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2 years ago

91 more than the square of a number in verbal form

1 answer:

3 0

91 more than the square of a number can be expressed algebraically as:
let the number be x;
square of x is x^2
thus our answer is x^2+91

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Luke can paint 91 portraits in 7 weeks. What is has average rate of painting portraits per week?

lina2011 [118]

Answer:

13.

Step-by-step explanation:

91 divided by 7.

8 0

2 years ago

Rewrite the expression with rational exponents as a radical expression.

Alik [6]

Answer:

$\displaystyle 4\sqrt[7]{x^3}$

General Formulas and Concepts:

Algebra I

Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle 4x^\bigg{\frac{3}{7}}$

Step 2: Rewrite

Split: $\displaystyle 4(x^\bigg{\frac{1}{7}})^3$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle 4\sqrt[7]{x^3}$

3 0

2 years ago

Find the area of a square if a side measures 15 mm

tankabanditka [31]

Formula for the area of a square = side^2
So, 15^2 will result in the area.
Thus, the answer is 225 mm^2 (when squaring, units are always given the square as well.)

5 0

3 years ago

HELPPPPP ASP PLZZZZZ

-Dominant- [34]

Answer:

$(f-g)(x)$

$f(x)-g(x)$

$x^{2} -6x-27-x+9$

$x^{2} -7x-18$

----------------------

$(f*g)(x)$

$=f(x)g(x)$

$(x^{2} -6x-27)(x-9)$

$=x^{3} -15x^{2}+27x+243$

----------------------

$\frac{f}{g} (x)$

$\frac{x^{2} -6x-27}{x-9}$

$\frac{(x-9)(x+3)}{x-9}$

$x+3$

-----------------------

$(f+g)(x)$

$f(x)+g(x)$

$=x^{2} -6x-27+x-9$

$=x^{2} -5x-36$

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OAmalOHopeO

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5 0

2 years ago

State the number of real zeros and what they are for "3x^2+5x-2" pleas help me:'(

Lady bird [3.3K]

This can be solved by factoring.

First, set the expression equal to zero.

$3x^2+5x-2=0$

Then, find two the factors of $ac$ whose sum is $b$ .

$6, -1$

Split $b$ into these two factors.

$3x^2+6x-x-2=0$

Next, factor by grouping.

$3x(x+2)-1(x+2) = (3x-1)(x+2) = 0$

By the Zero Product Property, set each factor equal to zero.

$3x-1 = 0 \\ x = 1/3$

$x+2=0 \\ x = -2$

These are the solutions. The Complex Conjugate Root Theorem and the Fundamental Theorem of Algebra both state that, in essence, real and imaginary solutions come in pairs of two and every polynomial of degree $n$ has exactly $n$ complex roots, but real roots are also complex roots. That sounds confusing, but this just means that you're done. Your answers are -2 and 1/3. There are two real roots.

3 0

3 years ago