Please help!!!!!!!!!!

What is the value of x in the equation (5^x)^5=5^35

Arada [10]

$\bf (5^x)^5=5^{35}\implies 5^{x\cdot 5}=5^{35}\impliedby \begin{array}{llll} \textit{same bases}\\ \textit{same exponents} \end{array}\implies 5x = 35 \\\\\\ x = \cfrac{35}{5}\implies x=7$

6 0

2 years ago

48 L = ____ ML steps plz

Oliga [24]

Ok there are 48000 mL in 48L.

7 0

2 years ago

-1/15 +(-3/5)=, show ur work plz

sergij07 [2.7K]

Answer: -2/3

-1/15 + (-3/5)
-1/15 - 3/5

-3/5 = -9/15

-1/15 - 9/15
-10/15
-2/3

5 0

3 years ago

Read 2 more answers

The ratio of pencil to rulers is 6:10. the are 21 pencils. how many rulers are there

Reika [66]

There would be 35 rulers.

In order to solve this, use cross multiplication.

6/10 = 21/x -----> cross multiply

10*21 = 6*x

210 = 6x

35 = x

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

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Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

2 years ago

Please help!!!!!!!!!!​

Please help!!!!!!!!!!