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KatRina [158]
2 years ago
15

Use the Multiplication Law of Exponents to solve the expression below. 6^1/2 6^1/2 = _____

Mathematics
1 answer:
klemol [59]2 years ago
8 0

The base of 6 is the same, so add 1/2+1/2 =1

6^1 = 6

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frozen [14]

Pick any number for n        n(y) = n(3/2x + 4)        n(y) = n(3/2x) + n(4) 
7 0
3 years ago
25pts awarded and brainliest awarded, plz help asap!!!!!!
ivann1987 [24]

Answer:

B. f(-1)=6

C.  The domain of f(x) is the set {-2.-1,0,1,2,3,4,5,6}

Step-by-step explanation:

To find f(5) from the table means, the y-value that corresponds to x=5.

This value is 12.

This implies that:

f(5)=12

Also the y-value that corresponds to -1 is 6.

Hence f(-1)=6

The domain of f(x) are the set of all the x-values.

The domain is : {-2.-1,0,1,2,3,4,5,6}

The range for f(x) is the set of all the corresponding y-values.

From the table, the range is: {5,6,7,8,9,10,11,12,13}

4 0
3 years ago
A rocket car on the bonneville salt flats is traveling at a rate of 640 miles per hour. How much time would it take for the car
luda_lava [24]
D = r t 384 = 640 * t 384/640 = t .6 = t .6 of an hour .6 hr * 60min/1 hr = 36 minutes (just in case it needs to be in minutes)
4 0
2 years ago
Read 2 more answers
4) The path of a satellite orbiting the earth causes it to pass directly over two
Naily [24]

Answers:

  • Satellite is approximately <u>2446.43 km</u> from station A.
  • Satellite is approximately <u>2441.61 km</u> above the ground.

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

Explanation:

I'm assuming tracking stations A and B are at the same elevation and are on flat ground. In reality, this is likely not the case; however, for the sake of simplicity, we'll assume this is the case.

The diagram is shown below. Points A and B describe the two stations, while point C is the satellite's location. Point D is on the ground directly below the satellite. We have these lengths

  • AB = 60 km
  • AD = x
  • CD = h

Focusing on triangle ACD, we can apply the tangent rule to isolate h.

tan(angle) = opposite/adjacent

tan(A) = CD/AD

tan(86.4) = h/x

x*tan(86.4) = h

h = x*tan(86.4)

We'll use this later in the substitution below.

--------------------

Now move onto triangle BCD. For the reference angle B = 85, we can use the tangent rule to say

tan(angle) = opposite/adjacent

tan(B) = CD/DB

tan(B) = CD/(DA+AB)

tan(85) = h/(x+60)

tan(85)*(x+60) = h

tan(85)*(x+60) = x*tan(86.4) .............  apply substitution; isolate x

x*tan(85)+60*tan(85) = x*tan(86.4)

60*tan(85) = x*tan(86.4)-x*tan(85)

60*tan(85) = x*(tan(86.4)-tan(85))

x*(tan(86.4)-tan(85)) = 60*tan(85)

x = 60*tan(85)/(tan(86.4)-tan(85))

x = 153.612786190499

--------------------

We'll use this approximate x value to find h

h = x*tan(86.4)

h = 153.612786190499*tan(86.4)

h = 2441.60531869599

h = 2441.61 km  is how high the satellite is above the ground.

Return to triangle ACD. We'll use the cosine rule to determine the length of the hypotenuse AC

cos(angle) = adjacent/hypotenuse

cos(A) = AD/AC

cos(86.4) = x/AC

cos(86.4) = 153.612786190499/AC

AC*cos(86.4) = 153.612786190499

AC = 153.612786190499/cos(86.4)

AC = 2446.43279498247

AC = 2446.43 km is the distance from the satellite to station A.

6 0
3 years ago
Find an equation to the line tangent to y = 5 + |x – 2| at the coordinate (2, 5) pls answer
myrzilka [38]

Answer:

Step-by-step explanation:

Finding an equation of a tangent line to that function requires that we find the derivative of the function at that point. Since this is an absolute value function with its cusp at (2, 5), the function is not differentiable here.

3 0
3 years ago
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