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3 years ago

What are two ways to write 3f?

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

7 0

Answer:

3 × f (f is a variable or a substitute for a missing number)

3f (Means the same thing as 3 times f "or missing number")

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insens350 [35]

That equation just equals ln(e^(-4/3)) which can be simplified to -4/3 because ln and e cancel each other out only leaving the power

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3 years ago

Pringles is planning their sales force spending for the next year. Each of their sales people spends about 3400 hours annually a

LUCKY_DIMON [66]

Answer:

459 sales people.

Step-by-step explanation

Time per call (t) = 45 min = 0.75 h

Hours per sales person (H) = 3,400 hours

Number of customers (n) = 40,000 customers

Call frequency (f)= 52 calls per year

The total number of sales people (S) needed, is given by the total time spent on calls for the year, divided by the amount of hours each person spends on sale:

$S=\frac{40,000*0.75*52}{3,400}\\ S=458.8$

Rounding up to the next whole person, Pringles needs 459 sales people.

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4 years ago

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Look at the image. (calculus)

mash [69]

<h3>Answer: Choice H) 2</h3>

=============================================

Explanation:

Recall that the pythagorean trig identity is $\sin^2 x + \cos^2x = 1$

If we were to isolate sine, then,

$\sin^2 x + \cos^2x = 1\\\\\sin^2 x = 1-\cos^2x\\\\\sin x = \sqrt{1-\cos^2x}\\\\$

We don't have to worry about the plus minus because sine is positive when 0 < x < pi/2.

Through similar calculations, $\cos x = \sqrt{1-\sin^2x}\\\\$

Cosine is also positive in this quadrant.

-------------

So,

$\frac{\sqrt{1-\cos^2x}}{\sin x}+\frac{\sqrt{1-\sin^2x}}{\cos x}\\\\\frac{\sin x}{\sin x}+\frac{\cos x}{\cos x}\\\\1+1\\\\2$

Therefore,

$\frac{\sqrt{1-\cos^2x}}{\sin x}+\frac{\sqrt{1-\sin^2x}}{\cos x}=2$

is an identity as long as 0 < x < pi/2

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2 years ago

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What is the area, in square centimeters, of the figure below? 2.4cm, 5.8cm

nirvana33 [79]

Answer:

13.92

Step-by-step explanation:

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4 years ago

If x>2,then x2-x-6/x2-4=

statuscvo [17]

$\frac{x^2-x-6}{x^2-4}=(*)\\------------\\x^2-x-6=0\\a=1;\ b=-1;\ c=-6\\\Delta=b^2-4ac;\ \Delta=(-1)^2-4\cdot1\cdot(-6)=1+24=25\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\x_1=\frac{1-\sqrt{25}}{2\cdot1}=\frac{1-5}{2}=\frac{-4}{2}=-2\\\\x_2=\frac{1+\sqrt{25}}{2\cdot1}=\frac{1+5}{2}=\frac{6}{2}=3\\\\x^2-x-6=(x+2)(x-3)$
$---------------------\\x^2-4=x^2-2^2=(x-2)(x+2)\\-------------------\\\\(*)=\frac{(x+2)(x-3)}{(x-2)(x+2)}=\frac{x-3}{x-2}$

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4 years ago

What are two ways to write 3f?​

What are two ways to write 3f?