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kherson [118]
3 years ago
12

A rectangle has an area of 28 ft2. If the width of the rectangle is 5 ft, what is the length?

Mathematics
1 answer:
Jobisdone [24]3 years ago
5 0
Area of a Rectangle = Length x Width
28 ft² = L x 5ft
28 ft² / 5ft = Length
5.6 ft = length.  CHOICE A.

Perimeter of a Rectangle = 2(L + W)
20 ft = 2(L + 4)
20 ft = 2L + 8
20 - 8 = 2L
12 = 2L
12/2 = 2L/2
6 ft= Length  CHOICE A.
You might be interested in
Chase consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Chase's body
PIT_PIT [208]

Answer:

(a) The 5-hour decay factor is 0.5042.

(b) The 1-hour decay factor is 0.8720.

(c) The amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.

Step-by-step explanation:

The amount of caffeine in Chase's body decreases exponentially.

The 10-hour decay factor for the number of mg of caffeine is 0.2542.

The 1-hour decay factor is:

1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720

(a)

Compute the 5-hour decay factor as follows:

5-hour\ decay\ factor=(0.8720)^{5}\\=0.504176\\\approx0.5042

Thus, the 5-hour decay factor is 0.5042.

(b)

The 1-hour decay factor is:

1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720

Thus, the 1-hour decay factor is 0.8720.

(c)

The equation to compute the amount of caffeine in Chase's body is:

A = Initial amount × (0.8720)<em>ⁿ</em>

It is provided that initially Chase had 171 mg of caffeine, 1.39 hours after consuming the drink.

Compute the amount of caffeine in Chase's body 2.39 hours after consuming the drink as follows:

A = Initial\ amount \times (0.8720)^{2.39} \\=[Initial\ amount \times (0.8720)^{1.39}] \times(0.8720)\\=171\times 0.8720\\=149.112

Thus, the amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.

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6 0
3 years ago
3y + 2/5 = -1/5 what is y?
choli [55]

Answer:

y = -1/5 = -0.200

Step-by-step explanation:

Step  1  :

           1

Simplify   —

           5

Equation at the end of step  1  :

        2           1

 (3y +  —) -  (0 -  —)  = 0

        5           5

Step  2  :

           2

Simplify   —

           5

Equation at the end of step  2  :

        2     -1

 (3y +  —) -  ——  = 0

        5     5

Step  3  :

Rewriting the whole as an Equivalent Fraction :

3.1   Adding a fraction to a whole

Rewrite the whole as a fraction using  5  as the denominator :

          3y     3y • 5

    3y =  ——  =  ——————

          1        5  

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

3.2       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

3y • 5 + 2     15y + 2

——————————  =  ———————

    5             5  

Equation at the end of step  3  :

 (15y + 2)    -1

 ————————— -  ——  = 0

     5        5

Step  4  :

Adding fractions which have a common denominator :

4.1       Adding fractions which have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

(15y+2) - (-1)     15y + 3

——————————————  =  ———————

      5               5  

Step  5  :

Pulling out like terms :

5.1     Pull out like factors :

  15y + 3  =   3 • (5y + 1)

Equation at the end of step  5  :

 3 • (5y + 1)

 ————————————  = 0

      5      

Step  6  :

When a fraction equals zero :

6.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

 3•(5y+1)

 ———————— • 5 = 0 • 5

    5    

Now, on the left hand side, the  5  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

  3  •  (5y+1)  = 0

Equations which are never true :

6.2      Solve :    3   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

6.3      Solve  :    5y+1 = 0

Subtract  1  from both sides of the equation :

                     5y = -1

Divide both sides of the equation by 5:

                    y = -1/5 = -0.200

One solution was found :

                  y = -1/5 = -0.200

Processing ends successfully

plz mark me as brainliest :)

7 0
3 years ago
Read 2 more answers
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