Factoring review. factor the following polynomials by determining a gcf. 21x-15y
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3(7x -5y)
Related Questions
Two cards are randomly selected, without replacement, from a standard deck of playing cards. What is the probability of first picking a five and then picking a card other than a five? I'm having a time with determining which formula to use?
Approximately how many miles are in a light-year?a. 6 × 106 b. 6 × 109 c. 6 × 1012 d. 6 × 1015
Help on this pleasee
Money is invested into an account earning 4.24% interest compounded annually. If the accumulated value after 18 years will be $25,000, approximately how much money is presently in the account?
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Final answer:
James and Anna walk their dogs every 3 and 4 days respectively. The Least Common Multiple of 3 and 4 is 12, so they will walk their dogs together every 12 days. Since they walk their dogs on a Monday, they will walk them together on a Monday.
Explanation:
This question is about finding the Least Common Multiple (LCM) of two numbers. In this case, the two numbers are the days James and Anna walk their dogs: 3 days and 4 days. The LCM of 3 and 4 is 12. Therefore, James and Anna will walk their dogs together every 12 days.
If they both walk their dogs on a Monday, 12 days later will also be a Monday. So, they will walk their dogs together on the same day of the week, which is Monday.
Learn more about Least Common Multiple here:
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you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before. But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.
Consider this: (2)^a negative power = (1/2)^the same power but positive.
So:
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.
What I just said in that paragraph was: log₂ of(N) = - log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.
Now let's look at the problem:
log₂(x-1) + log(base 1/2) (x-2) = log₂(x)
Subtract log₂(x) from each side:
log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0
Subtract log(base 1/2) (x-2) from each side:
log₂(x-1) - log₂(x) = - log(base 1/2) (x-2) Notice the negative on the right.
The left side is the same as log₂[ (x-1)/x ]
==> The right side is the same as +log₂(x-2)
Now you have: log₂[ (x-1)/x ] = +log₂(x-2)
And that ugly [ log to the base of 1/2 ] is gone.
Take the antilog of each side:
(x-1)/x = x-2
Multiply each side by 'x' : x - 1 = x² - 2x
Subtract (x-1) from each side:
x² - 2x - (x-1) = 0
x² - 3x + 1 = 0
Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .
I think you have to say that x=2.618 is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.
There,now. Doesn't that feel better.
A) $20; They paid $20 for the food.
B) (4h + 3f); Find the total cost by adding the price of 4 hamburgers to the price of 3 fries
C) 4h + 3f + 6; Find the total cost by adding the price of 4 hamburgers to the price of 3 fries to the price of 2 milkshakes.
D) c = 20 - (4h + 3f + 6); Find the total cost by subtracting the price of 4 hamburgers to the price of 3 fries to the price of 2 milkshakes.
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For this case we have the following variables:
h: Cost of each hamburger bought by Jana and her friend
f: Cost of each order of potato chips bought by Jana and her friend
b: Cost of each milkshake
If you bought 4 hamburgers, 3 orders of potatoes and 2 milkshakes, you have a cost of:
It is known that each shake costs $ 3, so$
Substituting we have:
Thus, the total cost is given by:
Answer:
Option C
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Do the opposite operation on 9 so you can get the like terms one side.
3a = 45
Then divide 3 on both side to get the variable by itself.
a = 15
Answer:
fifteen
Step-by-step explanation:
Got it right on Edg