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1 year ago

2+2 I will give brainlyest

Mathematics

2 answers:

AleksandrR [38]1 year ago

8 0

Answer:

Step-by-step explanation:

its super simple, i know u can do it!

Dafna11 [192]1 year ago

6 0

Answer: 2 + 2 = 4

Step-by-step explanation: imagine not knowing this.

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A box of matching game pieces contain a red pieces five green pieces seven yellow pieces you reach in and randomly take out one

Anit [1.1K]

Answer:

5/156

Step-by-step explanation:

First, how many pieces?

1 red+ 5 green + 7 yellow = 13 pieces

We draw a piece

The probability it is green

P(green) = number of green/ total

= 5/13

Now there are only 12 pieces left

Probability the second piece drawn is red

P(red with 1 piece removed) = number of red/ total left

= 1/12

P(green, then red) = P(green) * P(red with 1 piece removed)

= 5/13 *1/12

=5/156

3 0

3 years ago

Solve for the variable -x=-12

gayaneshka [121]

Answer:

x=12

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Sarah is making a cake using a recipe that requires

Alex73 [517]

Answer:

4/1

because 3+1 is 4 then put 1 so

3 0

2 years ago

I am confused help plz

Wewaii [24]

Answer:

Step-by-step explanation:

a 6(m+3)

b expression

equation

5 0

3 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