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Dennis_Churaev [7]

2 years ago

12

Barbara wants to buy some glittery heart-shaped beads to make friendship bracelets. The beads she wants come in a small pack wit

h 90 beads or a large pack with 300 beads. The small pack costs $3.60, while the large pack costs $9.00. Which pack is the better value?

1 answer:

AleksAgata [21]2 years ago

5 0

Answer:

I feel u

Step-by-step explanation:u need help too

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Evaluate ∣∣256+y∣∣ for y=74. A. 225 B. 315 C. 345 D. 4712

aleksklad [387]

Answer:

678

Step-by-step explanation:

5 0

2 years ago

Select the correct answer.

Maslowich

Answer:

x is the correct answer

4 0

2 years ago

Help please!!! I just seriously need help I'm bad at math>

azamat

There is only 1 real number solution

8 0

2 years ago

Suppose the cost C(x), to build a football stadium of x thousand square feet is approximated by C(x) = 7,250,000/ x + 60 . How m

Luba_88 [7]

Answer:

Step-by-step explanation:

Suppose the cost C(x), to build a football stadium of x thousand square feet is approximated by C(x) = 7,250,000/x + 60. Given the function, we can substitute values for x to determine the cost of a particular size of stadium or we can substitute values for C(x) to determine the number of square feet.

if the cost of the stadium was $8,000, the, we would determine the size of the stadium, x by substituting x $8,000 for C(x). It becomes

8000 = 7250,000/x + 60

8000 - 60 = 7250000/x

7940 = 7250000/x

7940x = 7250000

x = 7250000/ 7940

x = 913 ft^2

7 0

2 years ago

Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

$114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)$

Applying the following property of exponentials numbers in (II):

$e^{a}.e^{b}=e^{a+b}$

Therefore $e^{-28k}$ can be written as $e^{-14k}.e^{-14k}$

$152=190-Ae^{-14k}.e^{14k}$

Replacing (I) in the previous equation:

$152=190-76e^{-14k}$

Solving for k:

Subtracting 190 both sides, dividing by -76:

$0.5=e^{-14k}$

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

$Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76$

Dividing by 0.5

A=152

7 0

2 years ago