Answer:
First you do what is inside the parentheses. 10 x 5 - 3. 50 - 3 =47. Then you multiply 47 x 5 = 235.
Step-by-step explanation:
Answer:
15.07 Euro
Step-by-step explanation:
Let me express the equation clearly:
lim x→-9 (x²-81)/(x+9)
Initially, we solve this by substituting x=-9 to the equation.
((-9)²-81)/(-9+9) = 0/0
The term 0/0 is undefined. This means that the solution is not see on the number line because it is imaginary. Other undefined terms are N/0 (where N is any number), 0⁰, 0×∞, ∞-∞, 1^∞ and ∞/∞. One way to solve this is by applying L'Hopitals Rule. This can be done by differentiating the numerator and denominator of the fraction independently. Then, you can already substitute the x=-9.
(2x-0)/(1+0) = 2x = 2(-9) = -18
The other easy way is to substitute x=-8.999 to the original equation. Note that the term x→-9 means that x only approaches to -9. Thus, you substitute a number that is very close to -9. Substituting x=-8.999
((-8.999)²-81)/(-8.999+9) = -18
Multiply the numerators and denominators together:
-24/150
Divide the numerator and denominator by 6:
-24 / 6 = -4
150 / 6 = 25
-4/25
Hi there!
I can see why you've posted a question like this here - For adding, subtracting, multiplying, or dividing fractions by whole numbers can be difficult to come out with a precise answer, so let me make it look easier for you!
So, the steps are:
Smush the 2 (whole number) and the 2 (the numerator of the fraction) together on top so its (2)(2) over 6. I can show you what I mean, if you'd like.
This is what it should look like.
Now, what is 2 times 2? You can use a calculator for this part.
Now, we can get to the last part, my favorite one!
Simplify 
down to 
.
How did we do it? EASY!
See what number does evenly into both numerators and denominators. In this case, 2 goes into 4 twice, and 2 goes into 6 three times, so we're left with our beautiful answer .
That wasn't so bad, was it? You can use this as a guide for figuring out math problems like these, in case you're stuck again. But just in case, message me if you need anymore help! :D
