Answer:
carnation $2.50; rose $3.00
Step-by-step explanation:
Let c = price of 1 carnation.
Let r = price of 1 rose.
3c + 5r = 22.5
3c + 2r = 13.50
Subtract the se3cond equation from the first equation.
3r = 9
r = 3
Now substitute r = 3 in the first equation and solve for c.
3c + 5r = 22.5
3c + 5(3) = 22.5
3c = 7.5
c = 2.5
Answer: carnation $2.50; rose $3.00
One nice thing about this situation is that you’ve been given everything in the same base. To review a little on the laws of exponents, when you have two exponents with the same base being:
– Multiplied: Add their exponents
– Divided: Subtract their exponents
We can see that in both the numerator and denominator we have exponents *multiplied* together, and the product in the numerator is being *divided* by the product in the detonator, so that translates to *summing the exponents on the top and bottom and then finding their difference*. Let’s throw away the twos for a moment and just focus on the exponents. We have
[11/2 + (-7) + (-5)] - [3 + 1/2 + (-10)]
For convenience’s sake, I’m going to turn 11/2 into the mixed number 5 1/2. Summing the terms in the first brackets gives us
5 1/2 + (-7) + (-5) = - 1 1/2 + (-5) = -6 1/2
And summing the terms in the second:
3 + 1/2 + (-10) = 3 1/2 + (-10) = -6 1/2
Putting those both into our first question gives us -6 1/2 - (-6 1/2), which is 0, since any number minus itself gives us 0.
Now we can bring the 2 back into the mix. The 0 we found is the exponent the 2 is being raised to, so our answer is
2^0, which is just 1.
We have:
Then we use trigonometric identities to change the negative sign of the trigonometric functions, so:
We clear f(x):
we simply what we can:
Thus, the correct answer is;
I'd say the 3rd one. I haven't done this stuff in 2 years ... me forgot. But if it's not right, then the 1st one!
I want you to know you're smart and you can do this! Good luck!
It is 9 and 8:3.It is because 8 and 3 do not have any factors that are the same especially because 3 is a prime number.A prime number is a number that has only a factor of itself and 1
