Question

1).(1/8)^2 / 2^5*(1/2)^10=2).2x*(x^2-8x-2)=3).(-3y^2+0,6y)*(-1,5y^3)=4).5x+3(x-1)=6x+11

Answer 1

Объяснение:

1)

$\frac{(\frac{1}{8})^2}{2^5*(\frac{1}{2})^{10} } =\frac{((\frac{1}{2})^3)^2 }{2^5*2^{-10}} =\frac{(\frac{1}{2})^6 }{2^{-5}} =\frac{2^{-6}}{2^{-5}} =2^{-1}=\frac{1}{2} =0,5.$

2)

$2x*(x^2-8x-2)=2x^3-16x^2-4x.$

3)

$(-3y^2+0,6y)*(-1,5y^3)=4,5y^5-0,9y^4.$

4)

$5x+3*(x-1)=6x+11\\5x+3x-3=6x+11\\8x-3=6x+11\\2x=14|:2\\x=7.$

Answer 2

Let's solve each equation step by step:

1. $\left(\frac{1}{8}\right)^2 \div 2^5 \times \left(\frac{1}{2}\right)^{10}$

First, simplify the exponents: $\left(\frac{1}{8}\right)^2 = \frac{1}{64}$ $2^5 = 32$ $\left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$

Now substitute the values back into the equation: $\frac{1}{64} \div 32 \times \frac{1}{1024}$

Divide $\frac{1}{64}$ by 32: $\frac{1}{64} \div 32 = \frac{1}{2048}$

Now multiply $\frac{1}{2048}$ by $\frac{1}{1024}$: $\frac{1}{2048} \times \frac{1}{1024} = \frac{1}{2097152}$

Therefore, $\left(\frac{1}{8}\right)^2 \div 2^5 \times \left(\frac{1}{2}\right)^{10} = \frac{1}{2097152}$.

2. $2x \cdot (x^2 - 8x - 2)$

Multiply $2x$ by $x^2 - 8x - 2$: $2x \cdot (x^2 - 8x - 2) = 2x^3 - 16x^2 - 4x$.

Therefore, $2x \cdot (x^2 - 8x - 2) = 2x^3 - 16x^2 - 4x$.

3. $(-3y^2 + 0.6y) \cdot (-1.5y^3)$

Multiply $(-3y^2 + 0.6y)$ by $(-1.5y^3)$: $(-3y^2 + 0.6y) \cdot (-1.5y^3) = 4.5y^5 - 0.9y^4$.

Therefore, $(-3y^2 + 0.6y) \cdot (-1.5y^3) = 4.5y^5 - 0.9y^4$.

4. $5x + 3(x - 1) = 6x + 11$

Distribute $3$ to $x - 1$: $5x + 3x - 3 = 6x + 11$

Combine like terms on both sides of the equation: $8x - 3 = 6x + 11$

Move $6x$ to the left side by subtracting $6x$ from both sides: $8x - 6x - 3 = 11$