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2 years ago

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A man divided $9,000 among his wife, son, and daughter. The wife received twice as much as the daughter, and the son received $1

,000 more than the daughter. How much did each receive?

1 answer:

mestny [16]2 years ago

7 0

Answer:4000

Step-by-step explanation: becuase each would get 3000 and he got 1000 so 3000+1000=4000

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saw5 [17]

B.30

cause 60 is 30 halted so yeah

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3 years ago

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If B=3x-1 and C=2x^2+x+9, find an expression that equals 2B+2C in standard form.

Anestetic [448]

Answer:

$4x^2 +8x + 16$

Step-by-step explanation:

2 (3x - 1) + 2 (2x^2 + x + 9)

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3 years ago

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 <span>log(x)=ln(x)</span><span>.

The limit result can be proven if the base of </span><span>log(x)</span><span> is 10.
</span>
$\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}$

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

</span> $\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1$
<span>
Therefore,

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

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3 years ago

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Alisiya [41]

Answer:

C

Step-by-step explanation:

A function cannot have any repeated x-coordinates; therefore C is the correct answer.

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Ratling [72]

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