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3 years ago

Is 0.2971 irrational?

Mathematics

1 answer:

Vladimir [108]3 years ago

7 0

Answer:

Yes

Step-by-step explanation:

An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point.

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1/3(3+2x)−1=10 PLEASE HURRY I NEED It IN LIKE 2 MINUTES

Sholpan [36]

Answer:

x=15

Step-by-step explanation:

crown plz

3 0

3 years ago

Read 2 more answers

For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim

ICE Princess25 [194]

Answer:

Part a) The area of the figure is $\frac{9}{2}(4+\pi )\ cm^{2}$

Part b) The perimeter of the figure is $3(2+2\sqrt{2}+ \pi)\ cm$

Step-by-step explanation:

Step 1

Find the area of the figure

In this problem we have that

The figure ABC is the half of a square and the other figure is a semicircle

Find the area of the figure ABC

we have

$AB=6\ cm, BC=6\ cm$

The area of the half square ABC is equal to find the area of triangle ABC

$A1=\frac{1}{2}*6*6=18\ cm^{2}$

Find the area of the semicircle

The area of the semicircle is equal to

$A2=\pi r^{2}/2$

we have that

$r=6/2=3\ cm$

substitute

$A2=\pi (3)^{2}/2$

$A2=(9/2) \pi\ cm^{2}$

The area of the figure is equal to

$18\ cm^{2}+(9/2) \pi\ cm^{2}= \frac{9}{2}(4+\pi )\ cm^{2}$

Step 2

Find the perimeter of the figure

The perimeter of the figure is equal to

$P=AB+AC+length\ CB$

we have

$AB=6\ cm$

Applying Pythagoras theorem

$AC=\sqrt{6^{2}+6^{2}}\\AC=6\sqrt{2}\ cm$

Remember that

the circumference of a semicircle is equal to

$C=\frac{1}{2}2\pi r=\pi r$

$r=6/2=3\ cm$

$C=\pi(3)$

$C=3 \pi\ cm$

The perimeter of the figure is equal to

$P=6\ cm+6\sqrt{2}\ cm+3 \pi\ cm$

Simplify

$P=3(2+2\sqrt{2}+ \pi)\ cm$

5 0

3 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 Power Rule 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

What conversion factor would you use to solve thikproblem?

ira [324]

Conversion factor is the term that is to be multiplied to a variable when converting between units. The conversion factor is: $2.20lbs /kg$

Given that:

$54kg \to x\ lbs$

From standard unit of conversion:

$1kg = 2.20lbs$

So, we have:

$54kg \to x\ lbs$

$1kg = 2.20lbs$

Cross multiply

$54kg \times 2.20lbs = xlbs \times 1kg$

Divide both sides by 1kg

$54kg \times 2.20lbs /kg= xlbs$

Rewrite as:

$x = 54kg \times 2.20lbs /kg$

Hence, the conversion factor is: $2.20lbs /kg$

Read more about conversion factors at:

brainly.com/question/6988886

8 0

2 years ago

Use elimination to solve each system of equations. 1. 3x+4y=29 6x+5y=43

kakasveta [241]

Answer:

Step-by-step explanation:

3x+4y=29 .......(1)

6x+5y=43 .......(2)

Multiply equation (1) by 2

6x + 8y = 58......(3)

Substracting (2) from (3)

8y - 5y = 58 - 43

3y = 15

Dividing by 3

3y/3 = 15/3

y = 5

Substituting y = 5 into (2)

6x+5y=43

6x + 5(5) = 43

6x + 25 = 43

6x = 43 - 25

6x = 18

Dividing by 6

6x/6 = 18/6

x = 3

4 0

3 years ago