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3 years ago

The sum of two consecutive odd integers is 156. Which is an equation that can be used to solve for x? Please

Mathematics

1 answer:

OLEGan [10]3 years ago

4 0

Answer:

x+(x+2) = 156

Step-by-step explanation:

Let x = 1st odd integer

x+2 = next odd integer

x+(x+2) = 156

2x+2 =156

Subtract 2

2x= 154

Divide by 2

x = 77

x+2 = 79

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Max is stacking logs at his campground for firewood. After his first load of logs, he has 12 logs on the stack. After his sevent

Natasha2012 [34]

Answer:

Let L(n) be the function that gives the amount of logs stacked after n loads.

L(n) = 12 + 8(n-1)

Step-by-step explanation:

Let L(n) be the function that gives the amount of logs stacked after n loads.

Let's call for the moment the first load as L(0)

L(0)= 12

Let r be the number of logs carried in each load, then

L(n) = 12 + nr

Since L(6) (the seventh load) equals 60, we have

60 = 12 + 6r, and r = 8.

So a function for the number of loads starting from n=0 would be

L(n) = 12 + 8n

If we want to start with n=1, we simply change the variable

L(n) = 12 + 8(n-1) (n=1,2,3,...).

So L(1) = 12, L(2) = 20, L(3) = 28,...L(7) = 60 and so on.

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3 years ago

2/3 of a class are girls what is the ratio to boys in the class

adoni [48]

Ratio 1:2 from boys to girls

6 0

3 years ago

10-5(9n-9) simplify the expressions

eduard

Answer:

−45n+55

Step-by-step explanation:

10-5(9n-9)

distribute -5 into (9n-9)

10-45n+45

combine like terms

10+45=55

-45n+55

hope this helps

8 0

3 years ago

Factor

AfilCa [17]

Answer:

The answer is 6ab³c(3ab + 4c)

Step-by-step explanation:

You have to collect the like-term :

18a²b⁴c + 24ab³c²

= 6(3a²b⁴c + 4ab³c²)

= 6a(3ab⁴c + 4b³c²)

= 6ab³(3abc + 4c²)

= 6ab³c(3ab + 4c)

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3 years ago

Write an augmented matrix of each system. Then, solve each system using matrix notation. Describe the solution set in vector not

Naddik [55]

Answer:

The answer is below

Step-by-step explanation:

Given the system of equations:

x + y + 2z = 9

2x + 4y - 3z = 1

3x + 6y - 5z = 0

This system of equation can be solved using matrix. This done by first representing the equations as matrix and then solving:

The matrix form is:

$\left[\begin{array}{ccc}1&1&2\\2&4&-3\\3&6&-5\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}9\\1\\0\end{array}\right] \\\\\\ \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}1&1&2\\2&4&-3\\3&6&-5\end{array}\right] ^{-1} \left[\begin{array}{c}9\\1\\0\end{array}\right] \\\\\\$

$\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2&-17&11\\-1&11&-7\\0&3&-2\end{array}\right] \left[\begin{array}{c}9\\1\\0\end{array}\right] \\\\\\ \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}1\\2\\3\end{array}\right]$

5 0

2 years ago