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3 years ago

(HELP PLEASE) if f(x)=x/3-2 and g(x)=3x^2+2x-6 , find (f+g)(x)

Mathematics

2 answers:

lukranit [14]3 years ago

8 0

Identify and sum up the x^2 terms: Result: 3x^2
Next, ident. and sum up the x terms: Result: (1/3 + 2) = (7/3)x
Last, ident. and sum up the const. terms: Result: -2 - 6 = -8

The sum of these two functions is 3x^2 + (7/3)x - 8.

konstantin123 [22]3 years ago

7 0

Answer:

(A)

Step-by-step explanation:

The given functions are:

$f(x)=\frac{x}{3}-2$ and $g(x)=3x^2+2x-6$

Now, $(f+g)(x)=f(x)+g(x)$

therefore, $(f+g)(x)=\frac{x}{3}-2+3x^2+2x-6$

= $3x^2+2x+\frac{x}{3}-8$

= $\frac{9x^2+6x+x-24}{3}$

= $\frac{9x^2+7x-24}{3}$

= $3x^2+\frac{7}{3}x-8$

Thus, $(f+g)(x)=3x^2+\frac{7}{3}x-8$

Hence, Option A is correct.

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If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

Hence, they encircle the planet $3.76\times 10^{11}$ times.

8 0

3 years ago

(1.1.8 in the book) Let P be the proposition ’I bought a lottery ticket this weekend’ and Q be the proposition ’I won the millio

Karolina [17]

Answer:

~ is for negation

^ is for "and"

v is for "or"

=> for "if then"

<=> for "if and only if"

Step-by-step explanation:

(a) ~P (negation of P)

I didn't buy a lottery ticket this weekend.

(b) P v Q (P is in disjunction Q)

I have either bought a lottery ticket this weekend or won the million dollar jackpot.

I won the million dollar jackpot because I bought a lottery ticket this weekend.

(d) P ^ Q (P is in conjunction with Q)

I bought a lottery ticket this weekend, and I won the million dollar jackpot.

(e) P <=> Q (P and Q are dependent on each other)

If only I had bought a lottery ticket this weekend, I would have won the million dollar jackpot.

(f) ~P => ~Q (negation of Q is a consequence of negation of P)

I didn't win the million dollar jackpot because I didn't buy a lottery ticket this weekend.

(g) ~P ^ ~Q (negation of P is in conjunction with negation of Q)

I neither bought a lottery ticket this weekend nor won the million dollar jackpot.

(h) ~P v (P ^ Q)

This is logically equivalent to (~P v P) ^ (~P v Q) (negation of P is in disjunction with P, and also with disjunction with Q), and can be best expressed as:

It didn't matter that I bought a jackpot ticket or not, I won the million dollar jackpot.

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4 years ago

What is the value of C-F

photoshop1234 [79]

Answer:

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4 0

3 years ago

NEED HELP PLEASE WILL MARK BRAINLIEST !!!

Leona [35]

The answer you need is b

3 0

3 years ago

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kolbaska11 [484]

<u>Answer:</u>

$\frac{8\sqrt{y} }{x}$