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1 year ago

Can I take a picture of the problem and upload for you to see?

Mathematics

1 answer:

eduard1 year ago

6 0

Yes you can …………………………

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Look at the subtraction expression below. ( Please. Help. Quick.)

dimulka [17.4K]

Answer:

Step-by-step explanation:

Because they are both subtracting, both lines have to go to the left

4 0

2 years ago

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SOMEONE HELP ME this is due in like less than 15 mins

zmey [24]

Answer:

for the first one I think it is -5/-2 second is 1/-4 third one is 2/1 this may not be correct

Step-by-step explanation:

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3 years ago

the texas state legislature is composed of state senators and state representatives. the sum of the number of senators and repre

Thepotemich [5.8K]

Answer:

Step-by-step explanation:

s + r = 181

r = 119 + s

s + 119 + s = 181

2s + 119 = 181

2s = 181 - 119

2s = 62

s = 62/2

s = 31 <=== there are 31 senators

r = 119 + s

r = 119 + 31

r = 150 <=== there are 150 representatives

4 0

2 years ago

-7 times 8 = 7 times (-8) true or false

12345 [234]

Answer:

Yaa it's true

Step-by-step explanation:

-56=-56

from both side u get the same answer

5 0

2 years ago

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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