2:1 = ?:8 Write the missing term to make these ratios equal

Mathematics

1 answer:

Bogdan [553]3 years ago

3 0

Answer:

Step-by-step explanation:

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Help :') What integer does the arrow indicate? -9 -14 -3 6

gayaneshka [121]

it is -9 thats is the answer

4 0

3 years ago

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Solve this equations for all real solutions 3t^2+4t+8=-6t

ra1l [238]

Answer:

t= -1.33 recurring and -2

Step-by-step explanation:

3t²+4t+8=-6t

3t²+4t+6t+8

3t²+10t+8

This is a quadratic equation, it can be solved by using either factorisation, completing the square method or formulae method. For the purpose of this question, I would be using the formulae method.

-b±√(b²-4ac)/2a

Where a=3, b= 10 and c=8

-10±√(10²-4(3)(8))/2(3)

-10±√(100-96)/6

(-10±√4)/6

(-10±2)/6

-10+2/6 or -10-2/6

-8/6 or -12/6

-1.3333 recurring or -2

8 0

3 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