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3 years ago

N/-1.5=-1.2 what is n?

Mathematics

1 answer:

Nonamiya [84]3 years ago

6 0

N/-1.5=-1.2

-1.5*N/-1.5=-1.2*-1.5
n = 1.8

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Hasil dari 9 x (12 + (-5)) : (-8 - 13) Tolong dengan caranya

kifflom [539]

Answer:

-3

Step-by-step explanation:

9×(12+(-5)÷(-8-13)

9×7÷(-8-13)

63÷-21

=-3

5 0

3 years ago

Given that Cos0=5/13,find a)Tan0,b)Sin0,c)Sec0,d)Co sec0,e)Cot0,

qwelly [4]

cos θ = Adjacent/ hypotenuse

cosθ= 5/13

a²+b²= c²

a² + 5² = 13²

a² = 13² - 5²

a² = 144

a=√144

a= 12

a is the opposite = 12

b is the Adjacent = 5

c is the hypotenuse = 13

a) tan θ= opposite/Adjacent

tan θ = 12/5

b) sin θ= opposite/ hypotenuse

sinθ= 12/13

C) sec θ= hypotenuse / Adjacent

sec θ= 13/5

d) cscθ= hypotenuse /opposite

cscθ= 13/12

e) cotθ=Adjacent/ opposite

cotθ= 5/12

3 0

2 years ago

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

Mashcka [7]

Answer:

f'(x) > 0 on $(-\infty ,\frac{1}{e})$ and f'(x)<0 on $(\frac{1}{e},\infty)$

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

$f'(x)>0$

To find its decreasing interval :

$f'(x)$

2) Then let's find the critical point of this function:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0$

2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x'' $=\frac{1}{e}$ ≈0.37 for e≈2.72