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3 years ago

Simplify 12(6b+2). 96b 72b+24 18b+14 2b+6

Mathematics

1 answer:

alekssr [168]3 years ago

7 0

Distributive formula:
a(b+c) = ab + ac

12(6b+2) = 12(6b) + 12(2) = 72b + 24

Your answer is B.

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3 years ago

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3 years ago

PLEASE ANSWER ASAP

Nastasia [14]

Answer:

y = 4 $(2)^{x}$

Step-by-step explanation:

For an exponential function in standard form

y = a $(b)^{x}$

To find the values of a and b, use ordered pairs from the table

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y = 4 $(2)^{x}$ ← exponential function

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3 years ago

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Korvikt [17]

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3 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
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But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

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3 years ago