Answer:
Step-by-step explanation:
a= b^c << === >> logb(a) = c
a b c
Convert 45 = 9^√x << === >> log9 (45) = √x
1.7324867604 = √x
3.0 = x squared both sides
I used a log calculator on the internet when a Gaggled log calculators
Answer:
x = 13 ; x = 5
Step-by-step explanation:
The easiest related equation to get is simply solving for x:
a) x + 5 = 18
Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Subtract 5 from both sides:
x + 5 (-5) = 18 (-5)
x = 18 - 5
x = 13*
b) 66x = 330
Isolate the variable, x. Divide 66 from both sides of the equation:
(66x)/66 = (330)/66
x = 330/66
x = 5*
*Note: An equation simply is a expression that has an equal sign. This means that as long as there is an equal sign, it counts as an equation.
~
8 feedings because she recorded it two times and giant humming Bird fed it 4x so 4×2=8
The x-value at the point, denoted by d, is 18
<h3>Slope of a line </h3>
From the question, we are to determine the x-value at point B on the line
Given the points (x₁, y₁) and (x₂, y₂) on a line.
The slope, m, of a line with two points (x₁, y₁) and (x₂, y₂) on the line is given by the formula
m = (y₂ - y₁)/(x₂ - x₁)
From the given information,
Point A is (2, 1)
Point B is (d, 81)
That is,
x₁ = 2
y₁ = 1
x₂ = d
y₂ = 81
and
slope, m = 5
Thus,
5 = (81 - 1) / (d - 2)
5 = 80/(d -2)
5(d - 2) = 80
5d - 10 = 80
5d = 80 + 10
5d = 90
d = 90/5
d = 18
Hence, the x-value at the point, denoted by d, is 18
Learn more on Slope of a Line here: brainly.com/question/19012287
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Answer: 3.61×10^5 A
Step-by-step explanation: Since the brain has been modeled as a current carrying loop, we use the formulae for the magnetic field on a current carrying loop to get the current on the hemisphere of the brain.
The formulae is given below as
B = u×Ia²/2(x²+a²)^3/2
Where B = strength of magnetic field on the axis of a circular loop = 4.15T
u = permeability of free space = 1.256×10^-6 mkg/s²A²
I = current on loop =?
a = radius of loop.
Radius of loop is gotten as shown... Radius = diameter /2, but diameter = 65mm hence radius = 32.5mm = 32.5×10^-3 m = 3.25×10^-2m
x = distance of the sensor away from center of loop = 2.10 cm = 0.021m
By substituting the parameters into the formulae, we have that
4.15 = 1.256×10^-6 × I × (3.25×10^-2)²/2{(0.021²) + (3.25×10^-2)²}^3/2
4.15 = 13.2665 × 10^-10 × I/ 2( 0.00149725)^3/2
4.15 = 1.32665 ×10^-9 × I / 2( 0.000058)
4.15 × 2( 0.000058) = 1.32665 ×10^-9 × I
I = 4.15 × 2( 0.000058)/ 1.32665 ×10^-9
I = 4.80×10^-4 / 1.32665 ×10^-9
I = 3.61×10^5 A
