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uranmaximum [27]
3 years ago
14

11) 19, 13, 7, 1...

Mathematics
1 answer:
MariettaO [177]3 years ago
4 0

Answer:happy

Step-by-step explanation:

Vg

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3 (2x - 5) + 20 + 5x i need help
igomit [66]

Answer:

<h2>11x - 5</h2><h2>   or</h2><h2>x = -5 / 11</h2>

Step-by-step explanation:

3 (2x - 5) + 20 + 5x

= 6x - 15 + 20 + 5x

= 6x + 5x = 15 - 20

= 11x - 5

or if you are looking for x:

then x = -5 / 11

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4 years ago
Solve with steps<br>8y-6.9=3y+3.6
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Marquis used the steps below to find the solution to the inequality Negative 5. 6 greater-than-or-equal-to x 3. 6. Step 1 Subtra
NISA [10]

Marquis solutions are;

  • The value of x is less than or equal to -9.2.
  • Negative 9.2 is greater than or equal to x.
  • The closed circle is at -9.6 that is everything to the left of the circle is shaded.
<h3>What is inequality?</h3>

Inequality is simply a type of equation that does not have an equal sign in it. Inequality is defined as a statement about the relative size as we will as is used to compare two statements.

Marquis used the steps below to find the solution to the inequality Negative 5.6 greater than or equal to x + 3.6 that can be written as

\rm -5.6 \geq  x + 3.6

A.  Then the value of x will be

      - 5.6 ≥ x + 3.6

-5.6 - 3.6 ≥ x

       -9.2 ≥ x

B.  Negative 9.2 is greater than or equal to x.

C.  The number line is shown. The closed circle is at -9.6 that is everything to the left of the circle is shaded.

More about the inequality link is given below.

brainly.com/question/19491153

7 0
2 years ago
What is the slope intercept form equation of the line that passes through (3, 4) and (5, 16)?
luda_lava [24]

It is convenient to start with the 2-point form of the equation for a line.

... y - y1 = (y2 - y1)/(x2 - x1)×(x - x1)

Either point can be (x1, y1), and the other can be (x2, y2). If we take them in order, we get

... y - 4 = (16 - 4)/(5 - 3)×(x - 3) . . . . . fill in the two points

... y = 12/2(x -3) +4 . . . . . . . . . . . . . . add 4, simpliffy a bit

... y = 6x -18 +4 . . . . . . . . . . . . . . . . . eliminate parentheses

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5 0
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