Write an equation to find two consecutive odd integers whose product is 143. (A). x(x+2)+143=0 (B). x(x+1)=143 (C.) x2+2x‒143=0 (D.) x2+x=143

Answer:

500 kilobytes/ 1 min

Step-by-step explanation:

2,500 kilobytes/ 5 min

2,500 divide by 5 = 500 kilobytes

500 kilobytes/ 1 min

from 5 to 0 is a 5 degree change

then from 0 to -2 is a 2 degree change

5+2 = 7 degree change overall

We used De Morgan's laws to find the complement of the set formed by the intersection of the complement of A and B. We first found the complement of A, then the intersection of this set with B, and finally took the complement of this new set.

In set theory, De Morgan's laws relate the intersection and union of sets in a very elegant way. Here, (A' ∩ B)' represents the complement of the set formed by the intersection of the complement of A and B.

First, let's find A'. This is the set of all elements in U that are not in A. U = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } and A = { 0, 1, 4, 7, 9 } So, A' = { 2, 3, 5, 6, 8 }.

Now, let's find the intersection of A' and B, denoted by A' ∩ B. We have B = { 0, 4, 5, 6, 7 } so A' ∩ B = { 5, 6 }.

Finally, let's find the complement of this set, denoted by (A' ∩ B)'. This is the set of all elements in U that are not in A' ∩ B. So, (A' ∩ B)' = { 0, 1, 2, 3, 4, 7, 8, 9 }.

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The area of the region bounded by y = 9x2 ln(x) and y = 36 ln(x) is found by integrating the absolute difference of the two functions from x = 1 to x = 2, which yields ∫ from 1 to 2 (|36 ln(x) - 9x2 ln(x)|) dx.

To find the area of the region bounded by the curves y = 9x2 ln(x) and y = 36 ln(x), you first need to find the points where they intersect. This means setting the two functions equal to each other and solving for x, i.e., 9x2 ln(x) = 36 ln(x).

This simplifies to x2 = 4 which yields two solutions x = 2 and x = -2. However, since the natural logarithm ln(x) is not defined for x < 0, we discard x = -2. So, the two curves intersect at x = 2.

The area between the curves from x = 1 to x = 2 is then obtained by integrating the absolute difference of the two functions from x = 1 to x = 2, which gives:

∫ from 1 to 2 (|36 ln(x) - 9x2 ln(x)|) dx. You can evaluate this integral with standard calculus techniques.

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 20. Let x represent the amount of books he had:
x + 6 = 1.12x

Subtract x from both sides:
6 = 0.12x

Divide both sides by 0.12; x = 50

28. $6 + 0.025*$6 = $6 + $0.15 = $6.15

In this algebraic word problem, equations were set up based on the prices and relationships between the numbers of wrenches bought, leading to the discovery that Mr. Abernathy purchased 12 wrenches each at $1.50 and $2.50, 6 wrenches at $4, and 7 wrenches at $3 to make a total of $78.

Mr. Abernathy's allocation of his $78 budget on wrenches can be formulated as an algebraic word problem where we define the number of wrenches bought at each price point and the total cost. Let us denote x as the number of $1.50 wrenches and also the number of $2.50 wrenches, so the number of $4 wrenches is x/2 since it is half of that number.

The number of $3 wrenches is x/2 + 1 because it is one more than the number of $4 wrenches. The total amount spent can be expressed as the sum of the products of the numbers of wrenches and their respective prices, which must equal $78.

The equation for the total amount spent on wrenches is therefore:

1.50x + 2.50x + 4(x/2) + 3(x/2 + 1) = 78

Solving this equation for x gives us the number of $1.50 and $2.50 wrenches, and then we can infer the quantities of the $4 and $3 wrenches.

After simplifying the equation, we solve for x, and by plugging x back into the previous expressions, we find out that Mr. Abernathy bought 12 wrenches at $1.50 each, 12 wrenches at $2.50 each, 6 wrenches at $4 each, and 7 wrenches at $3 each.

Given:

An open-top box with a square base has a surface area of 100 square inches.

Question:

The Process:

Part-1: The surface area

Let us arrange the equation to get the surface area of ​​the box with a square base. Recall that the box is without a lid and its surface area is 100 square inches.

[tex]\boxed{ \ Surface \ area = area \ of \ base + (4 \times area \ of \ rectangle) \ }[/tex]

[tex]\boxed{ \ x^2 + (4 \cdot x \cdot h) = 100 \ }[/tex]

[tex]\boxed{ \ x^2 + 4xh = 100 \ }[/tex]

From the above equation, we set it again so that "xh" is the subject on the left.

Both sides are subtracted by x².

[tex]\boxed{ \ 4xh = 100 - x^2 \ }[/tex]

Both sides are divided by 4.

[tex]\boxed{ \ xh = \frac{100 - x^2}{4} \ }[/tex] ... (Equation-1)

That is the strategy we have prepared.

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Part-2: The volume

[tex]\boxed{ \ Volume \ of \ the \ box = length \times width \times height \ }[/tex]

[tex]\boxed{ \ Volume = x \cdot x \cdot h \ }[/tex]

[tex]\boxed{ \ Volume = x^2h \ }[/tex] ...(Equation-2)

Substitution Equation-1 into Equation-2.

[tex]\boxed{ \ Volume = x(xh) \ }[/tex]

[tex]\boxed{ \ Volume = x \bigg( \frac{100 - x^2}{4} \bigg) \ }[/tex]

[tex]\boxed{ \ Volume = \frac{100x - x^3}{4} \ }[/tex]

[tex]\boxed{ \ Volume = 25x - \frac{1}{4}x^3 \ }[/tex]

Thus, an expression of the volume of the box as a function of the length of the edge of the base is [tex]\boxed{\boxed{ \ Volume = 25x - \frac{1}{4}x^3 \ }}[/tex]

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Part-3: The domain of volume

The value of volume must always be positive, i.e., V > 0.

[tex]\boxed{ \ 25x - \frac{1}{4}x^3 > 0 \ }[/tex]

Both sides are multiplied by 4.

[tex]\boxed{ \ 100x - x^3 > 0 \ }[/tex]

Both sides are multiplied by -1, notice the change in the sign of the inequality.

[tex]\boxed{ \ x^3 - 100x < 0 \ }[/tex]

[tex]\boxed{ \ x(x^2 - 100) < 0 \ }[/tex]

[tex]\boxed{ \ x(x - 10)(x + 10) < 0 \ }[/tex]

We get [tex]\boxed{ \ x = 0, \ x = 10, \ and \ x = - 10 \ }[/tex].

Consider the test of signs:

x(x - 10) (x + 10) is negative to the left of x = 10, and positive to the right of x = 10 on the number line.

Examples of tests:

Remember this form above, [tex]\boxed{\ x(x - 10)(x + 10) < 0 \ }[/tex], the value of the test result must be negative (because < 0).

Thus, the domain of the volume is [tex]\boxed{ \ 0 < x < 10 \ }[/tex] or [tex]\boxed{ \ (0, 10) \ }[/tex]

Keywords: an open-top box, with, a square base, has a surface area, 100 square inches, express, the volume, as a function, the length, edge, base, what, its domain, test of signs, substitution

The large rectangle's dimensions are 3 times the dimensions of the small rectangle. If the small rectangle's dimensions are W for width and L for length, then the large rectangle's dimensions are 3W and 3L.

The subject of this question is Mathematics, specifically relating to geometry and ratios. You're asked to compare two rectangles: a large one and a small one. The large rectangle's dimensions (both the width and length) are three times the dimensions of the small rectangle. Let's assume the dimensions of the small rectangle are W (width) and L (length). Then, the dimensions of the large rectangle would be 3W and 3L, respectively. This is because the large rectangle is three times the size of the small rectangle in both width and length.

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Answer:

x=6

Step-by-step explanation:

Answer:

The slope is -3/2 and contains the point (0, −2). Which is b.

Step-by-step explanation:

(-3,-4) (3, 0)

slope = (0 +4) / (3 + 3) = 4/6 = 2/3

perpendicular so slope = -3/2

line on graph

y = mx + b

0 = 2/3 (3) + b

0 = 2+ b

b = -2

y intercept (0,-2)