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2 years ago

Apply the distributive property to simplify the expression -5 (-3x-6)

Mathematics

1 answer:

VARVARA [1.3K]2 years ago

8 0

Answer:

15x+30

Step-by-step explanation:

-5*-3x= 15x

-5*-6= 30

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Gas mileage actually varies slightly with the driving speed of a car (as well as with highway vs. city driving). Suppose your

natima [27]

Answer:

Step-by-step explanation:

Suppose the car mileage actually varies slightly with the driving speed of the car

Suppose your car averages 26 miles per gallon on the highway if your average speed is 45 miles per hour,

and it averages 18 miles per gallon on the highway if your average speed is 65 miles per hour.

The driving time for a 2500-mile trip if you drive at an average speed of 54 miles per hour

$= \frac{Distance}{Speed} = \frac{2500 }{45} \\\\= 55.56 hours.$

The driving time for a 2500-mile trip if you drive at an average speed of 65 miles per hour

$= \frac{Distance}{Speed} = \frac{2500 }{65} \\\\= 38.46 hours.$

7 0

3 years ago

Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]

Schach [20]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

$\rm \: = \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

$\rm \: = \: \dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }$

Now, Consider Denominator, we have

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

Now, we know that,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}$

So, using this, we get

$\rm \: = \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

$\rm \: = \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \: = \: \dfrac{e - 1}{e}$

$\rm \: = \: 1 - \dfrac{1}{e}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} } = \frac{e - 1}{e} = 1 - \frac{1}{e}}}$

3 0

2 years ago

Find the first three iterates of the function f(z) = 2z + (3 - 2i) with an initial value of z0 = 5i.

Agata [3.3K]

3 because the 2z minus the 3 gives you 1 rounded

4 0

3 years ago

Read 2 more answers

A triangle has sides with the following lengths: AB=3, BC=4, CA=5

scoundrel [369]

Complete the question properly

3 0

3 years ago

In planning her retirement, Liza deposits some money at 2% interest, twice as much deposited at 2.5%. What is the amount deposit

harkovskaia [24]

She deposited two amounts, let's say "a" and "b"
"a" earning 2% interest
"b" earning 2.5% interest

the 2.5% one, is twice as much as the 2% one
so.. one can say that, whatever "a" is, "b" is twice as much,
or b = 2*a -> b = 2a

now.. .the sum of both earned interest, was $1190

so.. one can say that (2% of a) + (2.5% of b) = 1190
let' us use the decimal notation then
$\frac{2}{100}a+\frac{2.5}{100}b=1190\implies 0.02a+0.025b=1190
\\ \quad \\
\textit{however, we know "b" is 2a thus}
\\ \quad \\
0.02a+0.025\boxed{b}=1190\implies 0.02a+0.025(\boxed{2a})=1190$

solve for "a" to find what the "a" amount is,
to find "b", well, "b" is "2a", so just do a 2*a to get "b"

5 0

3 years ago

Apply the distributive property to simplify the expression -5 (-3x-6)​

Apply the distributive property to simplify the expression -5 (-3x-6)