Translate the phrase into an algebraic expression. the sum of b and 4

Use the rule provided to generate the first five terms of the sequence. The rule is 4n + 5.

lesya692 [45]

Answer:

9, 13, 17, 21,25

Step-by-step explanation:

n=1, 4×1+5=9

n=2, 4×2 +5=13

::

:

n=5, 4×5 + 5= 25

4 0

2 years ago

Whats the answer to this ?

8090 [49]

Answer: $312 ft^{2}$

Step-by-step explanation:

Area of Rectangle = bh = 16 ft x 18 ft = 288 $ft^{2}$

Area of Triangle = 1/2 bh = 1/2 x (16-12) ft x (6) ft x 2 (There's 2 triangles) = 24 $ft^{2}$

24 ft^2 + 288 ft^2 = 312 ft^2

8 0

2 years ago

A triangular frame has sides that measure 15'-7",20'-4" and 26'-2". What is the total length of three sides?

tatuchka [14]

we know that

$1\ foot=12\ inches$

$1'=12"$

Part 1)

we know that

side 1=15'-7"

side 2=20'-4"

side 3=26'-2"

To find the total length of three sides sum the three sides

so

total length=side 1+side 2 +side 3

substitute

total length=15'-7"+20'-4"+26'-2"

total length=(15'+20'+26')+(7"+4"+2")

total length=(61')+(13")

remember that

12"=1'

13"=12"+1"=1'+1"

substitute

total length=(61')+(1'+1")

total length=(62')+(1")---------> 62'-1"

therefore

the answer is

the total length of three sides is 62'-1"

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

2 years ago

What is the best estimate for the sum of 3/4 and 1/12

TEA [102]

Well first you would find a common denominator which would be 1w and then since you multiplied 4 by 3 to get to 12 you multiply 3 by 3 which is 9. So then you would have 9/12 + 1/12 which = 10/12

3 0

2 years ago