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vlabodo [156]
4 years ago
5

What is the solution to the linear equation? 4b + 6 = 2 – b + 4

Mathematics
2 answers:
Ray Of Light [21]4 years ago
8 0

Answer:

b = 0

Step-by-step explanation:

4b + 6 = 2 – b + 4

4b + b = 2 + 4 - 6

5b = 0

b = 0/5

b = 0

mars1129 [50]4 years ago
5 0

b = 0 the pictures should help you to learn how to do it

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The cost of 24 peincle is 4.08 find the price in filler per peincle
choli [55]
.17 i believe by doing 4.08/24
8 0
4 years ago
The rectangle graphed has a length of 3 and a width of 2. How does the perimeter change if the width is increased by 4? A) The p
storchak [24]

Answer:

The answer to your question is the letter D.

Step-by-step explanation:

Data

First rectangle                       Second rectangle

length = 3                               length = 3

width = 2                                 width = 2 + 4 = 6

Process

1.- Calculate the Perimeter of the first rectangle

Perimeter = 2l + 2w

Perimeter = 2(3) + 2(2)

                 = 6 + 4

                 = 10

2.- Calculate the perimeter of the second rectangle

Perimeter = 2(3) + 2(6)

Perimeter = 6 + 12

Perimeter = 18

3.- Compare both perimeters

                Perimeter 1 is smaller than Perimeter 2 by 8 units

4 0
4 years ago
if sintheta is greater than 0 and costheta less than 0 then the terminal point determined by theta is in:
mihalych1998 [28]

Answer:

Quadrant  II

Step-by-step explanation:

Sin is positive in Quad I and II

Cos is negative in  Quad II and II

Therefor ∅ is in Quadrant  II

6 0
3 years ago
1) Se o 5o e o 9-o termos de uma PA são, respectivamente, 40 e 68, então a razão r da progressão é: a) r = 6 b) r = 7 c) r = 9 d
Maslowich

Answer:

1) b) r = 7

2) e) 82

Step-by-step explanation:

Progressão aritmética:

Em uma progressão aritmética, a diferença entre termos consecutivos é sempre a mesma, chamada de razão.

O n-ésimo termo de uma PA é dado por, tomando o primeiro termo como referência:

a_n = a_1 + (n-1)r

Tomando o m-ésimo termo como referência, tem-se que:

a_n = a_m + (n-m)r

1) Se o 5o e o 9-o termos de uma PA são, respectivamente, 40 e 68, então a razão r da progressão é:

a_5 = 40, a_9 = 68. Então:

a_n = a_m + (n-m)r

68 = 40 + 4r

4r = 28

r = \frac{28}{4} = 7

Então a resposta correta é dada pela alternativa b.

2) Inserindo-se 6 números entre 72 e 107, de modo que a sequência (72, a2, a3, a4, a5, a6, a7 ,107) seja uma progressão aritmética, tem-se a3 igual a:

O primeiro termo é a_1 = 72 e o oitavo termo é a_8 = 107. Com isso, é possível encontrar a razão.

a_n = a_1 + (n-1)r

107 = 72 + 7r

7r = 35

r = \frac{35}{7} = 5

Então, o terceiro termo é:

a_3 = a_1 + 2r = 72 + 2(5) = 82

Logo, a resposta correta é dada pela alternativa e.

5 0
3 years ago
Paco's cell phone carrier charges him $0.20 for each text message he sends or receives, $0.15 per minute for calls, and a $15 mo
Katarina [22]

Given:

Paco's cell phone carrier charges him $0.20 for each text message he sends or receives, $0.15 per minute for calls, and a $15 monthly service fee. Paco is trying to keep his bill for the month below $30.

To Find:

The number of texts 't' Paco can send/receive in a month.

Answer:

0\leq t

best describes the number of texts he can send or receive to keep his bill less than $30 in a month.

Step-by-step explanation:

Paco wants to keep is monthly bill below $30.

We see that he has to pay a foxed monthly service fee of $15. This means he is only left with a limit of $30 - $15 = $15 for his monthly calls and texts.

That is, the amount he has to pay for texting and calling has to be less than $15.

For texts, the cell phone carrier charges $0.20 for sending/receiving texts.

For calls, he is charged $0.15 per minute.

Let the number of text messages Paco can send or receive in a month be denoted by 't'.

Let the number of minutes Paco can call in a month be denoted by 'c'.

Then, the total cost of text messages he can send or receive per month would be 0.20t and the total cost of the minutes he spends on calls would be 0.15c. Together, the sum of these has to be less than $15 if his monthly bill has to be kept less than $30 (accounting for the monthly service fee).

So,

0.20t+0.15c

The number of texts he can send will dpend on the number of minutes he spends on his calls. For Paco to spend maximum number of texts, he has to spend 0 minutes on calls.

So, putting c = 0, the aboce equation can be written as

0.20t+0

That is, Paco has to send and receive less than 75 texts.

So,

0\leq t

best describes the number of texts he can send or receive to keep his bill less than $30 in a month.

3 0
3 years ago
Read 2 more answers
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