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2 years ago

Which property is used in the first step to solve this equation?

Mathematics

2 answers:

SOVA2 [1]2 years ago

6 0

Answer:

distributive

Step-by-step explanation:

Nat2105 [25]2 years ago

4 0

Answer:

distributive property

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Answer should be 3 I’m pretty sure

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What's the sum of 5 feet 9 inches in 12 ft and 3 in

galben [10]

First convert everything into inches.

1 foot = 12 in
5 feet = 5*12 = 60 in
12 feet = 12*12 = 144 in

Now add them together

60 + 9 + 144 + 3 = 216 in

or

18 feet

Hope this helps :)

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2 years ago

How many strings of eight uppercase English letters are there if no letter can be repeated?

zepelin [54]

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3 years ago

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

On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove

gogolik [260]

Velocity, distance and time:

This question is solved using the following formula:

$v = \frac{d}{t}$

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time $t_1$ , distance of $d_1$ , v = 40. So

$v = \frac{d}{t}$

$40 = \frac{d_1}{t_1}$

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is $v = 60$ .

On the first day, he drove 2 more hours, which means that for the second day, the time is: $t_1 - 2$

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: $d_1 - 20$

Thus

$v = \frac{d}{t}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

System of equations:

Now, from the two equations, a system of equations can be built. So

$40 = \frac{d_1}{t_1}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

Find the total distance traveled in the two days:

We solve the system of equation for $d_1$ , which gets the distance on the first day. The distance on the second day is $d_2 = d_1 - 20$ , and the total distance is: