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2 years ago

Solve the logarithmic equation . Round the answer to four decimal places . 2^x =25

Mathematics

2 answers:

levacccp [35]2 years ago

8 0

Answer:

x = 4.6439

Step-by-step explanation:

2^x =25

Take the log of each side

log 2^x = log 25

We know that log a^b = b log a

x log 2 = log 25

Divide each side by log 2

x = log 25 / log 2

x =4.643856

Rounding to 4 decimal places

x = 4.6439

Korvikt [17]2 years ago

4 0

Answer:

x = 4.6439

Step-by-step explanation:

$2^{x} =25log 2^x = log 25\\log 2 = log 25x = log 25 / log 2\\x =4.643856\\x = 4.6439$

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Caiden and Josh are practicing at a batting cage. Josh has hit nine more than one-half the number of balls that Caiden has. If t

OverLord2011 [107]

Answer:

The Answer is: Josh hit 47 balls.

Step-by-step explanation:

Let j = Josh number of hits and c = Caiden number of hits:

j = c/2 + 9

Total:

j + c = 123.

Substitute:

(c/2 + 9) + c = 123

c/2 + c = 123 - 9

c/2 + c = 114

1/2c + 2/2c = 114

3/2c = 144

c = 144 (2/3) = 76, Number of hits for Caiden

Solve for j:

j = 76/2 + 9

j = 38 + 9 = 47, number of hits for Josh

Proof:

j + c = 123

47 + 76 = 123

123 = 123

Hope this helps! Have an Awesome Day!! :-)

6 0

3 years ago

Question 17 i need help answer is -16 but don’t know how

Mila [183]

Answer:

-16

Step-by-step explanation:

-k² -(3k-5n)+4n k=-1 , n= -2

= -(-1)² - ( 3(-1) -5(-2) )+4(-2)

= -1-(-3+10)-8

= -1-(7)-8

= -1-7-8

= -16

4 0

2 years ago

What is the following sum? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0

Free_Kalibri [48]

Answer:

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}$

Hence, ption B is true.

Step-by-step explanation:

Given the expression

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

solving the expression

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

$\sqrt{x^2y^3}=xy\sqrt{y}$

$2\sqrt{x^3y^4}=2xy^2\sqrt{x}$

so the expression becomes

$\:\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}$

Group like terms

$=2xy^2\sqrt{x}+xy\sqrt{y}+xy\sqrt{y}$

Add similar elements

$=2xy^2\sqrt{x}+2xy\sqrt{y}$

Therefore, we conclude that:

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}$

Hence, option B is true.

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2 years ago

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Quick Question! Please help!

8090 [49]

Answer:

C --- b^16

Step-by-step explanation:

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2 years ago

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Objects A and B are in harmonic motion modeled by

Zepler [3.9K]

Answer:

The phase difference between yA and yB is $\frac{-\pi }{12}$

Step-by-step explanation:

Given harmonic modeled as :

yA = 8 sin(2t - $\frac{\pi }{3}$ ) And

yB = 8 sin(2t - $\frac{\pi }{4}$ )

The function as written as :

y = a sin(ωt - Ф) where Ф is phase difference

So , phase difference between yA and yB = ( Ф_1 - Ф_2 )

Or phase difference between yA and yB = ( - $\frac{\pi }{3}$ + $\frac{\pi }{4}$ )

Or, phase difference between yA and yB = $(\frac{-4\pi +3\pi }{12})$

I.e phase difference between yA and yB = $\frac{-\pi }{12}$

Hence The phase difference between yA and yB is $\frac{-\pi }{12}$ Answer

4 0

3 years ago

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