PLS HELP WILL GIVE BRAINLY!!!!!

If a rectangle has a length of 9.00 and a width of 4.25 .what is the area in square inches?

Lubov Fominskaja [6]

Answer:

Area=9.00 inches * 4.25 inches

= 38.25 square inches

Step-by-step explanation:

maybe this can help! just maybe?

3 0

3 years ago

Assume that the two polygons are similiar, with the same orientation.

spin [16.1K]

Answer:

4/6=t/(t+1)

Step-by-step explanation:

4/6=t/(t+1)

8 0

3 years ago

Pls answer thank you

Rashid [163]

Answer:

<h2>ANSWER </h2>

 

 $as \: the \: diagram \: given \: \\ surface \: area \: = (9 \times 17 \times 2 + 4 \times 17 + 4 \times 9 + 4 \times 17 + 9 \times 4) {in}^{2}$ 

 $(153 \times 2+ 68 + 36 + 68 + 36 \\ 306 + 104 + 104 ) {in}^{2}$ 

 $544 \:{ in}^{2}$ 

So the answer here = 544 in^2

4 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

$x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37$

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0

3 years ago

HELP HELP HELP HELP!!!!

Veseljchak [2.6K]

C.) Coordinate of x is (10, 8)

In short, Your Answer would be: Option C

Hope this helps!

7 0

3 years ago

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