To solve problem 19, we must remember the order of operations. PEMDAS tells us that we should simplify numbers in parentheses first, exponents next, multiplication and division after that, and finally addition and subtraction. Using this knowledge, we can begin to simplify the problem by working out the innermost set of parentheses:
36 / [10 - (3-1)²]
36 / [10 - (2)²]
Next, we should still simplify what is inside the parentheses but continue to solve the exponents (the next letter in PEMDAS).
36/ (10-4)
After that, we should compute the subtraction that is inside the parentheses.
36/6
Finally, we can solve using division.
6
Now, we can move onto problem 20:
1/4(16d - 24)
To solve this problem, we need to use the distributive property, which allows us to distribute the coefficient of 1/4 through the parentheses by multiplying each term by 1/4.
1/4 (16d-24)
1/4(16d) - 1/4(24)
Next, we can simplify further by using multiplication.
4d - 6
Therefore, your answer to problem 19 is 6 and the answer to problem 20 is 4d -6.
Hope this helps!
Lightly draw a vertical line at -3 (x= -3 is the divider) on the left side graph x + 1, on the right side graph 1/2x + 2.
because these equations don't match up at the -3 coordinate, you have to draw an open (not colored in) circle at the beginning at the 1/2x-2 equation
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.
So k in the first problem would =1/2
and i guess in the second problem just put k in place of the first space and in the second space put 8 for x. y would then =4. Idk if I did this right, though, I'm honestly a little confused too.
Answer:
19.4
Step-by-step explanation:
16^2+11^2=377
then square 377 which will equal to
19.416 which can be rounded up to 19.4
