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il63 [147K]
3 years ago
15

The sum of two consecutive even integers is -22 . Find the integers.

Mathematics
2 answers:
Doss [256]3 years ago
6 0
Remember that consecutive means right after the other.
In other words, an example would be:
8 + 10
This is because the next even number after 8 is 10. If we were to use 4, we'd do 4 + 6, and so on.
We're looking for two consecutive even numbers that sum up to a number with a 2 in the ones place.

Let's try 0 + 2 (as they sum up to 2).
Considering that 22 (pretend it's positive) is the sum of two consecutive even numbers, it must be in the range of half of that number.
22 / 2 = 11

So let's try the two even numbers close to 11.
10 and 12 have a sum of 2 in the ones place.
10 + 12 = 22

The two consecutive numbers are -10 and -12.

I hope this helps!
GenaCL600 [577]3 years ago
6 0
1) The sum of two consecutive even integers is -22. Find the integers.
Answer:
(-10) + (-12) = (-22)
Hope this helps. Please mark my answer brainliest so that I can move up to my next rank, and have a nice night.
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3 is a factor of__ A)282 B)187 C)385 D)412​
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Answer:

A. 282

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187:

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2 years ago
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2 years ago
Help ASAP show work please thanksss!!!!
Llana [10]

Answer:

\displaystyle log_\frac{1}{2}(64)=-6

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

log_b(1) = 0

Logarithm of the base:

log_b(b) = 1

Product rule:

log_b(xy) = log_b(x) + log_b(y)

Division rule:

\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)

Power rule:

log_b(x^n) = n\cdot log_b(x)

Change of base:

\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

\displaystyle log_\frac{1}{2}(64)

Factoring 64=2^6.

\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)

Since

\displaystyle 2=(1/2)^{-1}

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})

Applying the logarithm of the base:

\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}

5 0
2 years ago
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