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netineya [11]
3 years ago
10

You randomly draw a marble from a bag of marbles that contains 888 blue marbles, 555 green marbles, and 888 red marbles.

Mathematics
1 answer:
ella [17]3 years ago
5 0

Answer:

whats the question?

Step-by-step explanation:

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17+p=7p-13 Plz help me asap
Hunter-Best [27]

First you had to subtract by seventeen from both sides of equation form.

17+p-17=7p-13-17

Then simplify.

p=7p-30

Next, subtract by seven-p from both sides of equation form.

p-7p=7p-30-7p

Simplify.

-6p=-30

Then you divide by negative six from both sides of equation form.

\frac{-6p}{-6}= \frac{-30}{-6}

And finally, simplify by equation.

-30/-6=5

Final answer: \boxed{p=5}

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

-Charlie


4 0
3 years ago
A water cooler holds 15 liters of sports drink. Approximately how many gallons is this
Nadya [2.5K]
The correct Anwser it is about 4 gallons because in 1 gallon tgere is 3.75 liters
7 0
3 years ago
PLEASE HELP ME I WILL MARK YOU BRAINLIEST FOR THIS ANY ANSWERS!!!)Mr. Perez designs a probability model to help him predict whic
Nesterboy [21]

Answer:

Number A. is True. B. is False. C. is IDK D. is False. E. is true.

Step-by-step explanation:

If you read it you can tell plus I broke it down and solved it on paper.

6 0
3 years ago
Read 2 more answers
All of the following expressions have the same value, when x = -2 and y = 4, except _____.
irinina [24]

Answer:

C

Step-by-step explanation:

When plugged in it is -4(-2)² which equals -16 while the others equal 16

4 0
4 years ago
Read 2 more answers
Please find the derivative of that function...
vodka [1.7K]
You could apply a clever Algebra trick to avoid using the quotient rule,

\rm y=\dfrac{log(x)-1+1}{log(x)-1}=\dfrac{log(x)-1}{log(x)-1}+\dfrac{1}{log(x)-1}

\rm y=1+\dfrac{1}{log(x)-1}

and apply power rule into chain rule from that point.
But this problem was likely designed to teach you about quotient rule so let's do it that way.

Let's start by "setting up" our quotient rule:

\rm y'=\dfrac{(log x)'(log x-1)-log x(log x-1)'}{(log x-1)^2}

If this log notation is not intended to be natural log, then we'll have a little bit of extra work. Our change of base identity allows us to rewrite log base 10 in terms of the natural log,

\rm log(x)=\dfrac{ln x}{ln10}

so let's apply this to our problem,

\rm y'=\dfrac{\left(\frac{ln x}{ln 10}\right)'(log x-1)-log x\left(\dfrac{ln x}{ln 10}-1\right)'}{(log x-1)^2}

Derivative of ln(x) gives us 1/x in each case,

\rm y'=\dfrac{\left(\frac{1}{x ln 10}\right)(log x-1)-log x\left(\dfrac{1}{x ln 10}\right)}{(log x-1)^2}

Factor the 1/(x ln10) out of each term in the numerator,

\rm y'=\left(\dfrac{1}{x ln 10}\right)\dfrac{log x-1-log x}{(log x-1)^2}

and combine like-terms,

\rm y'=\dfrac{-1}{x(log x-1)^2ln 10}

Lemme know if you're confused with any of the steps.
6 0
4 years ago
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