1 yogurt = 45p
1p = ? yogurt
1 ÷ 45 = 0.02
5 × 0.02 = 0.10
0.10 yogurt can be bought with 5p (pounds)
Answer:
10 / 12
simplified:
5/6
since 5 is a prime number, we can't simplify this ratio any further
X²+3x-21=0
1) we solve this square equation:
x=[-3⁺₋√(9+84)] / 2=(-3⁺₋√93)/2
We have two solutions:
x₁=(-3-√93)/2
x₂=(-3+√93)/2
2) we compute the product of the 2 solutions found.
[(-3-√93)/2][(-3+√93)/2] =(-3-√93)(-3+√93) / 4=
=(9-93)/4=-84/4=-21
Answer: the product of the 2 solutions of this equation is -21
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.
__
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.
__
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.
10 is the answer
why are you struggling
so simple
<h2>
<em><u>hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>helps</u></em><em><u> </u></em><em><u>you</u></em><em><u> </u></em><em><u>❣❣</u></em><em><u> </u></em></h2>
<em><u>Mark</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>brainliest</u></em><em><u> </u></em>