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kakasveta [241]
3 years ago
7

Look at this graph:

Mathematics
1 answer:
Rama09 [41]3 years ago
3 0

Answer:

2

Step-by-step explanation:

slope=rise/run=2/1=2

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Pls help with linear equations!!!
artcher [175]

Answer:

266

Step-by-step explanation:

Let no. of main floor tickets = x

Balcony tickets = x + 100

4(x+100) + 12(x) = 3056

4x + 400 + 12x = 3056

16x = 3056-400

16x = 2656

x = 2656/16

x = 166

No. of Balcony tickets = x + 100 = 166 + 100 = 266

4 0
3 years ago
Read 2 more answers
Next week your math teacher is giving a
Westkost [7]

Answer:

15

Step-by-step explanation:

If 20 questions are worth 2 points ,the rest will be worth 4 point questions ....its 35-20 which is 15

4 0
3 years ago
An entry level civil engineer's annual pay is $63,648 based on 52 weeks per year. Due to the economy, his company is having to c
Sindrei [870]

Answer:

$2448 in a year

Step-by-step explanation:

$63,648 ÷ 52 = $1224 a week (this would be how much he earns per week)

Since he is not getting paid for two week we do 52 - 2 = 50 week of work.

Since per week salary is $1224 we multiply by 2 for two weeks = $2448

Before $63,648 - $2448 = $61200 per year (Now he's making $61200/year)

Hope this helps,

if so please mark Brainlist,

Thanks!

4 0
3 years ago
Can someone answer this please
Ira Lisetskai [31]
I got 75 and I’m pretty sure that right?
4 0
3 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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