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Karolina [17]
3 years ago
11

Write a cosine function that has a midline of 5 an amplitude of 2 and a period of 5/2

Mathematics
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

Uuy

Step-by-step explanation:

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I need help with this im currently brainded
makvit [3.9K]
> your welcome :) and tell me if it’s right
4 0
3 years ago
Read 2 more answers
Please help! Thank you!
Aloiza [94]

Answer:

X=sigma xi/no of observation

5= (?) /4

5*4=sum of all numbers

20=sum of all the observation

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4 0
3 years ago
Suppose that the functions s and t are defined for all real numbers x as follows.
jeka94

Answer:

(s.t)(x) = 4x^2+24x^2\\(s-t)(x) = x+6-4x^2\\(s+t)(-3) = 39

Step-by-step explanation:

Given functions are:

s(x)= x+6\\t(x)= 4x^2

We have to find:

(s.t)(x) => this means we have to multiply the two functions to get the result.

So,

(s.t)(x) = s(x)*t(x)\\= (x+6)(4x^2)\\=4x^2.x+4x^2.6\\=4x^3+24x^2

Also we have to find

(s-t)(x) => we have to subtract function t from function s

(s-t)(x) = s(x) - t(x)\\= (x+6) - (4x^2)\\=x+6-4x^2

Also we have to find,

(s+t)(-3) => first we have to find sum of both functions and then put -3 in place of x

(s+t)(x) = s(x)+t(x)\\= x+6+4x^2

Putting x = -3

= -3+6+4(-3)^2\\=-3+6+4(9)\\=3+36\\=39

Hence,

(s.t)(x) = 4x^2+24x^2\\(s-t)(x) = x+6-4x^2\\(s+t)(-3) = 39

8 0
3 years ago
Suppose f(x) and g(x) are differentiable functions satisfying f(1) = f′(1) = 2. Let
just olya [345]

Answer:

1777777777888876332222233

3 0
2 years ago
A cardboard box without a lid is to be made with a volume of 4 ft 3 . Find the dimensions of the box that requires the least amo
Blababa [14]

Answer:

<h2><em>2ft by 2ft by 1 ft</em></h2>

Step-by-step explanation:

Total surface of the cardboard box is expressed as S = 2LW + 2WH + 2LH where L is the length of the box, W is the width and H is the height of the box. Since the cardboard box is without a lid, then the total surface area will be expressed as;

S  = lw+2wh+2lh ... 1

Given the volume V = lwh = 4ft³ ... 2

From equation 2;

h = 4/lw

Substituting into r[equation 1;

S = lw + 2w(4/lw)+ 2l(4/lw)

S = lw+8/l+8/w

Differentiating the resulting equation with respect to w and l will give;

dS/dw = l + (-8w⁻²)

dS/dw = l - 8/w²

Similarly,

dS/dl = w  + (-8l⁻²)

dS/dw = w - 8/l²

At turning point, ds/dw = 0 and ds/dl = 0

l - 8/w² = 0 and w - 8/l² = 0

l = 8/w²  and w =8/l²

l = 8/(8/l² )²

l = 8/(64/I⁴)

l = 8*l⁴/64

l = l⁴/8

8l = l⁴

l³ = 8

l = ∛8

l = 2

Hence the length of the box is 2 feet

Substituting l = 2 into the function l = 8/w² to get the eidth w

2 = 8/w²

1 = 4/w²

w² = 4

w = 2 ft

width of the cardboard is 2 ft

Since Volume = lwh

4 = 2(2)h

4 = 4h

h = 1 ft

Height of the cardboard is 1 ft

<em>The dimensions of the box that requires the least amount of cardboard is 2ft by 2ft by 1 ft</em>

8 0
3 years ago
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