Answer:
72x + 288
Step-by-step explanation:
You would multiply 72 with (x + 4), which would leave you with 72x + 288. The value of x would be 72 i think
dance class. She
needs to have $600
save. How much has
she saved so far?
Answer:
$180
Step-by-step explanation:
The image of the point after a reflection over the y-axis is (-2, -1)
From the question, we have the following parameters that can be used in our computation:
Point = (2, -1)
Transformation rule
A reflection over the y-axis
The rule of a reflection over the y-axis is
(x, y) = (-x, y)
Using the above as a guide, we have the following:
Image = (-2, -1)
Hence the image of the point after a reflection over the y-axis is (-2, -1)
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There are no images included, but it would be -2,-1.
²-1
We will investigate the behavior of
both the numerator and denominator of h(x) near the point where x = 1. Let
f(x)= x³ + x -2 and g(x)=x²-1. Find the local linearizations of f and g at a = 1,
and call these functions Lf(x) and Lg(x), respectively.
Lf(x) =
L₂(x) =
Explain why h(x) ≈
Lf(x)
Lg(x)
for a near a = 1.
The local linearizations of f(x) and g(x) at a = 1 are Lf(x) = 4x - 5 and Lg(x) = 2x - 2 respectively. The function h(x) ≈ Lf(x)/Lg(x) because the local linearizations provide a good approximation of the numerator and denominator of h(x) near x = 1.
The local linearization of a function at a given point is an approximation of the function using a linear equation. To find the local linearization of a function f at a = 1, we need to find the slope of the tangent line at a = 1, which is equivalent to finding the derivative of f at x = 1. By taking the derivative of f(x) = x³ + x - 2, we get f'(x) = 3x² + 1. Evaluating f'(1), we find that the slope of the tangent line at a = 1 is 4. Therefore, the local linearization of f at a = 1, denoted as Lf(x), is given by Lf(x) = f(a) + f'(a)(x - a), which becomes Lf(x) = -1 + 4(x - 1) = 4x - 5.
Similarly, to find the local linearization of g(x) = x² - 1 at a = 1, we need to find the slope of the tangent line at a = 1. The derivative of g(x) is g'(x) = 2x. Evaluating g'(1), we find that the slope of the tangent line at a = 1 is 2. Therefore, the local linearization of g at a = 1, denoted as Lg(x), is given by Lg(x) = g(a) + g'(a)(x - a), which becomes Lg(x) = 0 + 2(x - 1) = 2x - 2.
When investigating the behavior of the function h(x) = (f(x))/(g(x)) near the point x = 1, we can approximate h(x) using the local linearizations of f and g at a = 1. Near the point a = 1, h(x) ≈ Lf(x)/Lg(x) because Lf(x) and Lg(x) provide a good approximation of the numerator and denominator, respectively, of h(x). This approximation holds as long as x is close to 1.
Answer:
(-11,-9)
Step-by-step explanation:
translate 10 units down: y value: 1-10=-9
reflect over y-axis: x-value becomes opposite: 11 becomes -11
rolling a multiple of 3
rolling an odd number
rolling a number greater than 6
Find each probability of
P(5)
P(# less than 3)
P(odd)
P(13)
Answer:
1/12 , 1/3 , 1/2 , 1/2
Step-by-step explanation:
P(x) = favorable of x / total
plug in your conditions and you get the probablility
Answer:
1/12 for rolling a 5.
4/12 for multiples of 3. (3,6,9,12)
6/12 for rolling an odd number.(1,3,5,7,9,11)
6/12 for rolling a number greater than 6. (7,8,9,10,11,12)
Step-by-step explanation:
Answer:
Pine board side = 16.4 ft
Steel fencing side = 57.5 ft
Step-by-step explanation:
Let 'B' be the length of each side constructed of pine boards, and 'S' be the length of the side with the steel fencing, the area (A) and cost (C) functions are:
The value of S for which the derivate of the cost function is zero, minimizes cost:
The value of B is:
Pine board side = 16.4 ft
Steel fencing side = 57.5 ft
To minimize the construction costs for the enclosure, the dimensions should be calculated using the calculus optimization technique. By incorporating the cost and area requirements into calculated equations and solving, you will find x = 2 times y. This is how you minimize the cost.
This problem involves the application of calculus and optimization techniques. Given that the area of the enclosure needs to be 945 ft2, and that it is adjacent to an external wall of the department store, we can infer that its shape is rectangular.
Let the width of the enclosure parallel to the department store be x (feet), and its length perpendicular to the store be y (feet). According to the area requirement, we have the equation x*y = 945 ft2.
The cost of the enclosure is the sum of the cost of the pine board fences and the steel fence. Since 2 sides are made of pine boards, and 1 side made of steel, the cost can be expressed as C = 2xy p + y s, where 'p' is the cost of pine board per foot ($7), and 's' is the cost of steel per foot ($4).
Since we are looking for the minimum cost, we derive this equation and set it equal to zero to find the dimensions x and y. After substituting and simplifying, we find that the minimum cost is obtained when x = 2 y. By substituting this into the area equation, we can solve for the dimensions of the enclosure.
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