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nikklg [1K]
2 years ago
10

Need help due in 10 mins

Mathematics
1 answer:
kap26 [50]2 years ago
4 0

Answer:

X = 3

Step-by-step explanation:

3x + 4x - 12 = - 10x + 30 + 9

7x - 12 = - 10x + 39

7x + 10x = 12 + 39

17x = 51

x = 3

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I just need #21 and #23 answered please.
kykrilka [37]
#21: Even
#23: Neither
7 0
3 years ago
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What is the sum?
BaLLatris [955]
4(x + 6)
4 × x = 4x
4 × 6 = 24

3(3x + 4)
3 × 3x = 9x
3 × 4 = 12

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So your answer is B) 13x + 36. I hope this helps!
4 0
3 years ago
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3 teachers each bought a box of markers and 2 notebooks. The box of markers cost $2.50. The total was $12.00. Write an equation
larisa [96]

Answer:

$1.12

Step-by-step explanation:

12÷3=4

1 box of markers is $2.50, Multiply $2.50 by 3 and you get $7.50. 12- $7.50= $4.50. $4.50÷ 2= $2.25. $2.25÷2= $1.12

Not 100% but i'm pretty sure this is the answer. Hope this helps. ( if i'm wrong sorry its 2:30 am)

4 0
2 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
Question 19 please........
MrRa [10]
The answer is B. The shape is staying the exact same, it’s just moving.
7 0
3 years ago
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