A) Give the line whose slope is m=4m=4 and intercept is 10.The appropriate linear function is y=
B) Give the line whose slope is m=3 and passes through the point (8,−1).The appropriate linear function is y=

Answers

Answer 1

The slope is m = 4 and the y-intercept is 10, so the linear function becomes:y = 4x + 10 and the appropriate linear function is y = 3x - 25.

A) To find the linear function with a slope of m = 4 and y-intercept of 10, we can use the slope-intercept form of a linear equation, y = mx + b, where m is the slope and b is the y-intercept.

In this case, the slope is m = 4 and the y-intercept is 10, so the linear function becomes:

y = 4x + 10

B) To find the linear function with a slope of m = 3 and passing through the point (8, -1), we can use the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

In this case, the slope is m = 3 and the point (x1, y1) = (8, -1), so the linear function becomes:

y - (-1) = 3(x - 8)

y + 1 = 3(x - 8)

y + 1 = 3x - 24

y = 3x - 25

Therefore, the appropriate linear function is y = 3x - 25.

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Answer 2

A)  The y-intercept of 10 indicates that the line intersects the y-axis at the point (0, 10), where the value of y is 10 when x is 0.

The line with slope m = 4 and y-intercept of 10 can be represented by the linear function y = 4x + 10.

This means that for any given value of x, the corresponding y-value on the line can be found by multiplying x by 4 and adding 10. The slope of 4 indicates that for every increase of 1 in x, the y-value increases by 4 units.

B) When x is 8, the value of y is -1.

To find the equation of the line with slope m = 3 passing through the point (8, -1), we can use the point-slope form of a linear equation, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line.

Plugging in the values, we have y - (-1) = 3(x - 8), which simplifies to y + 1 = 3x - 24. Rearranging the equation gives y = 3x - 25. Therefore, the appropriate linear function is y = 3x - 25. This means that for any given value of x, the corresponding y-value on the line can be found by multiplying x by 3 and subtracting 25. The slope of 3 indicates that for every increase of 1 in x, the y-value increases by 3 units. The line passes through the point (8, -1), which means that when x is 8, the value of y is -1.

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Related Questions

For the given function, find (a) the equation of the secant line through the points where x has the given values and (b) the equation of the tangent line when x has the first value. y=f(x)=x^2+x;x=−4,x=−1

Answers

The equation of the tangent line passing through the point (-4, 12) with slope -7: y = -7x - 16.

We are given the function: y = f(x) = x² + x and two values of x:

x₁ = -4 and x₂ = -1.

We are required to find:(a) the equation of the secant line through the points where x has the given values (b) the equation of the tangent line when x has the first value (i.e., x = -4).

a) Equation of secant line passing through points (-4, f(-4)) and (-1, f(-1))

Let's first find the values of y at these two points:

When x = -4,

y = f(-4) = (-4)² + (-4)

= 16 - 4

= 12

When x = -1,

y = f(-1) = (-1)² + (-1)

= 1 - 1

= 0

Therefore, the two points are (-4, 12) and (-1, 0).

Now, we can use the slope formula to find the slope of the secant line through these points:

m = (y₂ - y₁) / (x₂ - x₁)

= (0 - 12) / (-1 - (-4))

= -4

The slope of the secant line is -4.

Let's use the point-slope form of the line to write the equation of the secant line passing through these two points:

y - y₁ = m(x - x₁)

y - 12 = -4(x + 4)

y - 12 = -4x - 16

y = -4x - 4

b) Equation of the tangent line when x = -4

To find the equation of the tangent line when x = -4, we need to find the slope of the tangent line at x = -4 and a point on the tangent line.

Let's first find the slope of the tangent line at x = -4.

To do that, we need to find the derivative of the function:

y = f(x) = x² + x

(dy/dx) = 2x + 1

At x = -4, the slope of the tangent line is:

dy/dx|_(x=-4)

= 2(-4) + 1

= -7

The slope of the tangent line is -7.

To find a point on the tangent line, we need to use the point (-4, f(-4)) = (-4, 12) that we found earlier.

Let's use the point-slope form of the line to find the equation of the tangent line passing through the point (-4, 12) with slope -7:

y - y₁ = m(x - x₁)

y - 12 = -7(x + 4)

y - 12 = -7x - 28

y = -7x - 16

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a) perform a linear search by hand for the array [20,−20,10,0,15], loching for 0 , and showing each iteration one line at a time b) perform a binary search by hand fo the array [20,0,10,15,20], looking for 0 , and showing each iteration one line at a time c) perform a bubble surt by hand for the array [20,−20,10,0,15], shouing each iteration one line at a time d) perform a selection sort by hand for the array [20,−20,10,0,15], showing eah iteration one line at a time

Answers

In the linear search, the array [20, -20, 10, 0, 15] is iterated sequentially until the element 0 is found, The binary search for the array [20, 0, 10, 15, 20] finds the element 0 by dividing the search space in half at each iteration, The bubble sort iteratively swaps adjacent elements until the array [20, -20, 10, 0, 15] is sorted in ascending order and The selection sort swaps the smallest unsorted element with the first unsorted element, resulting in the sorted array [20, -20, 10, 0, 15].

The array is now sorted: [-20, 0, 10, 15, 20]

a) Linear Search for 0 in the array [20, -20, 10, 0, 15]:

Iteration 1: Compare 20 with 0. Not a match.

Iteration 2: Compare -20 with 0. Not a match.

Iteration 3: Compare 10 with 0. Not a match.

Iteration 4: Compare 0 with 0. Match found! Exit the search.

b) Binary Search for 0 in the sorted array [0, 10, 15, 20, 20]:

Iteration 1: Compare middle element 15 with 0. 0 is smaller, so search the left half.

Iteration 2: Compare middle element 10 with 0. 0 is smaller, so search the left half.

Iteration 3: Compare middle element 0 with 0. Match found! Exit the search.

c) Bubble Sort for the array [20, -20, 10, 0, 15]:

Iteration 1: Compare 20 and -20. Swap them: [-20, 20, 10, 0, 15]

Iteration 2: Compare 20 and 10. No swap needed: [-20, 10, 20, 0, 15]

Iteration 3: Compare 20 and 0. Swap them: [-20, 10, 0, 20, 15]

Iteration 4: Compare 20 and 15. No swap needed: [-20, 10, 0, 15, 20]

The array is now sorted: [-20, 10, 0, 15, 20]

d) Selection Sort for the array [20, -20, 10, 0, 15]:

Iteration 1: Find the minimum element, -20, and swap it with the first element: [-20, 20, 10, 0, 15]

Iteration 2: Find the minimum element, 0, and swap it with the second element: [-20, 0, 10, 20, 15]

Iteration 3: Find the minimum element, 10, and swap it with the third element: [-20, 0, 10, 20, 15]

Iteration 4: Find the minimum element, 15, and swap it with the fourth element: [-20, 0, 10, 15, 20]

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