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4 years ago

Graph the logarithmic equation

Mathematics

2 answers:

harkovskaia [24]4 years ago

6 0

x=1, y=0,

x=5, y=1,

x=25, y=2,

x=1/5, y= -1

PtichkaEL [24]4 years ago

3 0

BTW: it would be called a log with the base of 5, or log base 4

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4.10.1 checkup, answer questions about the unit circle

zavuch27 [327]

Using the unit circle, the correct answer regarding the comparison of angles w and t is given by:

a. t > w.

<h3>What is the unit circle?</h3>

For an angle $\theta$ the unit circle is a circle with radius 1 containing the following set of points: $(\cos{\theta}, \sin{\theta})$ .

From this, we get that the sine is represented by the vertical axis. As we advance into the second quadrant, as the angle increases, sine decreases. Hence, since sin w > sin t, we have that t > w and option a is correct.

More can be learned about the unit circle at brainly.com/question/16852127

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2 years ago

What is equavilent to (10x+7r-r^(2))+(-6r^(2)-18+5r)

mote1985 [20]

Answer:

−7r^(2)+12r+10x−18

Step-by-step explanation:

Grab the original equation: 10x+7r−r^(2)−6r^(2)−18+5r

For subtraction bits, treat them as negatives: 10x+7r+−r^(2)+−6r^(2)+−18+5r

Combine like terms: (−r^2+−6r^2)+(7r+5r)+(10x)+(−18)

Simplify that, and you get your final answer: −7r^(2)+12r+10x+−18

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3 years ago

Not shure how to do this someone help!!

Anika [276]

-- the greatest common factor of the 15, the 9, and the 24 is 3 .

-- the greatest common factor of the b⁴, the b⁵, and the b³ is b³.

-- the greatest common factor of the c³, the c², and the c¹, is c¹.

-- the greatest common factor of the d², the d¹, and the d³ is d¹.

-- SO, the greatest common factor of the whole thing is ( 3 b³ c d ) .

Can you factor that out of the big mess ?

Please try it.

You should get (3 b³cd) (5 bc²d - 3 b²c - 8 d²) .

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3 years ago

Combine like terms to write an equivalent expression.

Serjik [45]

Answer:

1. 6m+7

2. 5x^2+x+2

3. 9n-9

4. x^2+3x+4

Step-by-step explanation:

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3 years ago

A suspension bridge with weight uniformly distributed along its length has twin towers that extend 85 meters above the road surf

Dmitry_Shevchenko [17]

Answer:

The required height of the cable is 21.25 meter

Step-by-step explanation:

Consider the provided information.

A suspension bridge with weight uniformly distributed along its length has twin towers that extend 85 meters above the road surface and are 800 meters apart.

Let the road surface is the x-axis, and the point (0,0) represents the point that is on the road surface midway between the two towers.

Now it will form an upward parabola whose vertex is at (0,0), one on either side of vertex at a distance x= 400 or x= -400, and y for each of these points is 85.

The equation of parabola is: $y=ax^2$