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vredina [299]
3 years ago
6

Algebra help. Zoom in if you can

Mathematics
1 answer:
Artist 52 [7]3 years ago
7 0
Answers:

Pair-1: log_{2}x = 5 ↔ 32
Pair-2: log_{10}x = 3 ↔ 1000
Pair-3: log_{4}x = 2 ↔ 16
Pair-4: log_{5}x = 4 ↔ 625

Explanation:

Pair-1:
log_{2}x = 5 \\  \frac{ln(x)}{ln(2)} = 5 \\ ln(x) = 5 * ln(2) \\ x = e^{5*ln(2)} \\ x = 32

Pair-2:
log_{10}x = 3 \\ \frac{ln(x)}{ln(10)} = 3 \\ ln(x) = 3 * ln(10) \\ x = e^{3*ln(10)} \\ x = 1000

Pair-3:
log_{4}x = 2 \\ \frac{ln(x)}{ln(4)} = 2 \\ ln(x) = 2 * ln(4) \\ x = e^{2*ln(4)} \\ x = 16

Pair-4:
log_{5}x = 4 \\ \frac{ln(x)}{ln(5)} = 4 \\ ln(x) = 4 * ln(5) \\ x = e^{4*ln(5)} \\ x = 625
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Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
adoni [48]

Answer: The required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

Step-by-step explanation:

Since we have given that

y=\ln[x(2x+3)^2]

Differentiating log function w.r.t. x, we get that

\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}

Hence, the required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

3 0
3 years ago
Math help guys how wouls i work this out
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8 > 7 + \frac{x}{6}   Subtract 7 from both sides

1 > \frac{x}{6}   Multiply both sides by 6

6 > x   Flip it around so it's easier to read

x < 6

You can graph your answer by drawing an open circle at the 6 and coloring the line to the left. The circle should be open, because x is <em>less than 6</em>, not less than or equal to. You would color to the left to show that x can be anything less than 6.

4 0
3 years ago
Please help me ASAP. ​
igomit [66]
<h3><u>Solution: </u></h3>

Radius of Cylindrical Pillar , r = 28 Cm = 0.28 m.

  • Height, h= 4 m.

Curved surface area of a Cylindrical = 2πrh.

Curved surface area of a pillar -

=  >  \: 2 \times   \frac{22}{7}  \times 0.28 \times 4 = 7.04 {m}^{2}

Curved surface area of 24 such Pillar :-

(7.04 x 24 = 168.96m²)..

  • Cost of painting an area of 1 m² = Rs8...

• Therefore , cost of painting 1689.6m² :-

( 168.96 x 8 = Rs 1351.68)...

3 0
2 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
7 0
3 years ago
hi I'm Danica and it's sixth grade I would really like if you look over this problem and tell me the answer(sending it rn)!!!!!!
vaieri [72.5K]
Translation
explanation: it is shifting
6 0
3 years ago
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