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m_a_m_a [10]
3 years ago
10

1 Factor completely.

Mathematics
2 answers:
agasfer [191]3 years ago
8 0

Answer:

(1)The factor of  p²+ 7p + 10 are (p+5)(p+2) .

(2)The factor of x² + 10x + 16 are (x + 8)(x + 2) .

(3)The factors of x²+7x−18 are  (x-2)(x+9) .

(4) The factors of x² -5x - 24 are( x+3)(x-8).

(5) The factors of  x²− 3x − 40 are (x-8)(x+5).

Step-by-step explanation:

(1) As given the expression.

= p²+ 7p + 10

= p² + 5p + 2p + 10

= p (p+5)+2(p+5)

= (p+5)(p+2)

Therefore the factor of  p²+ 7p + 10 are (p+5)(p+2) .

(2) As given the expression

= x² + 10x + 16

= x² + 8x + 2x + 16

= x (x + 8)+2 (x + 8)

= (x + 8)(x + 2)

Therefore the factor of x² + 10x + 16 are (x + 8)(x + 2) .

(3)As given the expression

= x² + 7x − 18

= x² +9x -2x -18

=x (x+9)-2 (x +9)

= (x-2)(x+9)

Therefore the factors of x²+7x−18 are  (x-2)(x+9) .

(4)As given the expression

= x² -5x - 24

= x²-8x +3x -24

= x (x-8)+3(x-8)

= (x+3)(x-8)

Therefore the factors of x² -5x - 24 are( x+3)(x-8).

(5) As the expression given

= x²− 3x − 40

= x² -8x + 5x - 40

= x(x-8)+ 5(x-8)

=(x-8)(x+5)

Therefore the factors of  x²− 3x − 40 are (x-8)(x+5).

bulgar [2K]3 years ago
6 0

1 Factor completely.

p2+7p+10 = (p + 5)(p + 2)



 2 Factor completely.

x2+10x+16 = (x + 4)(x + 6)


3 Factor completely.

x2+7x−18 = (x + 9)(x - 2)



 4 Factor completely.

x2−5x−24 = (x - 8)(x + 3)



 5 Factor completely.

x2−3x−40 = (x - 8)(x + 5)

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I don't know if this is right... please someone help mee
worty [1.4K]
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\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

 so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.



now, let's check for second circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
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let's check the third circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
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Step-by-step explanation:

<u>Step 1</u>:-

Given log(9 x) -log 3 =3

we have to use formula  log a-log b=log\frac{a}{b}

log(9 x)- log 3 = log(\frac{9 x}{3} ) =3

Again by using formula log a =b

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so \frac{9 x}{3} =e^{3}

<u>step 2:</u>-

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Lapatulllka [165]

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