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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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The discriminant is ...

  b^2 -4ac = (-4)^2 -4(2)(16) = 16 -128 = -112

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If we assume the equation has a typographical error and is intended to be ...

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6 0
3 years ago
What are the solutions of the equation 9x4 – 2x2 – 7 = 0? Use u substitution to solve. tions of the equation 9
skelet666 [1.2K]

Answer:

Step-by-step explanation:

Let u^2=x^4\\u = x^2

Subbing in:

9u^2-2u-7=0

a = 9, b = -2, c = -7

The product of a and c is the aboslute value of -63, so a*c = 63.  We need 2 factors of 63 that will add to give us -2.  The factors of 63 are {1, 63}, (3, 21}, {7, 9}.  It looks like the combination of -9 and +7 will work because -9 + 7 = -2.  Plug in accordingly:

9u^2-9u+7u-7=0

Group together in groups of 2:

(9u^2-9u)+(7u-7)=0

Now factor out what's common within each set of parenthesis:

9u(u-1)+7(u-1)=0

We know this combination "works" because the terms inside the parenthesis are identical.  We can now factor those out and what's left goes together in another set of parenthesis:

(u-1)(9u+7)=0

Remember that u=x^2

so we sub back in and continue to factor.  This was originally a fourth degree polynomial; that means we have 4 solutions.

(x^2-1)(9x^2+7)=0

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis.  Factoring (x^2-1) gives us that x = 1 and -1.  The other set is a bit more tricky.  If

9x^2+7=0 then

9x^2=-7 and

x^2=-\frac{7}{9}

You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:

x=±\sqrt{-\frac{7}{9} }

which will simplify down to

x=±\frac{\sqrt{7} }{3}i

Those are the 4 solutions to the quartic equation.

5 0
3 years ago
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