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Oliga [24]
3 years ago
9

The sum of two numbers is 33. Their difference is 23. Find the numbers.

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer: The numbers are 28 and 5

Step-by-step explanation: The question gives a set of clues. In the first instance, the addition of two numbers equals 33. Let us assume the two numbers are A and B. What that means is that

A + B = 33.

We shall call this equation 1.

The other clue is given as “their difference is 23.” This can be expressed as

A - B = 23.

We shall call this equation 2.

Now we have a pair of simultaneous equations as follows

A + B = 33 —————-(1)

A - B = 23 —————-(2)

We shall use the substitution method

From equation (1), if we make A the subject of the equation we shall move B to the other side and we’ll have

A = 33 - B

Substitute for the value of A in equation (2)

A - B = 23 now becomes

(33 - B) - B = 23

33 - B - B = 23

33 - 2B = 23

By collecting like terms we now have 33 - 23 = 2B

(Remember that when a positive value crosses to the other side of an equation, it becomes negative, and vice versa)

33 - 23 = 2B

10 = 2B

Divide both sides of the equation by 2

5 = B

If we calculate B as 5

We can substitute for this value into equation 1, which is

A + B = 33

A + 5 = 33

Subtract 5 from both sides of the equation

A + 5 - 5 = 33 - 5

A = 28.

Therefore the numbers are 28 and 5.

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7 0
3 years ago
Determine the formula for the nth term of the sequence:<br>-2,1,7,25,79,...​
rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

x_n = a n^4 + b n^3 + c n^2 + d n + e

From the given known values of the sequence, we have

\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}

Solving the system yields coefficients

a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4

so that the n-th term in the sequence might be

\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}

Then the next few terms in the sequence could very well be

\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of \{x_n\} eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of \{x_n\} by \Delta^{k}\{x_n\}. Then

• 1st-order differences:

\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}

• 2nd-order differences:

\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}

• 3rd-order differences:

\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}

• 4th-order differences:

\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}

From here I made the assumption that \Delta^4\{x_n\} is the constant sequence {15, 15, 15, …}. This implies \Delta^3\{x_n\} forms an arithmetic/linear sequence, which implies \Delta^2\{x_n\} forms a quadratic sequence, and so on up \{x_n\} forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.

5 0
2 years ago
Does anyone know what the code is Ive been stuck on this for like ever.
timama [110]

Answer:

Binary code

Step-by-step explanation:

Code is a word or a symbol which is not disclosed to everyone and is considered as secretive. There are various codes which are required to run a software in order to minimize the risk of piracy. These codes are binary codes or java codes. These are complex so that not everyone can crack it.

3 0
3 years ago
Which is true about characterization?
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3 years ago
What is the value of x in the equation 2(x – 3) + 9 = 3(x + 1) + x?
Lana71 [14]

Answer:

Given the equation: 2(x-3)+9=3(x+1)+x

Distributive property of multiplication states that when a number is multiplied by the sum of the two numbers, the first number can be distributed to both of those numbers and multiplied by each of them separately i.e,

a \cdot (b+c) = a\cdot b+a\cdot c

Now, using distributive property to LHS and RHS, we get

2x-6+9=3x+3+x

Combine like terms on RHS;

2x-6+9 = 4x+3

or

2x+3 = 4x+3

Subtract 3 from both sides of an equation;

2x+3-3 =4x+3-3

Simplify:

2x = 4x

or

4x-2x =0

2x = 0

⇒ x =0

Therefore, the value of x in the given equation

2(x-3)+9=3(x+1)+x is 0.




8 0
4 years ago
Read 2 more answers
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