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dlinn [17]
3 years ago
8

of all the whole numbers that divide evenly into two or more numbers the one with the highest value is called the _____________.

Mathematics
1 answer:
Vadim26 [7]3 years ago
3 0
Greatest common factor (GCF)
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The tiles below represent the polynomial 2x+ 5x+ 3.
alekssr [168]

Answer:

(2x+3)(x+1)

Step-by-step explanation:

A P E X

6 0
3 years ago
you are assigned 25 spelling words to master you memorize 60% of the words how many do you have left to memorize
Volgvan
2.5 is 10% of 25. 2.5 * 6 = 15 = 60%

25 - 15 = 10. You have 10 words left to memorize.

I hope this helps.
5 0
4 years ago
Point Q’ is the image of Q (3,-4) under a reflection across the x-axis
7nadin3 [17]

Answer:

Q' (-3,4)

Step-by-step explanation:

A reflection across the x-axis would mean that only the x value of the point would change. This means that the y value would still be 4. To find the x value, just flip it to a negative. The negative of 3 is -3

5 0
3 years ago
Read 2 more answers
For all values of x
Yuliya22 [10]

Answer:

A.) gf(x) = 3x^2 + 12x + 9

B.) g'(x) = 2

Step-by-step explanation:

A.) The two given functions are:

f(x) = (x + 2)^2 and g(x) = 3(x - 1)

Open the bracket of the two functions

f(x) = (x + 2)^2

f(x) = x^2 + 2x + 2x + 4

f(x) = x^2 + 4x + 4

and

g(x) = 3(x - 1)

g(x) = 3x - 3

To find gf(x), substitute f(x) for x in g(x)

gf(x) = 3( x^2 + 4x + 4 ) - 3

gf(x) = 3x^2 + 12x + 12 - 3

gf(x) = 3x^2 + 12x + 9

Where

a = 3, b = 12, c = 9

B.) To find g '(12), you must first find the inverse function of g(x) that is g'(x)

To find g'(x), let g(x) be equal to y. Then, interchange y and x for each other and make y the subject of formula

Y = 3x + 3

X = 3y + 3

Make y the subject of formula

3y = x - 3

Y = x/3 - 3/3

Y = x/3 - 1

Therefore, g'(x) = x/3 - 1

For g'(12), substitute 12 for x in g' (x)

g'(x) = 12/4 - 1

g'(x) = 3 - 1

g'(x) = 2.

5 0
3 years ago
Evaluate the following limit:
Makovka662 [10]

If we evaluate the function at infinity, we can immediately see that:

        \large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \lim_{x \to \infty}{\frac{(x^2 + 1)^2 - 3x^2 + 3}{x^3 - 5}} = \frac{\infty}{\infty}} \end{gathered}$}

Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.

We can solve this limit in two ways.

<h3>Way 1:</h3>

By comparison of infinities:

We first expand the binomial squared, so we get

                         \large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{x^4 - x^2 + 4}{x^3 - 5}} = \infty \end{gathered}$}

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.

<h3>Way 2</h3>

Dividing numerator and denominator by the term of highest degree:

                            \large\displaystyle\text{$\begin{gathered}\sf L  = \lim_{x \to \infty}\frac{x^{4}-x^{2} +4  }{x^{3}-5  }  \end{gathered}$}\\

                                \ \  = \lim_{x \to \infty\frac{\frac{x^{4}  }{x^{4} }-\frac{x^{2} }{x^{4}}+\frac{4}{x^{4} }    }{\frac{x^{3} }{x^{4}}-\frac{5}{x^{4}}   }  }

                                \large\displaystyle\text{$\begin{gathered}\sf \bf{=\lim_{x \to \infty}\frac{1-\frac{1}{x^{2} } +\frac{4}{x^{4} }  }{\frac{1}{x}-\frac{5}{x^{4} }  }  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{0}=\infty } \end{gathered}$}

Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.

5 0
2 years ago
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