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Alik [6]
3 years ago
14

What is the value of f(-4) in the piecewise function f(x)= -x-5 for -5<= x <= -2; -x^2+1 for -2<=x<=2; (x-3)^2+2 for

2<=x<=4? If that makes any since
Mathematics
2 answers:
Black_prince [1.1K]3 years ago
6 0

Answer:

The value for f(-4) in the piecewise function is -1

Step-by-step explanation:

As the function is defined by pieces we must choose the correct interval when replacing. In this case, the value of the function needs to be calculated in x=4.  

This value of x is within the interval:

-5<=x<=-2

Therefore we replaced in the corresponding expression:

f(x)=-x-5

f(-4)=-(-4)-5=-1

Take into account this case is not particularly a function since, for example, in the value of x=-2 the expression can take two values corresponding to two different pieces.  The obtained result is only valid in the range for -5<=x<=-2.

Alik [6]3 years ago
5 0


We first need to pick the correct piece to use. In this case, we will use f(x) = -x-5 because -5<=-4.

Then, we use direct substitution:
f(-4) = -(-4) - 5
f(-4) = -1

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Answer:

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Step-by-step explanation:

If t represents the number of tents made, then 4.5t represents the amount of canvas needed. We have ...

  4.5t = 3330

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7 0
2 years ago
A) How many cups of energy drink does Jerome need to make?
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3 years ago
The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much
Y_Kistochka [10]

Answer:

x = ∛ 2*V/5  

y = ∛ 2*V/5

h  = V/ ∛ 4*V²/25

Step-by-step explanation:

Dimensions of the aquarium base is  x*y

We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.

let call h the height of the aquarium then volume of the aquarium is:

V = x*y*h      where   h =  V / x*y

As the base is a rectangular one there are 2 sides x*h .  and 2 sides  y*h

According to this:

Ct (cost of aquarium )  = cost of the base  + cost of the sides

cₐ  ( cost of the base) = 5*c₁*x*y

c₆ (cost of the sides ) = c₁*2*x*h   +   c₁*2*y*h

C(t)  =  5*c₁*x*y +2* c₁*x* V/x*y  +  2* c₁*y* V/x*y    or

C(t)  =  5*c₁*x*y  + 2*c₁*V/y   *2*c₁* V/x

Taking partial derivatives en x and y we have:

C´(x)  =  5*c₁*y - 2*c₁*V/x²

C´(y)  =  5*c₁*x - 2*c₁*V/y²

C´(x)  = C´(y)        ⇒  5*c₁*y - 2*c₁*V/x²  =   5*c₁*x - 2*c₁*V/y²

or    5*y - 2*V/x²  =   5*x - 2*V/y²

(5*y*x² - 2*V)/x²  = ( 5*y²x - 2*V) /y²

(5*y*x² - 2*V)*y²  = ( 5*y²x - 2*V)*x²

5*y³*x² - 2*V*y²  =  5*y²x³  - 2*V*x²

5*y³*x² - 5*y²x³  =  2*V * ( y² - x²)

by symmetry  x =  y

Then using x = y  and plugging that value on the derivatives

C´(x) =  5*c₁*y - 2*c₁*V/x²

C´(x) =  5*c₁*x - 2*c₁*V/x²

C´(x) = 0          ⇒     5*c₁*x - 2*c₁*V/x²  = 0

5*x  - 2*V/x² = 0      ⇒  5*x³ - 2*V = 0   ⇒   5*x³  = 2*V  ⇒ x³ = 2*V/5

x = ∛ 2*V/5       and   y = ∛ 2*V/5    and   h  =  V/ x*y    h  = V/ ∛ 4*V²/25

7 0
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3 years ago
Help:(, i’m not sure how to do this
Schach [20]

Answer:

i think you have to multiply the sides..

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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