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3 years ago

23X67 Estimate then find the prouduct

1 answer:

3 0

To estimate, you would usually round to a significant figure
so, 20* 70=1400

The actual answer is 1521, so we know we are close

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How to solve logarithmic equations as such

Serga [27]

$\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5$

$\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}$

5 0

3 years ago

A variable needs to be eliminated to solve the system of equations below. Choose the

sasho [114]

Step-by-step explanation:

-4x - 2y = 18

+ (4x - 4y = -36)

=> -6y = -18, y = 3

As we added the equations together, we have eliminated x and found y.

Hence the correct answer is "Add to eliminate x".

4 0

3 years ago

What is the coefficient in this expression? <br> 17 + 5a

babymother [125]

The answer is the a next to the 5

4 0

4 years ago

Read 2 more answers

Add 1} + (-5) using the number line.

gizmo_the_mogwai [7]

Answer:

1/2

Step-by-step explanation:

1 1/3+(-5/6)

1 1/3=4/3

4/3+(-5/6)

4/3-5/6

8/6-5/6

3/6

simplify

1/2

6 0

3 years ago

Find the volume <br><br> Help me please

Darina [25.2K]

Answer:

1372/3 π

that's the volume

4 0

3 years ago

Read 2 more answers