Answer:
Step-by-step explanation:
Answer:
A) l = w + 900
B) l * w = 4,000,000
B) l = 4,000,000 / w putting this into A)
A) 4,000,000 / w = w + 900
A) 4,000,000 = w^2 +900 w
A) w^2 +900 w - 4,000,000
Solving by quadratic formula
width = 1,600 length = 2,500
Double-Check
1,600 * 2,500 = 4,000,000
Length = width + 900
2,500 = 1,600 + 900
Answer is okay
Step-by-step explanation:
The distance the ships traveled are like the legs of a triangle and the question wants to know the hypotenuse. To find the hypotenuse, use the pythagorean theorem. this is a^2 + b^2 = c^2, with a and b being the legs and c being the <span>hypotenuse.
</span>Plug in known values:
84^2 + 62^2 = c^2
Solve:
84^2 = 7056
62^2 = 3844
7056 + 3844 = c^2
7056 + 3844 = 10900
10900 = c^2
Now you just need to isolate c by finding the square root of both sides.
√10900 = 104.403
√c^2 = c
So c = 104.403, or just 104.40 when rounded to the nearest tenth.
And if c is 104.40, then that means the hypotenuse is 104.40.
And all of that basically means that the distance between the ships is 104.40 miles.
Find area of square: 7^2 = 49
Find radius of circle: 7/2=3.5
find area of circle: 3.5^2*3.14=38.465
Remove area of circle from area of square:
49-38.465=10.535 cm^2
Answer A is the correct one :)
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
