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2 years ago

Find the greatest maximum y-value of the following equation:

1 answer:

5 0

A quick way to do this is by recalling the fact that $-1\le\cos x\le1$ , so the largest value that $\cos ax$ alone (for any real value of $a$ ) can take on is 1.

This means $-3\le3\cos2x\le3$ , and so $1\le3\cos2x+4\le7$ .

So the maximum value of this function is 7.

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Plz help me with this

Elodia [21]

Answer:

its green so you got it right

and if thats not the case you are correct

Step-by-step explanation:

3 0

2 years ago

(HELP ME 15 PTS)

kap26 [50]

Answer: $\bold{\dfrac{27x^2y^3}{16}}$

<u>Step-by-step explanation:</u>

$\dfrac{(3x^2y^5)^3}{(2xy^3)^4}\\\\\\=\dfrac{3^3\cdot x^{2\cdot 3}\cdot y^{5\cdot 3}}{2^4\cdot x^4\cdot y^{3\cdot 4}}\\\\\\=\dfrac{27\cdot x^6\cdot y^{15}}{16\cdot x^4\cdot y^{12}}\\\\\\=\dfrac{27\cdot x^{6-4}\cdot y^{15-12}}{16}\\\\\\=\dfrac{27\cdot x^2\cdot y^3}{16}$

8 0

2 years ago

What is the interquartile range (IQR) of the data set represented by this box<br> plot?

Zinaida [17]

Answer:

Step-by-step explanation:

The interquartile range is the difference between the upper quartile and the lower quartile.

The upper quartile is the right side value of the box

The lower quartile is the left side value of the box

upper quartile = 61 and lower quartile = 39

IQR = 61 - 39 = 22 → B

7 0

2 years ago

Read 2 more answers

A cylinder has a volume of 384 cubic inches and a height of 8 inches, what is the radius

Triss [41]

The answer is: 3.91 inches .
___________________________________________

Note: Volume of cylinder: V = (base area) * (height);

in which: V = volume = 384 in.³ ;
h = height = 8 in. ;
Base area = area of the base (that is; "circle") = π r² ;
in which; "r" = radius;
___________________________________________
Solve for "r" :
___________________________________________
V = π r² * (8 in.) ;

384 in.³ = (8 in.) * (π r²) ;
___________________________________________
Divide EACH SIDE of the equation by "8" ;
___________________________________________
(384 in.³) / 8 = [ (8 in.) * (π r²) in.] / 8 ;
___________________________________________
to get:
___________________________________________
48 in.³ = (π r²) in.² * in. ;
___________________________________________
↔ (π r²) in.² * in. = 48 in.³ ;
___________________________________________
Rewrite this equation; using "3.14" as an approximation for: π ;
__________________________________________________
(3.14 * r²) in.² * in. = 48 in.³
_______________________________________
Divide EACH SIDE of the equation by:

"[(3.14)*(in.²)*(in.)]" ; to isolate "r² " on one side of the equation;
(since we want to solve for "r") ;
_____________________________________________________
→ [(3.14 * r²) in.² * in.] / [(3.14)*(in.²)*(in.)] = 48 in.³ / [(3.14)*(in.²)*(in.)] ;
__________________________________________________
→ to get: r² = 48/3.14 ;
________________________
→ r² = 15.2866242038216561 ;
_______________________________________
To solve for "r" (the radius; take the "positive square root" of EACH side of the equation:
__________________________________________________
→ +√(r²) = +√(15.2866242038216561)
__________________________________________________
→ r = 3.9098112747064475286 ; round to 3.91 inches .
___________________________________________________

4 0

3 years ago

Read 2 more answers

A pool is surrounded by a rectangular fence with a

PtichkaEL [24]

Answer:

width is 33

Step-by-step explanation:

width (b) =?

length(l)=3

perimeter (p)=?

now,

perimeter = 2(l+b)

or,72= 2(3+ b)

or,72/2= 3+b

or,36= 3+b

or,36-3= b

or,33= b

therefore width is 33

7 0

2 years ago