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3 years ago

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wilhelmina has 8.3 ounces of peanut butter.if she makes 5 sandwijes with an equal amount ofbpeanut butter on each,how much peanu

t butter does she put on each one?

1 answer:

kotykmax [81]3 years ago

3 0

Peanut butter: 8.3 ounces

Sandwich: 5

Divide 8.3 by 5, for equal amount of peanut butter on each sandwich.

8.3/5 = 1.66

She puts 1.66 ounces of peanut butter on each sandwich.

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Uninhibited growth can be modeled by exponential functions other than A(t) =Upper A 0 e Superscript kt. For example, if an i

laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than $A(t)=A_{0}e^{kt}$ . for example, if an initial population P₀ requires n units of time to triple, then the function $P(t)=P_{0}(3)^{\frac{t}{n} }$ models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form $P(t)=P_{0}(3)^{\frac{t}{n} }$ that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form $A(t)=A_{0}e^{kt}$ .

Answer: a) $P(t)=50(3)^{\frac{t}{30} }$

b) P(t) = 280 insects

c) t = 74 days

d) $A(t)=50e^{0.037t}$

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

$P(t)=50(3)^{\frac{t}{30} }$

b) In t = 47 days:

$P(t)=50(3)^{\frac{t}{30} }$

$P(47)=50(3)^{\frac{47}{30} }$

$P(47)=50(3)^{1.567}$

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

$750=50(3)^{\frac{t}{30} }$

$\frac{750}{50}=(3)^{\frac{t}{30} }$

$(3)^{\frac{t}{n} }=15$

Using the property <u>Power</u> <u>Rule</u> of logarithm:

$log(3)^{\frac{t}{30} }=log15$

$\frac{t}{30}log(3)=log15$

$t=\frac{log15}{log3} .30$

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

$A(t)=A_{0}e^{kt}$

$3A_{0}=A_{0}e^{30k}$

$3=e^{30k}$

Use Power Rule again:

$ln3=ln(e^{30k})$

$ln3=30k$

$k=\frac{ln3}{30}$

k = 0.037

Equation for exponential growth will be:

$A(t)=50e^{0.037t}$

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3 years ago

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Step-by-step explanation:hope this helps

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2 years ago

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Romashka [77]

<h2>Explanation:</h2>

Here we need to solve some expressions. Remember the following rules:

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Then, from the left side:

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from the right side:

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3 years ago

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finlep [7]

Hey There!

8x^46=8x(40+

We move all terms to the left:

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3 years ago

The diffrence between 2 numbers is 1 and their product Is 132

il63 [147K]

Answer:

x = 12

y = 11

For any clarification, Please contact me in the comment section :)

I Hope It Helps .

Step-by-step explanation:

$Let -the-unkonown- numbers-be-x-and-y \\x-y = 1 (1)\\xy = 132 (2)\\Make-x-subject-of-formula-in-equation1 \\x = 1+y\\Substitute , 1+y for- x -in -equation ,2\\y(1+y) = 132\\y + y^{2} =132\\y^{2} +y-132=0\\Solve- the-quadratic-equation.\\y = 11\\substitute- 11 -for- y- in- equation -1- 0r- 2\\x -y = 1\\x-11 = 1\\x = 1+11\\x = 12$

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