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3 years ago

PLEASE HURRY!!! 30 POINTS

Mathematics

2 answers:

miskamm [114]3 years ago

8 0

Answer:

(Really really really sorry if this incorrect)... ok so it's 17 cm tall in length. That would be 17 blocks high, for example. The area is 102, so 102 cm2 divided in half equals 51, so the width would be 51 cm. So the perimeter is 17 + 17 + 51 + 51 = 136 cm.

prisoha [69]3 years ago

5 0

102/17 = 6

(6*2)+(17*2) = 46

Hope this helps! :)

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Solve for all values of x by factoring. x^2 + 4x – 30 = -3x

vichka [17]

Answer:

${x}^{2} + 4x - 30 = - 3x \\{x}^{2} + 7x - 30 = 0 \\ {x}^{2} - 3x + 10x - 30 = 0 \\ x(x - 3) + 10(x - 3) = 0 \\ (x - 3)(x + 10) = 0 \\ \boxed{x = 3} \\ \boxed{x = - 10}$

<h3>x=3 or x=-10 is the right answer.</h3>

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2 years ago

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ValentinkaMS [17]

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3 years ago

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3 years ago

Give g(x)=2x+3, find g(-3)

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Answer:

Step-by-step explanation:

g(x)=2x+3

La derivada de un polinomio es la suma de las derivadas de sus términos. La derivada de cualquier término constante es 0. La derivada de ax

es nax

n−1

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3 years ago

Help? I can't seem to understand arithmegic sequence.

katrin2010 [14]

A sequence $\{a_n\}$ is arithmetic if the difference between consecutive terms is some fixed number, regardless of which pair of consecutive terms you pick out of the sequence.

For example, the following sequences are arithmetic:

1, 2, 3, 4, 5, 6, ... (difference = 1)

-25, -20, -15, -10, -5, ... (difference = 5)

2. Carla's sequence is not arithmetic, because the differences between consecutive terms are all different:

13 - 11 = 2

17 - 13 = 4

25 - 17 = 8

She can adjust the sequence by changing the last two numbers to 15 and 17, since this makes the difference fixed:

13 - 11 = 2

15 - 13 = 2

17 - 15 = 2

and so on.

3. The sequence

45, 48, 51, 54, ...

is arithmetic with difference 3 between terms. Recursively, we can write the $n$ th term, $a_n$ , in terms of the previous, $(n-1)$ th term, $a_{n-1}$ :

$a_n=a_{n-1}+3$

By this definition, we can just as easily write the $(n-1)$ th term in terms of the $(n-2)$ th term:

$a_{n-1}=a_{n-2}+3$

Then, substituting this into the previous equation, we have

$a_n=(a_{n-2}+3)+3=a_{n-2}+2\cdot3$

We can continue this process to write $a_n$ in terms of $a_1$ :

$a_{n-2}=a_{n-3}+3\implies a_n=a_{n-3}+3\cdot3$

$a_{n-3}=a_{n-4}+3\implies a_n=a_{n-4}+4\cdot3$

and so on. (You might notice that the subscript of the term on the right side, and the number of 3s being added, together sum to $n$ .) The pattern continues down to

$a_n=a_1+(n-1)\cdot3$