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murzikaleks [220]
3 years ago
13

What is the answer for this question please

Mathematics
2 answers:
diamong [38]3 years ago
7 0

Answer:

2.7

Step-by-step explanation:

den301095 [7]3 years ago
6 0

Answer: 2.7

Step-by-step explanation:

Each line represents 0.2, so between 2.6 and 2.8 is 2.7

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Simplify <br> (3k^4) (7k^5)
Hunter-Best [27]

Answer:

21 k^9

Step-by-step explanation:

(3k^4) (7k^5)

Multiply the coefficients

3*7 = 21

Add the exponents

k^(4+5) = k^(9)

Put back together

21 k^9

4 0
3 years ago
mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi
chubhunter [2.5K]

Hello!  

We have the following data:  

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)  

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

a_n = a_1 + (n-1)*r

* second year salary

a_2 = a_1 + (2-1)*300

a_2 = 32000 + 1*300

a_2 = 32000 + 300

\boxed{a_2 = 32300}

* third year salary

a_3 = a_1 + (3-1)*300

a_3 = 32000 + 2*300

a_3 = 32000 + 600

\boxed{a_3 = 32600}

* fourth year salary

a_4 = a_1 + (4-1)*300

a_4 = 32000 + 3*300

a_4 = 32000 + 900

\boxed{a_4 = 32900}

* fifth year salary

a_5 = a_1 + (5-1)*300

a_5 = 32000 + 4*300

a_5 = 32000 + 1200

\boxed{a_5 = 33200}

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

a_6 = a_1 + (6-1)*300

a_6 = 32000 + 5*300

a_6 = 32000 + 1500

\boxed{a_6 = 33500}

* seventh year salary

a_7 = a_1 + (7-1)*300

a_7 = 32000 + 6*300

a_7 = 32000 + 1800

\boxed{a_7 = 33800}

*  eighth year salary

a_8 = a_1 + (8-1)*300

a_8 = 32000 + 7*300

a_8 = 32000 + 2100

\boxed{a_8 = 34100}

* ninth year salary

a_9 = a_1 + (9-1)*300

a_9 = 32000 + 8*300

a_9 = 32000 + 2400

\boxed{a_9 = 34400}

*  tenth year salary

a_{10} = a_1 + (10-1)*300

a_{10} = 32000 + 9*300

a_{10} = 32000 + 2700

\boxed{a_{10} = 34700}

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

*  eleventh year salary

a_{11} = a_1 + (11-1)*300

a_{11} = 32000 + 10*300

a_{11} = 32000 + 3000

\boxed{\boxed{a_{11} = 35000}}\end{array}}\qquad\checkmark

Respuesta:

In the eleventh year of salary he will earn more than 34700, in the case, this value will be 35000

________________________

¡Espero haberte ayudado, saludos... DexteR! =)

7 0
3 years ago
Moose plans to eat 8 of his mom's minty
antoniya [11.8K]

Answer:

75%

Step-by-step explanation:

\frac{6}{8}  \times 100\% = 75\%

6 0
2 years ago
If you punched 1 raffle ticket, what is the probability of winning of only 10 tickets are winners and 179 total tickets were sol
Licemer1 [7]

Answer:

1 out of 179

Step-by-step explanation:

8 0
3 years ago
A pieces of wire was 12.36 centimeters long. Thelma cut the wire into 6 pieces of equal length. What was the length of each piec
Zigmanuir [339]

Answer:

2.06cm

Step-by-step explanation:

Given data

Length of wire= 12.36cm

We are told that it was cut into 6 pieces

Hence, the length of a piece is  

=12.36/6

=2.06cm

Therefore, the answer is 2.06cm

3 0
3 years ago
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