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Ugo [173]
3 years ago
14

What is the solution to 6|3x+5| less than equal to 14?

Mathematics
1 answer:
zimovet [89]3 years ago
3 0
In this case there will be 2 solutions to solve for in this equation, both will equal plus and minus the answer. Which is 14.

6(3x + 5) = +_ 14

The solutions will be the following :

(14/6 - 5)/3

(-14/6 - 5)/3.
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Which is a factor of x2 – 9x + 14?
PtichkaEL [24]

Answer:

(x-7)(x-2)

Step-by-step explanation:

what multiples to 14, but add to -9 ?

-7*-2 = 14

-7*-2= -9

8 0
3 years ago
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2 years ago
Thе іmagе оf (-2, 5) is (1, 1).
dem82 [27]

Answer:

  (6, -2)

Step-by-step explanation:

To get to x=1 from x=-2, you must add 3 to the x-coordinate. Only one answer choice has an x-coordinate that is 3 more than that of the given point:

  3 + 3 = 6

The image point is (6, -2).

_____

<em>Check</em>

To get from a y-coordinate of 5 to an image point y-coordinate of 1, you must subtract 4. If you subtract 4 from the y-coordinate of (3, 2), you get 2-4 = -2, the y-coordinate in the chosen answer above.

8 0
3 years ago
Solve the following system of equations. Write each of your answers as a fraction reduced to lowest terms. In other words, write
morpeh [17]

Answer:

x = 57/28

y = -95/84

z = 97/168

Step-by-step explanation:

Use the application in the next link: https://www.zweigmedia.com/RealWorld/tutorialsf1/scriptpivotold.html

Start with the expanded array:

\left[\begin{array}{cccc}1&5&8&1\\3&2&2&5\\-2&-7&2&5\\\end{array}\right]

then using the tool provided, make row operations until you find the solution:

r2 = r2-3r1

\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\-2&-7&2&5\\\end{array}\right]

r3 = r3+2r1

\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\0&3&18&7\\\end{array}\right]

r2 = r2*(-1/13)

\left[\begin{array}{cccc}1&5&8&1\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]

r1 = r1- r2*5

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]

r3 = r3+ r2*-3

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&168/13&97/13\\\end{array}\right]

r3 = r3*13/168

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&1&97/168\\\end{array}\right]

r2 = r2- r3*22/13

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]

r2 = r2+ r3*6/13

\left[\begin{array}{cccc}1&0&0&57/28\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]

Here you have a reduced array an therefore the answers to each variable are on each row:

\left[\begin{array}{c}x\\y\\z\end{array}\right]

8 0
3 years ago
Help me on this math problem?
shusha [124]

Answer:

just add them

Step-by-step explanation:

4 0
2 years ago
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