6 (x+15) evaluate the expression for x=-4

2 answers:

8 0

The answer would be 66 by using the distributive property you multiply 6 times -4 then 6 times 15 which is -24+90 which is 66.

Gala2k [10]3 years ago

4 0

The answer would be 66.

6(-4 +15)
=
66

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How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12?

siniylev [52]

Count the number of multiples of 3, 4, and 12 in the range 1-2005:

⌊2005/3⌋ ≈ ⌊668.333⌋ = 668

⌊2005/4⌋ = ⌊501.25⌋ = 501

⌊2005/12⌋ ≈ ⌊167.083⌋ = 167

(⌊x⌋ means the "floor" of x, i.e. the largest integer smaller than x, so ⌊a/b⌋ is what you get when you divide a by b and ignore the remainder)

Then using the inclusion/exclusion principle, there are

668 + 501 - 2•167 = 835

numbers that are multiples of 3 or 4 but not 12. We subtract the number multiples of 12 twice because the sets of multiples of 3 and 4 both contain multiples of 12. Subtracting once removes the multiples of 3 and 4 that occur twice. Subtracting again removes them altogether.

7 0

3 years ago

How do you solve this problem?

sveticcg [70]

8/9

O=16H

H+H+O=water

18H =water molecules
so u can see ratio 16H/!8H=8/9

7 0

3 years ago

Find the indefinite integral by using the substitution x = 4 sec(θ). (Use C for the constant of integration.) x2 − 16 x d

guapka [62]

Answer:

$\frac{x^2-16}{2} + 16ln\frac{4}{x} +16C$

Step-by-step explanation:

Given the indefinite integral $\int\limits{\frac{x^2-16}{x} } \, dx$ , using the substitute

x = 4 sec(θ)...1

The integral can be calculated as thus;

First let us diffrentiate the substitute function with respect to θ

dx/dθ = 4secθtanθ

dx = 4secθtanθdθ... 2

Substituting equation 1 and 2 into the integral function we will have;

$\int\limits{\frac{(4sec \theta)^2-16}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\int\limits{\frac{16sec^2 \theta-16}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\int\limits{\frac{(16(sec^2 \theta-1)}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\\from \ trig \ identity,\ sec^2 \theta - 1 = tan^\theta\\\\\int\limits{\frac{16 tan^2 \theta}{4sec \theta} } \, 4sec \theta tan \theta d \theta\\\\\int\limits 16 tan^3 \theta d \theta\\\\$