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3 years ago

What is the range of the relation shown below? {(-1, -3), (5, 5), (2, 2), (3, 0)}

Mathematics

1 answer:

poizon [28]3 years ago

5 0

(-1,-3) is the smallest and (5,5) is the largest. you have to subtract them.

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2 3/4+1 1/5 i need this for math, im kinda stupid, please dont anwser if you dont know its for a test XD

Oksi-84 [34.3K]

Answer:

3 19/20

Step-by-step explanation:

So...

The first step is to find a common denominator between the 2. The smallest common denominator is 20. So you would do 3/4 times 5/5 which is 15/20. (2 15/20)

1/5 times 4/4 = 4/20(1 4/20)

So the final step is to add them together

2 15/20 + 1 4/20= 3 19/20.

This is the most simplified answer you can get.

5 0

2 years ago

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:

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

3 0

3 years ago

Please help me find y and show work/steps. I don't know how to do it

trasher [3.6K]

We will check if y and 60cm are parallel.

If the lengths 28cm, 56cm and 15cm, 30cm are in proportion, then y and the segment 60cm are parallel.

$\dfrac{56}{28}=\dfrac{30}{15}\\\\2=2\qquad CORRECT!$

OK. We know: y and 60cm are parallel. Therefore we have equation:

$\dfrac{y}{28+56}=\dfrac{60}{56}\\\\\dfrac{y}{84}=\dfrac{15}{14}\qquad|\text{cross multiply}\\\\14y=15\cdot84\qquad|\text{divide both sides by 14}\\\\y=90cm$