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Vsevolod [243]
4 years ago
6

How do you put 3x-3-(x-2) in Algebra Tiles on an Algebra Tiles mat.​

Mathematics
1 answer:
mrs_skeptik [129]4 years ago
3 0
I’m actually not sure but what are algebra tiles?
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Can someone help me please I really need better grades<br>​
Novosadov [1.4K]

Answer:

Choose answers: A, C, D

Step-by-step explanation:

It is simply arithmetic. Multiply out the exponents.

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3 years ago
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Convert this rational number into a decimal form 3/5 round to the nearest thousands
il63 [147K]
0.6 is the answer

I hope this helps.

Have a awesome day. :)
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3 years ago
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Correct answer will get brainliest.
Alecsey [184]

Answer:

  B)  1

Step-by-step explanation:

The smallest integer larger than 0.6 is 1.

___

The funny single-ended brackets signify the "ceiling" function. If the bent end is at the bottom, the "floor" function is indicated. The "ceiling" function returns the smallest integer not less than the given value. The "floor" function returns the largest integer not greater than the given value.

These are more commonly used than you might think. For example, the charge in a parking lot may be for a number of hours or fraction thereof. That is an expression of the ceiling function applied to hours. In this problem, that would mean that parking for 0.6 hours would be charged as 1 hour.

4 0
4 years ago
9^6 * 9^3 = <br><br><br> Give the answer in exponent form
Andrews [41]

Step-by-step explanation:

1,983593e11

5 0
3 years ago
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20 POINTS!!! Use the quadratic formula above to solve for h(t) = -4.9t^2 + 8t + 1 where h is the height of the ball in meters an
Elina [12.6K]

Answer:

Two solutions: -0.12 and 1.75.

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}. Assuming that the x² term is a, the x term is b, and the constant is c, we can plug the values into the equation.

\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {8^2 - 4\cdot-4.9\cdot1} }}{{2\cdot-4.9}}} \end{array}

\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {64 + 19.6} }}{{-9.8}}} \end{array}

\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {83.6} }}{{-9.8}}} \end{array}

\begin{array}{*{20}c}{\frac{{ - 8 \pm \sqrt {9.14} }}{{-9.8}}} \end{array}

\frac{-8 + 9.14}{-9.8} = -0.12

\frac{-8-9.14}{-9.8} =1.75

Hope this helped!

3 0
3 years ago
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