Answer:
sum of angles in a quadrilateral is 360
Step-by-step explanation:
3x-15 +3x+15+y+30+D=360
hope that explanation helps
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.
The standard equation of a circle is:

where (h,k) is the center and r is the radius.
We need to make an equation of a circle, whose center is (11,-5) with any radius. Therefore, substitute the value.

The bolded part is your answer. Since the radius is not given, we can put any value.
Answer:
3
Step-by-step explanation:
2x + 3 = -x + 12
subtract 3 from both sides
2x = - x + 9
add x to both sides
3x = 9
divide by 3
x = 3
Answer:
m∠MON = 15°
Step-by-step explanation:
The given parameters are;
m∠LON = 77°
m∠LOM = 9·x + 44°
m∠MON = 6·x + 3°
By angle addition postulate, we have;
m∠LON = m∠LOM + m∠MON
Therefore, by substituting the known values, we have;
∴ 77° = 9·x + 44° + 6·x + 3°
77° = 9·x + 44° + 6·x + 3° = 15·x + 47°
77° = 15·x + 47°
77° - 47° = 15·x
15·x = 77° - 47° = 30°
15·x = 30°
x = 30°/15 = 2°
x = 2°
Given that m∠MON = 6·x + 3° and x = 2°, we have;
m∠MON = 6 × 2° + 3° = 12° + 3° = 15°
m∠MON = 15°.