Kyle is buying gifts for Megan's birthday and needs to stay in budget

Cake = half his money + 36.50
Decorations = half of what he had left + 36.50
Sweets = half of what he had left + 18.25

Now he is out of money, what did he start with?

To get a 1 in matrix row 1, column 1 all we have to do is divide the row by 7 which gives us:

[tex]\left[\begin{array}{ccc} 1\frac {-4}{7}&\frac{8}{7} \\-12&7&-13\end{array}\right][/tex]

The solution provided above uses a method of matrix called Echelon.

Echelon matrix is a matrix that has been transformed using row operations into a specific form that makes it easier to solve systems of linear equations.

By applying these row operations, a matrix can be transformed into row echelon form, which is a weaker form of echelon form where the leading non-zero entry of each row is to the right of the leading non-zero entry of the row above it, but not necessarily strictly to the right.

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Answer:

h=.34

Step-by-step explanation:

1/2(bh)=A

1/2(5.5*h)=8.9375

5.5h=1.875

h=.34

Answer:

  see below

Step-by-step explanation:

Given a $3500 net income per month and a list of budget percentages, you want to know the budget amounts for each item, and the amount left over.

Each budget amount is the product of the net income and the associated budget percentage. Those products are ...

The sum of these amounts is $3500, so there will be $0 left over.

__

Additional comment

The list is stored in the variable 'b' by the first line in the calculator. Then these values are multiplied by the net income to get the budget amounts.

The Taylor polynomial t3(x) for f(x) = cos(x) centered at a = pi/2 is t3(x) = -(x-pi/2) + (x-pi/2)³/3!.

A polynomial is  described as an expression consisting of variables and coefficients, which involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents.

The Taylor polynomial of degree n for a function f(x) centered at a is given by the formula:

We get the derivatives :

pn(x)=f(c)+f′(c)(x−c)+f′′(c)2!....

f(x) = cos(x)

f'(x) = -sin(x)

f''(x) = -cos(x)

f'''(x) = sin(x)

We the  evaluate these derivatives at x = pi/2:

f(pi/2) = cos(pi/2) = 0

f'(pi/2) = -sin(pi/2) = -1

f''(pi/2) = -cos(pi/2) = 0

f'''(pi/2) = sin(pi/2) = 1

We substitute  these values into the formula for the Taylor polynomial, and simplify to get Taylor polynomial T3(x) for the function:

the Taylor polynomial t3(x) for f(x) = cos(x) centered at a = pi/2 is t3(x) = -(x-pi/2) + (x-pi/2)³/3!.

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The exact values of sinθ, tanθ, secθ, cscθ, and cotθ are -√8/9, √8/3, -3, -√9/8, and √9/8, respectively.

Given: cosθ = -1/3 and θ is in quadrant III.

This means that sinθ = ± √(1 - (cosθ)2).

We know that cosθ is negative, so sinθ will be negative.

Therefore, sinθ = -√(1 - (-1/3)2 )

sinθ = - √(1 - 1/9)

sinθ = -√8/9

tanθ = sinθ/cosθ

tanθ = (-√8/9)/-1/3

tanθ = √8/3

secθ = 1/cosθ

secθ = 1/-1/3

secθ = -3

cscθ = 1/sinθ

cscθ = 1/-√8/9

cscθ = -√9/8

cotθ = cosθ/sinθ

cotθ = -1/3/-√8/9

cotθ = √9/8

Therefore, the exact values of sinθ, tanθ, secθ, cscθ, and cotθ are -√8/9, √8/3, -3, -√9/8, and √9/8, respectively.

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The solution to each of the given polynomials are:

1)  x² + 3x - 4

2) x² - 4x + 4

3) x² - 7x + 8

4) x² - 13 - (15/(x + 3)

1) We want to divide the polynomial. Thus:

          x² + 3x - 4

x + 2| x³ + 5x² + 2x - 8

     -   x³ + 2x²

                 3x² + 2x

               - 3x² + 6x

                        - 4x - 8

                      -  -4x - 8

                                0

2) We want to divide the polynomial. Thus:

        x² - 4x + 4

x - 2| x³ - 6x² + 12x - 8

     -  x³ - 2x²

              - 4x² + 12x

              - -4x² + 8x

                         4x - 8

                      -  4x - 8

                                0

3) We want to divide the polynomial. Thus:

         x² - 7x + 8

x + 3| x³ - 4x² - 13x + 24

      -  x³ + 3x²

              - 7x² - 13x

              - -7x² - 21x

                         8x + 24

                      -  8x + 24

                                0

4) We want to divide the polynomial. Thus:

        x² - 13

x + 3| x³ + 3x² - 13x - 15

      -  x³ + 3x²

              - 0x² - 13x

              - -7x² - 13x

                       0 - 15

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The simplest form of log5+log6-log2​ is:  log15.

Let us simplify log5 + log6 - log2 by making use  logarithmic rules:

log a + log b = log (ab)

log a - log b = log (a/b)

So,

log5 + log6 - log2

= log (5 x 6) - log2

= log30 - log2

Now let make use of the  logarithmic rule:

log a - log b = log (a/b) = log a - log b

log30 - log2

= log (30/2)

= log15

Therefore the simplest form is log15.

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The equation of the line parallel to the graph of 2x - 3y = 7 and containing the point (-3,-3) is y = (2/3)x - 1.

The formula for equation of line is expressed as;

y = mx + b

Where m is slope and b is y-intercept.

Given the equation of the graph: 2x - 3y = 7.

First, we need to rearrange the given equation into slope-intercept form (y = mx + b) by solving for y:

2x - 3y = 7

-3y = -2x + 7

y = (2/3)x - 7/3

The slope of the given line is 2/3.

Hence, any line parallel to it will also have a slope of 2/3.

Using the point-slope form, we determine the equation of the line that passes through (-3,-3) and has a slope of 2/3:

y - y1 = m(x - x1)

y - (-3) = (2/3)(x - (-3))

y + 3 = (2/3)x + 2

y = (2/3)x - 1

Therefore, the equation of the line is y = (2/3)x - 1.

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The feasible region on the attached graph of the inequalities created with MS Excel indicates that the maximum value of g = 8·x + 12·y is 100

The feasible region is the region on the graph of the inequalities that satisfies the specified constraints.

The specified equations obtained from the constraints in slope intercept form are;

y = -0.5·x + 7; The region below the graph is shaded

y = -0.75·x + 9; The region below the graph is shaded

x = 0; The region to the right of the line x = 0 is shaded

y = 0; The region above the line y = 0 is shaded

The point of intersection of the graphs can be found from the equation; -0.5·x + 7 = -0.75·x + 9

Which indicates that at the intersection, x = 8, and y = -0.5×8 + 7 = 3

The corner points are; (0, 0), (0, 7), (8, 3), (12, 0)

The value of the function g for each of the feasible region are therefore;

g(0, 0) = 8 × 0 + 12 × 0 = 0

g(0, 7) = 8 × 0 + 12 × 7 = 84

g(8, 3) = 8 × 8 + 12 × 3 = 100

g(12, 0) = 8 × 12 + 12 × 0 = 96

Please find attached the graph of the inequalities showing the feasible region created with MS Excel

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A. Using the formula for area of rectangle, the length of the rectangular space is: 2x - 3

B. The expression to prove this is: (2x - 3) * 4x = 8x² - 12x

To find the area of a rectangle, multiply the width and the length together. This means: Area = length * width.

Part A: We are given that area of the rectangular space is expressed as 8x² - 12x, if the width is 4x, then:

Length of the rectangular space = (8x² - 12x) / 4x = 4x(2x - 3)/4x

Length = 2x - 3

Part B: To prove this, we have:

length * width = 8x² - 12x

Substitute:

(2x - 3) * 4x = 8x² - 12x

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The height of the school building is approximately 4.38 meters (rounded to the nearest hundredth).

JX / BX = JM / BT

Substituting the known values, we get:

(d1 + d2) / h = d1 / (h - 1.75)

(d1 + d2) * (h - 1.75) = d1 * h

d1 * h + d2 * h - 1.75 * d1 - 1.75 * d2 = d1 * h

d2 * h - 1.75 * d1 - 1.75 * d2 = 0

h = (1.75 * d1 + 1.75 * d2) / d2

h = (1.75 * 8.75) / 3.5

h = 4.375

The height of the school building is approximately 4.38 meters

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It will take approximately 12.85 hours until only 1 gram of the element remains.

If after 2 hours, 36% of the initial amount remains, then the remaining amount is 0.36 times 100 grams, which is 36 grams. Let's define the amount of the element remaining after time t in hours as R(t). Since the element is decaying exponentially, we can model it with the equation R(t) = 100[tex](0.36)^t[/tex].

We want to find how many hours it will take until only 1 gram remains. We can set R(t) equal to 1 gram and solve for t:

1 = 100[tex](0.36)^t[/tex]

Taking the logarithm of both sides, we get:

ln(1) = ln(100[tex](0.36)^t[/tex])

0 = ln(100) + t*ln(0.36)

t = -ln(100)/ln(0.36)

Using a calculator, we find that t ≈ 12.85 hours.

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My own multi-step combination problem is given below:

To solve this problem, Amanda  need to use the formula for combinations and it is:

nCr = n! / (r! x (n-r)!)

where:

n = total number of items to select from

r is the number of items to select.

First, we have to calculate the number of ways to select 3 fruit from 5, hence it will be:

5C3

=  5! / (3! x (5-3)!)

= 10

Next, we have to calculate the number of ways to select 2 meatpie from 4 and it will be

4C2

= 4! / (2! x  (4-2)!)

= 6

Lastly,, we need to calculate the number of ways to select 2 desserts from 3 and it will be:

3C2

= 3! / (2! x  (3-2)!)

= 3

To have the total number of dinner menus, we have to multiply these three numbers together:

= 10 x 6 x 3

= 180

Therefore, one can say that Amanda have 180 different dinner menus that she can be create.

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The equation of the line passing through the points (6, 5) and (8, -2) is[tex]y = -\frac{7}{2} x + 26[/tex].

The formula for equation of line is expressed as;

y = mx + b

Where m is slope and b is y-intercept.

Determine the slope of the line.

Given the two points are (-7, -3) and (1, 2)

We can find the slope of the line by using the slope formula:

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

Substituting the values:

m = (-2 - 5) / (8 - 6)

m = -7/2

Using the point-slope form, plug in one of the given points and slope m = -7/2 to find the equation of the line.

Let's use the point (6, 5):

y - y₁ = m(x - x₁)

[tex]y - 5 = -\frac{7}{2}( x - 6 )\\\\y - 5 = -\frac{7}{2} x + 21\\\\y = -\frac{7}{2} x + 21 + 5\\\\y = -\frac{7}{2} x + 26[/tex]

Therefore, the equarion of the line is [tex]y = -\frac{7}{2} x + 26[/tex].

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Check the picture below.

Some of the common order of operation mistakes that students make when evaluating expressions are the use of basic mathematical operations.

You should know that an error or fault results from defective judgment, deficient knowledge, or carelessness.

A combination of numbers, letters, and relational operators containing at least one operation is deemed a mathematical expression. Additionally, it must also include variables and/or numbers, excluding an equal sign.

Some prevalent errors include using both a positive and a negative sign, using two negative signs, and misusing brackets.

By being unable to comprehend the problems accurately, learners often commit significant errors while assessing formulations, which underscores the importance of employing fundamental mathematical operations.

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The systems of linear equations can be classified as "consistent dependent," "consistent independent," or "inconsistent" as follows:

A. System A = Inconsistent (No solution)

B. System B = Consistent dependent (Indefinitely many solutions)

C. System C = Consistent independent (A unique solution)

To identify the systems of linear equations, you first need to note the meaning of the different systems of equations. First, the Consistent dependent has two equations that represent the same line, so they intersect at infinitely many points and any point on the line represents a solution to the system.

Next, the Consistent independent has the two equations represented by two distinct non-parallel lines, so they intersect at exactly one point, which is the unique solution to the system.

Finally, the Inconsistent has two equations that are represented by two distinct parallel lines, so they do not intersect, and there is no solution to the system.

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Answer: $3120

Step-by-step explanation:

3500-3240

=260

260*12=3120

Using some logarithm properties we will get the expanded expression:

log(y^10) = 10*ln(y)/ln(10)

So there are two logarithm properties that we can use here, these are:

logₙ(x) = ln(x)/ln(n)

and:

logₙ(x^a) = a*logⁿ(x)

Now let's go to our expression, it is:

log(y^10)

Using the second property we can remove the exponent, then we will get:

log(y^10) = 10*log(y)

Now we can use the first rule, remember that log(x) has a base of 10, then we can write:

10*log(y) = 10*ln(y)/ln(10)

That is the expession expanded.

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Answer:

Step-by-step explanation:

No

Answer:

The answer is A.

Step-by-step explanation:

For this graph, you are using the linear equation formula: y = mx + b

__________________________________

__________________________________

Answer:

-------------------------

Given a rectangle with dimensions of:

Use area equation and find the value of x:

The second root is ignored as being negative.

The dimensions of the rectangle are:

The equation that models the balance y after t months is y = -75t + 1,500

To write an equation that models the balance y after t months, we need to use the given information to determine the rate at which the balance is decreasing. We can use the formula for a straight line, y = mx + b, where m is the slope and b is the y-intercept.

We can start by finding the change in the balance over the three-month period from four to seven months after the purchase:

Change in balance = $1,200 - $975 = $225

The rate of change can be calculated by dividing the change in the balance by the number of months:

Rate of change = $225 / 3 months = $75 per month

We can use this rate of change as the slope of the equation:

y = -75t + b

To find the y-intercept b, we can use the fact that the balance was $1,200 four months after the purchase:

1,200 = -75(4) + b

b = 1,500

Therefore, the equation that models the balance y after t months is:

y = -75t + 1,500

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Answer:

A!!!!!!!!!!!!!!!!!!!!!!!!!

Answer:

4mn + 3m + 8n + 6 = (m + 2)(4n + 3)

The missing term is 4n.

A graph of the parabola with five points on the parabola is shown below.

In Mathematics and Geometry, the vertex form of a quadratic equation is given by this formula:

y = a(x - h)² + k

Where:

For the given quadratic function, we have;

y = x² - 2x + 4

When x = 0, we have:

y(0) = 0² - 2(0) + 4

y(0) = 4

When x = 2, we have:

y(2) = 2² - 2(2) + 4

y(2) = 4

When x = -2, we have:

y(-2) = -2² - 2(-2) + 4

y(-2) = 12

When x = 4, we have:

y(4) = 4² - 2(4) + 4

y(4) = 12

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