Rewrite in simplest terms: 4(-5u +3) – 4(8u - 1)

Mathematics

2 answers:

denis-greek [22]4 years ago

6 0

Answer:

−52u+16

plz like my answer

viktelen [127]4 years ago

3 0

Answer:

You distribute then combine

your answer is <u><em>-52u+16</em></u>

Step-by-step explanation:

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How do you combine like terms of -8(r-2) + 6(r-2) ??? Will mark brainliest if the answer is right.

solmaris [256]

Answer:

$-2r+4$

Step-by-step explanation:

$-8(r-2)+6(r-2)$

Distribute:

$(-8)(r)+(-8)(-2)+(6)(r)+(6)(-2)\\-8r+16+6r+-12$

Combine like terms:

$(-8r+6r)+(16+-12)\\-2r+4$

-2r + 4 is your answer.

6 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