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laila [671]
3 years ago
6

I need the answer as fast as i can get it!

Mathematics
1 answer:
blondinia [14]3 years ago
7 0

Answer:

The answer should be choice D.

HOPE THIS HELPS! :)

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[8-(-12)]÷(2x-2)-(-8)
WITCHER [35]

Answer:

3

Step-by-step explanation:

6 0
3 years ago
Find the volume of the figure. Use 3.14 for pi
Natasha2012 [34]

Answer:

13

Step-by-step explanation:

6 0
3 years ago
Which of the following shows the extraneous solution(s) to the logarithmic equation? log4 (x) + log4(x - 3) = log4 (-7x + 21)
Salsk061 [2.6K]
First join the log4 on the left:

log4( x*(x-3) = log4(-7x+21)

Then x = -7, works: -7*(-10)=70 = -7*(-7)+21

x=-3, 18 = 42, does not work

x=3 0=0 works,

However, when one puts x = -7 in the *original* exression, log4(-7) or log4(-10) do not exist (you know why?). So x= -7 is extraneous.

Now x=3 gives log4(0) on the left and right, which does not exist.

So, C is the answer, both are extraneous. Seem to work but indeed don't work in the *original* equation
 
6 0
3 years ago
Read 2 more answers
Is Figure 1 similar to Figure 2? ​
tamaranim1 [39]

Answer:

no figure 1 isn't similar to figure 2

Step-by-step explanation:

because its sides aren't the same numbers nor do the numbers correlate with the numbers in figure one its 2 different numbers

6 0
3 years ago
A boat is pulled toward a dock by means of a rope wound on a drum that is located 6 ft above the bow of the boat. If the rope is
Monica [59]

Answer:

Step-by-step explanation:

Given that:

The height of the dock (h) = 6

Let represent d to be the distance between the boat and the dock

Let the length of the rope between the boat and the drum be denoted by (l)

Then, the rate of change for the length of the rope be:

dl/dt = -5 ft/s

Using Pythagoras rule to determine the relationship between these values, we have:

l^2 = h^2 +d^2

l^2 = 6^2 + d^2

l^2 = 36 + d^2

We relate to:  2l * \dfrac{dl}{dt} = 2d* \dfrac{dd}{dt}

From the question;

l = 34,

So to find \dfrac{dd}{dt}, we get;

d = \sqrt{l^2 - 36}

d = \sqrt{34^2 - 36}

d = \sqrt{1156- 36}

d = \sqrt{1120}

d = 33.46

So, we have:

\dfrac{dd}{dt}= \dfrac{l}{d} \times \dfrac{dl}{dt}

\dfrac{dd}{dt}= \dfrac{34}{33.46} \times - 5

\dfrac{dd}{dt}= 1.016 \times - 5

\dfrac{dd}{dt}=-5.08 \ ft/sec

4 0
3 years ago
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