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3 years ago

The sum of three consecutive multiples of 4 is 444. Find these multiples.

2 answers:

3 0

First multiple of 4 is x = 144

Second multiple of 4 is x + 4 = 144 + 4 = 148

Third multiple of 4 is x + 8 = 144 + 8 = 152

Archy [21]3 years ago

3 0

Step-by-step explanation:

Let the first multiple of 4 be x and second multiple be (x + 4) and third multiple be (x + 8) ...[Since, the number are consecutive multiple of 4]

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}\\$

$:\implies \sf x \: + \: x \: + \: 4 \: + \: x \: + \: 8 \: = \: 444 \\ \\ \\$

$:\implies \sf 3x \: + \: 12 \: = \: 444 \\ \\ \\$

$:\implies \sf 3x \: = \: 444 \: - \: 12 \\ \\ \\$

$:\implies \sf 3x \: = \: 432 \\ \\ \\$

$:\implies \sf x \: = \: \dfrac{432}{3} \\ \\ \\$

$:\implies\underline{ \boxed{\sf x \: = \: 144}} \\$

<h3>Therefore,</h3>

$\bullet\:\:\textsf{First multiple of 4 = x = \textbf{144}} \\$

$\bullet\:\:\textsf{Second multiple of 4 = x + 4 = 144 + 4 = \textbf{148}} \\$

$\bullet\:\:\textsf{Third multiple of 4 = x + 8 = 144 + 8 = \textbf{152}}$

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3 years ago

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Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

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3 years ago

A company produces a set of blocks. It costs them $5 to make. They mark it up 300% when they sell it to a toy distributor. The t

Nostrana [21]

Hello! So, the toy company makes blocks that cost $5 to make. Then, they mark it up by 300%, which is marking the price up by 4. 100% markup is doubling the price, 200% is tripling the price, and 300% marks the price up by 4. Then, the blocks would cost $20. After that, the price would be marked up by 150%, so that’s 2.5 times the price of the original. When you do $20 by 2.5, then the blocks cost $50. The local toy store will mark the block up by 200%, which as said before, is tripling the price. Then, you do 50 * 3, and then the price of the blocks is $150. Then, it gets marked off by 30% and 10% of 150 is 15, so 30% off 150 is 45. When you do 150 - 45, the difference is $105. Then, you take off an additional 15%, But you would still have to pay 85% of the original price. You would do 105 * 0.85, and then you would get $89.25 for the blocks. But that’s not all. You have to pay 8% sales tax for the blocks. You can find the tax by doing 89.25 * 0.08 and then adding the prices together. Or, you can do 89.25 * 1.08 to find the total price. Either way, the product should be 96.39. The total price you paid for the blocks is $96.39.

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3 years ago

The function g(x) is a transformation of the quadratic parent function, f(x)=x2. What function is g(x)?

larisa86 [58]

Answer:

Step-by-step explanation:

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3 0

2 years ago