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2 years ago

What is your estimate for the number of airline flights in a year?

Mathematics

1 answer:

Firdavs [7]2 years ago

8 0

The answer is between 45,000,000-65,000,000

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Let C be between D and E. Solve for x <br> DC = 5x + 8<br> CE = 3x + 5<br> DE = 61

morpeh [17]

DE/2 = 30.5
CE=30.5
DC=30.5

3x + 5 = 30.5
3x = 30.5 - 5
3x = 25.5
25.5/3 =8.5
x=8.5

8 0

2 years ago

A store owner recorded the number of tickets sold each hour. the graph shows some of the store owners data what is the slope of

Shalnov [3]

Answer: B 2

Step-by-step explanation:

If you look at the graph, you can see it goes up every hour by 2.

So therefore your answer will be 2.

Hope this helps!

7 0

2 years ago

Read 2 more answers

How do you identify a quadratic trinomial?

velikii [3]

Google it. ('>')...................................................

5 0

2 years ago

Translate then Solve for the variable:

Sergeu [11.5K]

The linear equation will be 2x + 10 = 5(x - 6). Then the value of the variable x will be 23.33.

<h3>What is the linear system?</h3>

A linear system is one in which the parameter in the equation has a degree of one. It might have one, two, or even more variables.

Translate then Solve for the variable:

The sum of two times a number and 10 is five times the difference between a number and six.

Let the number be x.

Then the equation will be

2x + 10 = 5(x - 6)

Then the value of x will be

2x + 10 = 5x - 60

5x - 2x = 60 + 10

3x = 70

x = 23.33

More about the linear system link is given below.

brainly.com/question/20379472

#SPJ1

8 0

1 year ago

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$

5 0

2 years ago