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3 years ago

I really need help with this question as fast as possible

Mathematics

2 answers:

valentinak56 [21]3 years ago

3 0

Answer:

-1625

Step-by-step explanation:

→ (-125) × 5 + (-125) × 8

Taking (-125) as in common,

→ (-125) × (8 + 5)

→ (-125) × 13

→ -1625

Stolb23 [73]3 years ago

3 0

The answer should be -1625

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ziro4ka [17]

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Step-by-step explanation:

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3 years ago

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2 years ago

Please solve thank you

Y_Kistochka [10]

9514 1404 393

Answer:

Step-by-step explanation:

The second step of the solution shows division by 2 on the left. Consequently, it must also show division by 2 on the right.

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3 years ago

The sum of the sequence of 685+678+671+664+...+6

lukranit [14]

Answer:

Sum of the sequence (Sn) = 33,859

Step-by-step explanation:

Given:

Sequence = 685+678+671+664+...+6

Find:

Sum of the sequence (Sn)

Computation:

a = 685

d = 678 - 985 = -7

an = 6

an = a+(n-1)d

6 = 685+(n-1)(-7)

-679 = (n-1)(-7)

97 = n-1

n = 98

So,

Sum of the sequence (Sn) = (n/2)[a+an]

Sum of the sequence (Sn) = (98/2)[685+6]

Sum of the sequence (Sn) = (49)(691)

Sum of the sequence (Sn) = 33,859

3 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

I really need help with this question as fast as possible​

I really need help with this question as fast as possible