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crimeas [40]
3 years ago
6

What is the solution to this equation 2/3x+1=1/6×-7

Mathematics
1 answer:
Alika [10]3 years ago
7 0

Answer:

A . x= -16

Step-by-step explanation:

\\\frac{2}{3}x+1=\frac{1}{6}x-7\\\\\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}\\\\\frac{2}{3}x+1-1=\frac{1}{6}x-7-1\\\\\frac{2}{3}x=\frac{1}{6}x-8\\\\\mathrm{Subtract\:}\frac{1}{6}x\mathrm{\:from\:both\:sides}\\\\\frac{2}{3}x-\frac{1}{6}x=\frac{1}{6}x-8-\frac{1}{6}x\\\\= \frac{1}{2}x=-8\\\\\mathrm{Multiply\:both\:sides\:by\:2}\\\\2\times \frac{1}{2}x=2\left(-8\right)\\\\x=-16

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Yo givin brainiest !!
VikaD [51]

Answer:

C arithmetic

Step-by-step explanation:

-2,0,2,4,6

To get from the first term to the second term, we add 2

To get from the second term to the third term, we add 2

To get from the third term to the fourth term, we add 2

The common difference is 2

This is an arithmetic sequence.

There is no single number we can multiply by to move from once term to the next, so it is not a geometric sequence.

8 0
3 years ago
Consider the graph.
Arlecino [84]

Answer:

C, f(x) = 2x + 6

Step-by-step explanation:

First, we need to plug in the values of the x coordinates and see if it matches with the y coordinate to determine if it is on the same line. Startin with 2x + 8, we have the point (1, 8) on the graph. Plugging in 1 gets you 10 for the y. This is wrong since 8 is the y coordinate. Moving on, we have 6.4(1.25)^x for the same point. Plugging in 1, we have 6.4 * 1.25 = 8, which is true. Moving on to the second point, (2, 10), we have 1.25 squared times 6.4. This is thus wrong. So, moving on to 2x + 6, we have the point (1, 8), and plugging in 1 for x, we have 8 as y. Since this satisfies the equation we move on to the next point, (2,10). Plugging in x, we have 2 * 2 + 6 = 10, which is also true. Moving on to our third point (3 , 12), we plug in 3 for x. We then get 3 * 2 + 6 = 12, which is correct. This, is our answer then.

5 0
3 years ago
Read 2 more answers
Caroline has some dimes and some quarters. She has a maximum of 15 coins worth at least $2.85 combined. If Caroline has 3 dimes,
gayaneshka [121]

Answer:

11, 12.

Step-by-step explanation:

Let q represent number of quarters.

We have been given that Caroline has a maximum of 15 coins worth at least $2.85 combined. We are also told that Caroline has 3 dimes. This means that total coins are less than or equal to 15.

We can represent this information in an inequality as:

q+3\leq 15...(1)

We are also told that the coins worth at least $2.85 combined. This means that the worth of all coins is greater than or equal to 2.85.

We know that each dime is worth $0.10 and each quarter is worth $0.25.

0.25q+3(0.10)\geq 2.85...(2)

Now, let us solve our system of inequalities.

From 1st inequality, we will get:

q+3-3\leq 15-3

q\leq 12

From 2nd inequality, we will get:

0.25q+0.30\geq 2.85

0.25q+0.30-0.30\geq 2.85-0.30

0.25q\geq 2.55

\frac{0.25q}{0.25}\geq \frac{2.55}{0.25}

q\geq 10.2

Upon combining our both inequalities, we will get:

10.2\leq q\leq 12

This means that numbers of quarters would be greater than or equal to 10.2 and less than or equal to 12.

Since we cannot have 0.2 of a coin, therefore, Caroline could have 11 or 12 quarters.

6 0
3 years ago
What is the value of 2 ^ 3
Setler79 [48]

Answer:

8

Step-by-step explanation:

2^3 is the same as 2 x 2 x 2

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so 2^3 is 8

3 0
3 years ago
Read 2 more answers
Please help with 1 & 2
yaroslaw [1]
First question:

I urge you to perform the division using the synthetic division method:

      ________________
-4  /   1   3   -6   -6   8
             -4    4     8  -8
       -----------------------
         1   -1   -2    2    0

Note that there is no remainder.  When this is the case, the divisor (here, that's -4) is a root of the given polynomial, and the value of that polynomial, g(-4), is 0.

If the remainder were not 0, then the remainder represents the value of the polynomial for that particular divisor.  For example, if x = -3, the remainder is -28.  We'd write that as g(-3) = -28.

But here, g(-4) = 0.
6 0
3 years ago
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