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3 years ago

7

Using the remainder theorem, which is not a factor of

xFormula1" title="x^4-3x^3-17x^2+39x-20?" alt="x^4-3x^3-17x^2+39x-20?" align="absmiddle" class="latex-formula">
A. x-1
B. x+4
C. x-5
D. x+1

1 answer:

True [87]3 years ago

4 0

Answer:

D.

$x + 1$

Step-by-step explanation:

$x^4-3x^3-17x^2+39x-20 = \\ = (x - 5) {(x - 1)}^{2} (x + 4)$

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The vertex of f(x) is (2, - 3) . is the vertex of f(x) + 3

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Answer:

D. (2 , 0)

Step-by-step explanation:

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3 years ago

Read 2 more answers

Divide 12x 4 y 4 by 3x 2 y. A. 4x 6y 5 B. 4x 2y 3 C. 9x 6 D. 9x 2y 3

andrew-mc [135]

Answer:

$\frac{12x^4y^4}{3x^2y}=4x^2y^3$

Step-by-step explanation:

Consider an element $a\neq 0$

We know that $\frac{a^m}{a^n}=a^{m-n}$

Here, m and n are exponents and a is a base. Exponent refers to the no. of times a term is multiplied by itself. Another word for exponents is powers.

Other Properties of exponents:

$a^m\times a^n=a^{m+n}\\a^0=1\\\left ( a^m \right )^n=a^{mn}$

Here, we are required to divide $12x^4y^4$ by $3x^2y$ .

Consider $\frac{12x^4y^4}{3x^2y}$

We will solve $\frac{12}{3}\,,\,\frac{x^4}{x^2}\,,\,\frac{y^4}{y}$ separately.

Here,

$\frac{12}{3}=4\\\frac{x^4}{x^2}=x^{4-2}=x^2\\\frac{y^4}{y}=y^3$

So, $\frac{12x^4y^4}{3x^2y}=4x^2y^3$

8 0

2 years ago

Help me with this please!!

valina [46]

<h2><u>Given:</u><u>-</u></h2>

Points C = (-7,2) → $\sf{(X_1,Y_1)}$
D = (3,12) → $\sf{(X_2,Y_2)}$

<h2><u>To </u><u>Find</u><u>:</u><u>-</u></h2>

The Midpoint of CD.

<h2><u>Required</u><u> </u><u>Response</u><u>:</u><u>-</u></h2>

Let,

Midpoint of CD be (x,y).

WKT,

$\boxed{\sf{(x,y) = \frac{X_1+X_2}{2},\frac{Y_1+Y_2}{2}}}$

$→\;{\sf{\frac{-7+3}{2},\frac{2+12}{2}}}$

$→\;{\sf{\frac{-4}{2},\frac{14}{2}}}$

$→\;{\sf{-2,7}}$

The Midpoint of CD ◕➜ $\Large{\red{\mathfrak{(-2,7)}}}$

Let,

The centre be O

Radius = CO & OD

Here, C = (-7,2) → $\sf{(X_1,Y_1)}$

O = (-2,7) → $\sf{(X_2,Y_2)}$

$\boxed{\sf{Distance = \sqrt{(X_2-X_1)²+(Y_2-Y_1)²}}}$

$→\;{\sf{\sqrt{(-2+7)²+(7-2)²}}}$

$→\;{\sf{\sqrt{5²+5²}}}$

$→\;{\sf{\sqrt{25+25}}}$

$→\;{\sf{\sqrt{50}}}$

$→\;{\sf{5√2 (or) 7.07}}$

Radius of Circle ◕➜ $\Large{\red{\mathfrak{7.07}}}$

<h2>Option D.</h2>

Hope It Helps You ✌️

3 0

2 years ago