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steposvetlana [31]
3 years ago
13

What equation is solved by the graphed systems of equations? Two linear equations that intersect at the point negative 1, negati

ve 4.
Mathematics
1 answer:
GuDViN [60]3 years ago
6 0

To solve this problem, we have to manually solve for the value of x for each choices or equations. The correct equation will give a value of -1 since the linear equations intersects at point (-1, -4).

<span>1st: 7x + 3 = x + 3</span>

7x – x = 3 – 3

6x = 0

<span>x = 0                (FALSE)</span>

 

<span>2nd: 7x − 3 = x – 3</span>

7x – x = 3 – 3

6x = 0

<span>x = 0                (FALSE)</span>

           

<span>3rd: 7x + 3 = x − 3</span>

7x – x = - 3 – 3

6x = -6

<span>x = -1               (TRUE)</span>

 

<span>4th: 7x − 3 = x + 3</span>

7x – x = 3 + 3

6x = 6

<span>x = 1                (FALSE)</span>

 

Therefore the answer is:

<span>7x + 3 = x − 3</span>

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siniylev [52]
The pertinent formula for the circumference of a circle with diameter d is

C =pi*d.

Thus, if the dia. is 6 inches, the circumf. is C = 6pi inches.  Alternatively, if we let pi = 3.14, the circumf. is   C = 6*3.14   (answer A).
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3 years ago
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What is the answer to x+4=6
sattari [20]

Answer:

x=2

Step-by-step explanation:

When solving an algebraic equation, you have to get x alone

you do this by doing the inverse of whatever is efecting x

in this case it is x+4 so you subtract 4 from x+4

x+4-4=x

but when doing something like this, you have to do the same thing to both sides

x+4<em><u>-4</u></em>=6<u><em>-4</em></u>

x=2

please mark me brainliest!

4 0
3 years ago
Runner A crosses the starting line of a marathon and runs at an average pace of 5.6 miles per hour. Half an hour later, Runner B
marishachu [46]
Time it took runner A to complete the marathon = 26.2 / 5.6 = 4 hrs 41 mins
Time it took runner B to complete the marathon = 26.2 / 6.4 = 4 hrs 6 mins

Time it took runner B to complete the marathon relative to when runner A started = 30 mins + 4 hrs 6 mins = 4 hrs 36 mins

Therefore, runner B will finnish ahead of runner A.

Let x be the time the two runners are at the same point, then
5.6x + 5.6(0.5) = 6.4x
6.4x - 5.6x = 2.8
0.8x = 2.8
x = 2.8/0.8 = 3.5
Therefore, runner B will catch up with runner A 3.5 hours after runner A starts the race.
<span>Runner B; Runner B will catch up to Runner A 3.5 hours after Runner A crosses the starting line.</span>

5 0
3 years ago
A restaurant serves custom-made omelets, where guests select meat, cheese, and vegetables to be added to their omelet. There are
kondor19780726 [428]

Answer: The number of different combinations of 2 vegetables are possible = 15 .

Step-by-step explanation:

In Mathematics , the number of combinations of selecting r values out of n values = ^nC_r=\dfrac{n!}{r!(n-r)!}

Given : Number of available vegetables = 6

Then, the number of different combinations of 2 vegetables are possible will be :

^6C_2=\dfrac{6!}{2!(6-2)!}=\dfrac{6\times5\times4!}{2\times4!}=15

Hence , the number of different combinations of 2 vegetables are possible = 15 .

5 0
3 years ago
Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

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