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3 years ago

PLZ HELP! FACTOR THEN SOLVE x2+17x+16=0

2 answers:

6 0

So most likely what this equation would end up being would be 2x + 17x added together t make 19x + 16 = 0

Hpe I helped =)

Marina86 [1]3 years ago

3 0

X2+17x+16=0
(x+1)(x+16)=0
x+1=0 x+16=0
x=-1, x=-16

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Write 44 as a product of primes using index notation

trasher [3.6K]

Answer:

Step-by-step explanation:

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2 years ago

Simplify the following expression as much as you can use exponential properties. (6^-2)(3^-3)(3*6)^4

gtnhenbr [62]

Answer:

Simplifying the expression $(6^{-2})(3^{-3})(3*6)^4$ we get $\mathbf{108}$

Step-by-step explanation:

We need to simplify the expression $(6^{-2})(3^{-3})(3*6)^4$

Solving:

$(6^{-2})(3^{-3})(3*6)^4$

Applying exponent rule: $a^{-m}=\frac{1}{a^m}$

$=\frac{1}{(6^{2})}\frac{1}{(3^{3})}(18)^4\\=\frac{(18)^4}{6^{2}\:.\:3^{3}} \\$

Factors of $18=2\times 3\times 3=2\times3^2$

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Replacing terms with factors

$=\frac{(2\times3^2)^4}{(2\times 3)^{2}\:.\:3^{3}} \\=\frac{(2)^4\times(3^2)^4}{(2)^2\times (3)^{2}\:.\:3^{3}} \\$

Using exponent rule: $(a^m)^n=a^{m\times n}$

$=\frac{(2)^4\times(3)^8}{(2)^2\times (3)^{2}\:.\:3^{3}} \\=\frac{2^4\times 3^8}{2^2\times 3^{2}\:.\:3^{3}}$

Using exponent rule: $a^m.a^n=a^{m+n}$

$=\frac{2^4\times 3^8}{2^2\times 3^{2+3}}\\=\frac{2^4\times 3^8}{2^2\times 3^{5}}$

Now using exponent rule: $\frac{a^m}{a^n}=a^{m-n}$

$=2^{4-2}\times 3^{8-5}\\=2^{2}\times 3^{3}\\=4\times 27\\=108$

So, simplifying the expression $(6^{-2})(3^{-3})(3*6)^4$ we get $\mathbf{108}$

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