lower interest card after paying off the higher interest card and making the monthly payment for that month. 4) 9 months 5) $168.79, because it is the difference between the total amount paid to each card when paying off the higher interest card first versus paying off the lower interest card first.

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Answer:

The number is → 56

Step-by-step explanation:

tens digit [tex]\Rightarrow x[/tex]

unit digit [tex]\Rightarrow y[/tex]

"The difference between the digits of a two-digit number is 1...", " ...the unit digit is greater than the tens digit..."

[tex]y-x=1 \qquad \textbf{ec.1}[/tex]

"The number itself is one more (unit) than five times the sum of its digits..."

[tex]10x+y=5(x+y)+1\\ 10x+y= 5x + 5y+1\\5x= 4y+1 \qquad \textbf{ec.2}[/tex]

we clear "y" in equation 1:

[tex]y=1+x \qquad \textbf{ec.3}[/tex]

then we substitute in equation 2:

[tex]5x=4(1+x)+1\\5x=5+4x\\\boxed{x=5}[/tex]

Finally, we substitute in equation 3:

[tex]y=1+5\\\boxed{y=6}[/tex]

With this we have solved the exercise.

[tex]\text{-B$\mathfrak{randon}$VN}[/tex]

The system of three inequalities that describes this region is:x > 0y > 0x + y < 5

In a coordinate plane, the region that consists of all points that have positive x-and y-coordinates whose sum is less than 5 is the triangular region in the first quadrant bounded by the x-axis, y-axis, and the line x + y = 5. The system of three inequalities that describes this region is:x > 0y > 0x + y < 5Explanation:To find the region that consists of all points that have positive x-and y-coordinates whose sum is less than 5, we need to first graph the line x + y = 5 on a coordinate plane. This line has a slope of -1 and passes through the points (0,5) and (5,0). The region that we are looking for is the triangular region in the first quadrant bounded by the x-axis, y-axis, and this line.To write a system of three inequalities that describes this region, we need to consider the following facts:- All points in this region have positive x-coordinates, so x > 0.- All points in this region have positive y-coordinates, so y > 0.- All points in this region have x-and y-coordinates whose sum is less than 5, so x + y < 5.Therefore, the system of three inequalities that describes this region is:x > 0y > 0x + y < 5

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The product of these two matrices is a 3x3 matrix, AB.
AB = [[-9, -7, -1]
     [-9, -42, -7]
     [63, -42, 7]]

To determine AB, we need to multiply matrix A and matrix B. The dimensions of matrix A are 3x1 and the dimensions of matrix B are 1x3. Since the number of columns in matrix A is equal to the number of rows in matrix B, we can multiply these matrices. The resulting matrix will have the dimensions of the number of rows in matrix A and the number of columns in matrix B, which is 3x3.

To multiply the matrices, we take the dot product of each row in matrix A with each column in matrix B. The dot product is the sum of the products of the corresponding entries in the row and column.

AB = [[(-1)(-9) + (-6)(-7) + (7)(-1)], [(-1)(-9) + (-6)(-7) + (7)(-1)], [(-1)(-9) + (-6)(-7) + (7)(-1)]]
AB = [[-9, -7, -1]
     [-9, -42, -7]
     [63, -42, 7]]

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Answer:

too blurry

Step-by-step explanation:

To obtain 40g of active ingredient from a 100:1000 solution, you would need 400mL of the solution. The correct answer is option c. 400mL. A 100:1000 solution means that there are 100g of active ingredient in 1000mL of the solution. To find out how many milliliters of the solution you need to obtain 40g of active ingredient, you can use the following proportion:

100g/1000mL = 40g/x mL

Cross-multiplying gives:

100g * x mL = 40g * 1000mL

Simplifying and solving for x gives:

x = (40g * 1000mL)/100g

x = 400mL

Therefore, you would need 400mL of the 100:1000 solution to obtain 40g of active ingredient.

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The equation of the graph of the function g(x) is -√(x-7).

Turns are frequently used to refer to reflection, which is when an object is flipped over a line without affecting its size or shape. The preimage is flipped over a line in a rigorous transition known as a reflection, but its size and shape are left unchanged. Flips is another name for reflections.

A figure can be translated if it is moved in any direction without altering its size, form, or orientation. A hard transformation called a translation alters the preimage's position but not its size, shape, or orientation. Slides are another name for translations.

Given that, f(x)= square root of x, that is:

f(x) = √x

Reflect the graph over x-axis we have:

Reflecting f(x) across the x-axis gives us -f(x) = -√x.

Translating -f(x) = -√x 7 units to the right gives us -f(x-7) = -√(x-7).

Hence, the equation of the graph of the function g(x) is -√(x-7).

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end{bmatrix} = \begin{bmatrix} 1 \\ \frac{1}{5} \\ 1 \end{bmatrix}.

(a) Let $a^1 = \begin{bmatrix} 1 \\ 1 \\ 2 \\ 1 \end{bmatrix}, a^2 = \begin{bmatrix} -1 \\ 2 \\ 0 \\ -2 \end{bmatrix},$ and $a^3 = \begin{bmatrix} 1 \\ 4 \\ 4 \\ 0 \end{bmatrix}.$ Write the matrix $A = \begin{bmatrix} a^1 & a^2 & a^3 \end{bmatrix}$ in the form $A = QR$ by using the Gram-Schmidt process. (b) Use the QR factorization of $A$ in part (a) to solve the equation $Ax = b,$ where $b = \begin{bmatrix} 3 \\ 1 \\ 2 \\ 1 \end{bmatrix}.$The Gram-Schmidt algorithm is a numerical method to produce orthonormal basis of a subspace in Hilbert space that spans the same space, which makes the basis more convenient to work with. As for the first part of the question, let us begin by applying the Gram-Schmidt algorithm to $a^1, a^2, a^3.$ We begin by defining $q_1 = a^1 / \|a^1\|.$ Hence,$$q_1 = \frac{1}{3}\begin{bmatrix} 1 \\ 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1/3 \\ 1/3 \\ 2/3 \\ 1/3 \end{bmatrix}.$$Next, we define $v_2 = a^2 - \langle q_1, a^2 \rangle q_1.$ Therefore,$$v_2 = a^2 - \frac{-1}{3}(1/3)q_1 = \begin{bmatrix} -7/9 \\ 8/9 \\ -2/9 \\ -4/9 \end{bmatrix}.$$Now, we can define $q_2 = v_2 / \|v_2\|.$ Thus,$$q_2 = \frac{1}{3}\begin{bmatrix} -7 \\ 8 \\ -2 \\ -4 \end{bmatrix}.$$Finally, we define $v_3 = a^3 - \langle q_1, a^3 \rangle q_1 - \langle q_2, a^3 \rangle q_2.$ Then,$$v_3 = a^3 - \frac{5}{9}q_1 - \frac{7}{27}q_2 = \begin{bmatrix} -1/27 \\ 5/9 \\ 22/27 \\ -5/27 \end{bmatrix}.$$Lastly, we can define $q_3 = v_3 / \|v_3\|,$ so$$q_3 = \frac{1}{3}\begin{bmatrix} -1 \\ 5 \\ 22 \\ -5 \end{bmatrix}.$$Now, we can write $A = QR$ as $$\begin{bmatrix} a^1 & a^2 & a^3 \end{bmatrix} = \begin{bmatrix} q_1 & q_2 & q_3 \end{bmatrix} \begin{bmatrix} r_{11} & r_{12} & r_{13} \\ 0 & r_{22} & r_{23} \\ 0 & 0 & r_{33} \end{bmatrix}.$$We can obtain the entries of the $R$ matrix by calculating the inner product of each $q_i$ with $a^j.$ Thus,$$r_{11} = \|a^1\| = \sqrt{7},$$$$r_{12} = \langle q_1, a^2 \rangle = \frac{-1}{3}\sqrt{7},$$$$r_{13} = \langle q_1, a^3 \rangle = \frac{5}{9}\sqrt{7},$$$$r_{22} = \|v_2\| = \frac{5}{3}\sqrt{2},$$$$r_{23} = \langle q_2, a^3 \rangle = \frac{-7}{9}\sqrt{2},$$$$r_{33} = \|v_3\| = \frac{2}{3}\sqrt{6}.$$Therefore,$$\begin{bmatrix} a^1 & a^2 & a^3 \end{bmatrix} = \begin{bmatrix} q_1 & q_2 & q_3 \end{bmatrix} \begin{bmatrix} \sqrt{7} & -\frac{1}{3}\sqrt{7} & \frac{5}{9}\sqrt{7} \\ 0 & \frac{5}{3}\sqrt{2} & -\frac{7}{9}\sqrt{2} \\ 0 & 0 & \frac{2}{3}\sqrt{6} \end{bmatrix}.$$Now, let us solve the equation $Ax = b$ by using the QR factorization of $A.$ We can write $Ax = QRx = b.$ Since $Q$ is orthogonal, we can multiply both sides of the equation by $Q^T$ to obtain $Rx = Q^Tb.$ Note that $Q^Tb$ is easy to compute since $Q^T$ is just the matrix with the $q_i$'s as rows. Thus,$$\begin{bmatrix} \sqrt{7} & -\frac{1}{3}\sqrt{7} & \frac{5}{9}\sqrt{7} \\ 0 & \frac{5}{3}\sqrt{2} & -\frac{7}{9}\sqrt{2} \\ 0 & 0 & \frac{2}{3}\sqrt{6} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{bmatrix}.$$This gives the system of equations$$\begin{cases} \sqrt{7}x_1 - \frac{1}{3}\sqrt{7}x_2 + \frac{5}{9}\sqrt{7}x_3 = \frac{2}{3}, \\ \frac{5}{3}\sqrt{2}x_2 - \frac{7}{9}\sqrt{2}x_3 = \frac{1}{3}, \\ \frac{2}{3}\sqrt{6}x_3 = \frac{2}{3}. \end{cases}$$Solving the last equation for $x_3,$ we obtain $x_3 = 1.$ Substituting this into the second equation, we obtain $x_2 = \frac{1}{5}.$ Finally, substituting these values into the first equation gives us $x_1 = 1.$ Therefore,$$x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{1}{5} \\ 1 \end{bmatrix}.$$

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This |x - 32| <= 2 means that the tire pressure can be anywhere between 30 psi and 34 psi.

In order for Ms. Sartain's car to have optimal gas mileage, the tire pressure should remain within 2 psi of 32 psi at all times. This can be modeled with an absolute value inequality.

The absolute value inequality that models this situation is |x - 32| <= 2. This inequality states that the difference between the tire pressure, x, and the optimal pressure, 32, should be less than or equal to 2.

In other words, the tire pressure can be 2 psi above or below the optimal pressure of 32 psi and still be within the acceptable range. This means that the tire pressure can be anywhere between 30 psi and 34 psi.

So the correct answer is |x - 32| <= 2.

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2. The given quadratic equation is in the general form:

ax² + bx + c = 0

therefore:

a = 2

b = -4

c = -3

The quadratic formula is thus:

[tex]x=\frac{-b(+-)\sqrt{b^2-4ac} }{2a}[/tex]

Substituting the values found for a, b, and c:

[tex]x=\frac{-(-4)+\sqrt{(-4)^2-4(2)(-3)} }{2(2)}[/tex] and [tex]x=\frac{-(-4)-\sqrt{(-4)^2-4(2)(-3)} }{2(2)}[/tex]

Therefore x = 2.58, x = -0.58

3. Using the same method as above, first, bring all values to one side, leaving the RHS = 0

a = 1

b = 2

c = -1

The quadratic formula is thus:

[tex]x=\frac{-b(+-)\sqrt{b^2-4ac} }{2a}[/tex]

Substituting the values found for a, b, and c:

[tex]x=\frac{-(2)+\sqrt{(2)^2-4(1)(-1)} }{2(1)}[/tex] and [tex]x=\frac{-(2)-\sqrt{(2)^2-4(1)(-1)} }{2(1)}[/tex]

Therefore, x = 0.41, x = -2.41

[tex]2 {x}^{2} - 4x - 3 = 0[/tex]

A Here ,

[tex]\boxed{a = 2 }\\\boxed{b = - 4} \\ \boxed{c = - 3}[/tex]

B Filling in the values of a , b and c in the Quadratic formula below , we get

[tex]x = \frac{- (b)\pm \sqrt{( {b}^{2}) - 4(a)(c) } }{2(a)} \\ [/tex]

C Simplifying each section , we get

[tex]x = \frac{ - ( - 4) + \sqrt{( { - 4}^{2} ) - 4(2)( - 3)} }{2 \times 2} [/tex]

or

[tex]x = \frac{ - ( - 4) - \sqrt{ {( - 4})^{2} - 4(2)( - 3) } }{2 \times 2} [/tex]

D Simplifying answers from Part C , we get

[tex]\boxed{x = \frac{2 + \sqrt{10} }{2}} \: \: \: \: or \: \: \: \: \boxed{ x = \frac{2 - \sqrt{10} }{2} } \\ [/tex]

Therefore ,

[tex]\boxed{x = 2.58} \: \: \: \: and \: \: \: \: \boxed{x = - 0.58}[/tex]

_____________________________________

[tex] {x}^{2} + 2x = 1 \\ \implies \: {x}^{2} + 2x - 1 = 0[/tex]

A Here ,

[tex]\boxed{a = 1} \\ \boxed{b = 2} \\ \boxed{c = - 1}[/tex]

B Filling in the values of a , b and c in the Quadratic formula below , we get

[tex]x = \frac{- (b)\pm \sqrt{( {b}^{2}) - 4(a)(c) } }{2(a)} \\ [/tex]

C Simplifying each section , we get

[tex]x = \frac{ - (2) + \sqrt{ ({2}^{2} ) - 4(1)( - 1)} }{2 \times 1} [/tex]

or

[tex]x = \frac{ - (2) - \sqrt{( {2}^{2}) - 4(1)( - 1) } }{2 \times 1} [/tex]

D Simplifying answers from Part C , we get

[tex]\boxed{x = - 1 + \sqrt{2} } \: \: \: \: or \: \: \: \: \boxed{x = - 1 - \sqrt{2} }[/tex]

Therefore

[tex]\boxed{x = 0.41} \: \: \: \: or \: \: \: \: \boxed{x = -2.41 }[/tex]

hope helpful! :)

Answer:

they finished the entire pie

Step-by-step explanation:

a. The general exponential growth equation is given by: y = abˣ

b. The general exponential decay equation is given by: [tex]y = a (1 - r)^x[/tex]

Exponential growth is a type of growth pattern in which a quantity grows at an increasing rate proportional to its current value. This means that the larger the quantity, the faster it grows.

a. The general exponential growth equation is given by:

 y = abˣ

Where y is the final value, 'a' is the initial value, b is the growth factor or base, and x is the time or number of periods.

Exponential decay is a type of decay pattern in which a quantity decreases at a decreasing rate proportional to its current value. This means that the larger the quantity, the slower it decays.

b. The general exponential decay equation is given by:

[tex]y = a (1 - r)^x[/tex]

Where y is the final value, a is the initial value, r is the decay rate, and x is the time or number of periods

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This would give us a modified statement of the limit, which would be "the limit of the rational function as x approaches 4".

A rational function is a type of mathematical function that can be expressed as the ratio of two polynomials. It can be written in the form f(x) = p(x)/q(x), where p(x) and q(x) are polynomials with q(x) not equal to zero. Rational functions are used to model many real-world phenomena, such as the rate of change of a quantity with respect to another. They can also be used to solve complex equations and to analyze the behavior of a system.

The statement of the limit of a rational function can be modified by substituting different values for the variable and determining the resulting limit. For example, if the limit of the rational function is as x approaches 3, then we can substitute x = 4 and determine the resulting limit. This would give us a modified statement of the limit, which would be "the limit of the rational function as x approaches 4".

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The standard matrix for the stated composition of linear operators on R2 is:

The standard matrix for the stated composition of linear operators on R2 can be found by multiplying the matrices for each individual operation.

First, let's find the matrix for a rotation of 270° counterclockwise:

The weighted mean is a type of average that takes into account the relative importance of each data point. In this case, each unit has a different weighting, so we need to take that into account when calculating the weighted mean. Here is how to calculate the weighted mean:

1. Multiply each unit's weighting by its corresponding value.

2. Add up all of the products from step 1.

3. Divide the sum from step 2 by the sum of all the weightings.

Using the chart provided, here is how to calculate the weighted mean for this question:

Unit Weighting Value Product

Trigonometry 25% x 0.25x

Algebra 15% y 0.15y

Statistics 10% z 0.1z

Financial Math 12% a 0.12a

Linear Functions 20% b 0.2b

Quadratic Functions 18% c 0.18c

Weighted mean = (0.25x + 0.15y + 0.1z + 0.12a + 0.2b + 0.18c) / (0.25 + 0.15 + 0.1 + 0.12 + 0.2 + 0.18)

Weighted mean = (0.25x + 0.15y + 0.1z + 0.12a + 0.2b + 0.18c) / 1

Weighted mean = 0.25x + 0.15y + 0.1z + 0.12a + 0.2b + 0.18c

So the weighted mean is a combination of the values of each unit, weighted by their relative importance.

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Given that the polynomial has a zero at 5 with multiplicity 2, its complete factorization is (x - 5)²(z + 9)(z - 2).

To factor the given polynomial z⁴ - 3z³ - 63z² + 355z - 450, given that it has a zero at 5 with multiplicity 2, we can use the fact that (z - 5)² is a factor of the polynomial. We can then use synthetic division to find the other factors.

First, we divide the polynomial by (z - 5) using synthetic division:
5   |   1  -3  -63  355  -450
    |       5    10  -265  450
        1   2  -53    90       0

The result of the division is z³ + 2z² - 53z + 90 . Divide it by (z - 5) again since the multiplicity is 2.
5   |   1   2  -53    90
    |       5    35   -90

        1   7   -18      0

The result of the second division is z² + 7z - 18. This polynomial can still be factorized as (z + 9)(z - 2).

So, the final answer is:

z⁴ - 3z³ - 63z² + 355z - 450 = (x - 5)²(z + 9)(z - 2).

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The remainder when -233 is divided by 11 is 9. To find the remainder when a is divided by b, we can use the formula:

r = a % b

Where % is the modulo operator, which gives the remainder when one number is divided by another.

In this case, we have a = -233 and b = 11. Plugging these values into the formula, we get:

r = -233 % 11

Using a calculator or doing the division by hand, we find that the remainder is -2. However, since we are looking for the positive remainder, we can add b to this value to get the correct answer:

r = -2 + 11 = 9

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Yes, if u1, u2, and u3 are linearly independent, then v1, v2, and v3 will also be linearly independent.  


To show this, assume that v1, v2, and v3 are linearly dependent. This means that there are scalars a,b, and c, such that:

a*v1 + b*v2 + c*v3 = 0

Since v1 = u1 + u2, v2 = u1 + u3, and v3 = u2 + u3, the equation above can be rewritten as:

a*(u1 + u2) + b*(u1 + u3) + c*(u2 + u3) = 0


Simplifying, this gives us:

(a + b + c)*u1 + (a + c)*u2 + (b + c)*u3 = 0


But since u1, u2, and u3 are linearly independent, the coefficients (a + b + c), (a + c), and (b + c) must all be equal to 0. This implies that a = b = c = 0, meaning that the original equation must be equal to 0. This means that v1, v2, and v3 are linearly independent.

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Answer:

for the first, the answers are 1/2, 1, 2, 4, and 8. for the second, 22[tex]\frac{1}{2}[/tex] sq. km.

Step-by-step explanation:

1/4 times 2 is 1/2, times 2 is 1, times 2 is 2, times two is 4, time 2 is 8.

for the second one, area = base times height. 6 3/4 times 3 1/3 is 22 1/2 km squared.

The graph is symmetric about the y-axis, so its a even function.

EVEN function:

ODD function:

Thus, the graph is symmetric about the y-axis, so its a even function.

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Answer: I believe its A.

Step-by-step explanation:

The measure of <TAV is [tex]92^{o}[/tex].

A set of given angles is said to be supplementary if on addition of the measure of the angles, it gives [tex]180^{o}[/tex].

In the given diagram, given that PQ is parallel to RS, it can be observed that;

<SCU ≅ <ACB (definition of vertically opposite angles)

Thus, <ACB = [tex]18^{o}[/tex]

<QAC = <BCA     (alternate angle property)

So that,

<QAC = [tex]18^{o}[/tex]

But,

<TAP ≅ <QAC  (definition of vertically opposite angles)

So that;

<BAP + <PAT + <TAV = [tex]180^{o}[/tex]  (sum of angles on a straight line)

Then,

70 + 18 + <TAV = 180

<TAV = 180 - 88

         = 92

<TAV = [tex]92^{o}[/tex]

The measure of <TAV = [tex]92^{o}[/tex].

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a. The formula for the upper limit (U) is U = 220 - x, where x is the age of the person. The formula for the lower limit (L) is L = 0.65(220 - x).

b. For a 40-year-old person, the lower limit is L = 0.65(220 - 40) = 117 beats per minute and the upper limit is U = 220 - 40 = 180 beats per minute.

c. For a 60-year-old person, the lower limit is L = 0.65(220 - 60) = 104 beats per minute and the upper limit is U = 220 - 60 = 160 beats per minute.

d. Let x be the age of the younger woman and y be the age of the older woman. Since the older woman is 34 years older than the younger woman, we have y = x + 34. Since the older woman is working at the upper limit of her target heart rate zone and the younger woman is working at the lower limit of her target heart rate zone, we have U = L. Substituting the formulas for U and L, we get 220 - y = 0.65(220 - x). Substituting y = x + 34, we get 220 - (x + 34) = 0.65(220 - x). Simplifying and solving for x, we get x = 38. Therefore, the age of the younger woman is approximately 38 years and that of the older woman is approximately 72 years. Their pulse is approximately U = 220 - 72 = 148 beats per minute.

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