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3 years ago

- 14x² + 7x - 6+ 6x- 9x + 16

Mathematics

2 answers:

zheka24 [161]3 years ago

8 0

Answer:

14xx2(square) + 7x - 6 + 6x - 9x + 16

14x2x(square) + 7x - 6 - 3x + 16

114x2(square) + 7x - 3x - 6 + 16

14x2(square) + 4x - 10

Dahasolnce [82]3 years ago

4 0

Solution
- 14 x²+7 x+6 x-9 x -6+16
-14 x²+4 x+10

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which statement correctly compares two values? a)the value of the 7 in 37.596 is 1/10 the value of the7 in 26.74 b) the value of

Andrews [41]

The value of 7 in 26..74 is 0.7

The value of 7 in 37.596 is 7.

1/10 the value of 7 in 26.74 is 0.07 and 0.07 ≠ 7.

So, statement a) is incorrect.

1/100 the value of 7 in 26.74 is 0.007 and 0.007 ≠ 7.

So, statement c) is incorrect.

100 times the value of 7 in 26.74 is 70 and 70 ≠ 7.

So, statement d) is incorrect.

10 times the value of 7 in 26.74 is 7 and 7 = 7.

So, statement b) is correct and it correctly compares two values.

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

Factors of $6=2\times 3$

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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