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nexus9112 [7]
3 years ago
13

What is three fourths the square of a number in an algebraic expression?

Mathematics
2 answers:
Effectus [21]3 years ago
5 0
The square of a number -> a²

three fourts the square = 3/4 *a ²=  3a²/4
Lemur [1.5K]3 years ago
4 0

Answer:

The algebraic expression is \frac{3}{4}x^2

Explanation:

An expression which contains constants, variables and algebraic operators is called an algebraic expression.

For example, 10x + y +2 is an algebraic expression.

In writing algebraic expression from a statement we use some keywords. For example the keyword combined represents addition, ratio represents division, e.t.c.

Further Explanation:

Here, we have to translate the given statement "three fourths the square of a number" to an algebraic expression.

Let the number is x.

Three fourths can be written as \frac{3}{4}

Square of the number x can be written as  x^2

The keyword 'of' means multiplication.

Thus, the algebraic expression for the statement "three fourths the square of a number" can be written as \frac{3}{4}x^2

Learn More:

brainly.com/question/12762448 (Answered by Calculista)

brainly.com/question/20988 (Answered by Taskmasters)

Keywords:

Algebraic expressions, Translating statement to Algebraic expressions.

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A box of pencils is 5 1/4 inches wide. Seven pencils, laid side by side, take up 2 5/8 inches of the width. How many inches of t
Andre45 [30]

Answer:   Width of box is not taken up by pencils =  2\dfrac58\text{ inches}

Width of each pencil =\dfrac38\text{ inches}

Step-by-step explanation:

Given: Width of pencil box = 5\dfrac14\text{ inches}=\dfrac{21}{4}\text{ inches}

Width of seven pencils = 2\dfrac58\text{ inches}=\dfrac{16+5}{8}\text{ inches}=\dfrac{21}{8}\text{ inches}

Width of box is not taken up by pencils =  Width of pencil box  - Width of seven pencils

\left \{ {{y=2} \atop {x=2}} \right. \dfrac{21}{4}-\dfrac{21}{8}\\\\=\dfrac{21\times2-21}{8}\\\\=\dfrac{21}{8}\text{ inches}=2\dfrac58\text{ inches}

Width of box is not taken up by pencils =  2\dfrac58\text{ inches}

Width of each pencil = (Width of seven pencils ) ÷ 7

=\dfrac{21}{8}\div7\\\\=\dfrac{21}{8}\times\dfrac17\\\\=\dfrac38\text{ inches}

Width of each pencil =\dfrac38\text{ inches}

3 0
3 years ago
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2x - 2y = 10 and x + 2y = 11
Rufina [12.5K]

Answer:

Step-by-step explanation:

What are u asking for x or y

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3 years ago
Which red triangle shows a 90° counterclockwise rotation of the blue triangle? Check all that apply. On a coordinate plane, a bl
Tcecarenko [31]

Question:

1) Blue triangle points; (-1, 4), (-5, 4), and (-1, 1)

Red triangle points; (-4, -1), (-1, -1). and (-4, -5)

2) Blue triangle points (1, 1), (4, 5), (4, 1)

Red triangle points; (-1, 1), (-1, 4). and (-5, 4)

3) Blue triangle points (0, 1), (4, 4), (4, 1)

Red triangle points; (-4, 1), (-4, 4). and (0, 1)

4) Blue triangle points (-5, 4), (-1, 4), (-1, 1)

Red triangle points; (1, 1), (4, 1). and (4, 5)

5) Blue triangle points (-2, 5), (4, 5), (4, 1)

Red triangle points; (-1, 4), (-5, 4). and (-5, -2)

Answer:

The correctly rotated red triangles are those of (1), (2), and (5)

Step-by-step explanation:

In a 90° counterclockwise rotation, every x and y points of the original triangle are switched while the y is turned negative, as shown in the following equation;

(x, y) to (-y, x)

Therefore, the triangles that undergo a 90° counterclockwise rotation are as follows;

1) Blue triangle points; (-1, 4), (-5, 4), and (-1, 1)

Red triangle points; (-4, -1), (-1, -1). and (-4, -5)

(-1, 4) → (-4, -1)

(-5, 4) → (-4, -5)

(-1, 1) → (-1, -1)

Correctly rotated 90° counterclockwise

2) Blue triangle points (1, 1), (4, 5), (4, 1)

Red triangle points; (-1, 1), (-1, 4). and (-5, 4)

(1, 1) → (-1, 1)

(4, 5) → (-5, 4)

(4, 1) → (-1, 4)

Correctly rotated 90° counterclockwise

3) Blue triangle points (0, 1), (4, 4), (4, 1)

Red triangle points; (-4, 1), (-4, 4). and (0, 1)

(0, 1) → (0, 1) ≠ (-1, 0)

(4, 4) → (-4, 4)

(4, 1) → (-4, 1)

Not correctly rotated 90° counterclockwise

4) Blue triangle points (-5, 4), (-1, 4), (-1, 1)

Red triangle points; (1, 1), (4, 1). and (4, 5)

(-5, 4) → (4, 5) ≠ (-4, -5)

(-1, 4) → (4, 1)

(-1, 1) → (1, 1)

Not correctly rotated 90° counterclockwise

5) Blue triangle points (-2, 5), (4, 5), (4, 1)

Red triangle points; (-1, 4), (-5, 4). and (-5, -2)

(-2, 5) → (-5, -2)

(4, 5) → (-5, 4)

(4, 1) → (-1, 4)

Correctly rotated 90° counterclockwise.

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Answer:

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Solve for r.<br> 4r – 39 &gt; -43<br> AND 8x + 31 &lt; 23
Vsevolod [243]

Answer:

4r-39>-43

4r-39+39>-43+39

4r>-4

\frac{4r}{4}>\frac{-4}{4}

\boxed{r>-1}

______________________

8x+31

8x+31-31

8x

\frac{8x}{8}

\boxed{x

7 0
2 years ago
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