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2 years ago

2 Points Which expression is equal to the expression below? (2 • 64

Mathematics

1 answer:

Verdich [7]2 years ago

4 0

Answer:

2(1 x 32)

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While traveling to Europe phlegm exancheged 250 US dollars he spent 150 euros on his trip after returning to the United States h

anzhelika [568]

1euro = 1.3687 USD

So 150 Euros = 150*1.3687 USD

=205.305 USD

So he spent $205.305 USD on his trip.

He started off with $250 so to find the amount left we just take

250 - 205.305

= $44.695

Or 44 dollars and 69.5 cents. It's a weirdly exact amount, sure, but the question had a very precise exchange rate, so we'll assume this is fine.

4 0

2 years ago

80,531 rounded to nearest thousand

Vlad [161]

Answer:

81,000

Step-by-step explanation:

go to whatever place it is asking you to round nearest to the if it is 5 or more you round up 4 or less it stays the same.

Saying for this -" 5 or more raise the score, 4 or less let it rest"

^round up^ ^stays the same^

4 0

2 years ago

Read 2 more answers

Eliza estimated she would spend $120 at the grocery store. She actually spent $95. What is the percent error?

Amanda [17]

Answer: 26.32% (rounded to the hundredths)

6 0

2 years ago

Any help please for my class

sesenic [268]

Answer:

Step-by-step explanation:

do 700/20

700/20 = $35 each

hope this helps <3

4 0

2 years ago

Read 2 more answers

Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))

e-lub [12.9K]

$sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0$

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,

$\large \lim_{ x \to 0 } ~ x^x = 1$

I will make the assumption that log(x)=ln(x).

The limit result can be proven if the base of log(x) is 10.

$\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}$

We get the indeterminate form 0/0, so we have to use Lhopitals rule

 $\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1$

Therefore,

 $\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } =\boxed{ -1}$

3 0

2 years ago