Answer:
4r ^2 +9r + 12
Step-by-step explanation:
Hope this helps! :)
Answer: To clear an equation of decimals, multiply each term on both sides by the power of ten that will make all the decimals whole numbers. In our example above, if we multiply .25 by 100, we will get 25, a whole number. Since each decimal only goes to the hundredths place, 100 will work for all three terms.
So let's multiply each term by 100 to clear the decimals:
(100)0.25x + (100)0.35 = (100)(-0.29)
25x + 35 = -29
Now we can solve the equation as normal:
25x + 35 - 35 = -29 - 35
25x = -64
x = -2.56 Since the original was in decimal form, the answer should most likely also be in decimal form.
Let's look at one more:
1.75x + 4 = 6.2
We have to think a little more carefully about what multiple of ten to use here. 6.2 only needs to be multiplied by 10, but 1.25 needs 100, so we will multiply every term by 100. Don't forget to multiply the 4 by 100 as well.
(100)(1.75x) + (100)(4) = (100)(6.2)
175x + 400 = 620
We had to be extra careful as we multiplied by 100. Now we can solve the equation as normal:
175x + 400 - 400 = 620 - 400
175x = 220
x = 1.26
Step-by-step explanation:
Answer:
x=−1 and y=−5
Step-by-step explanation:
hope this helps
Answer:
Yes. Explanation: 120%=1.2. If Maxine is correct, then she spent 1.2 times the hours she did homework than last week. 15⋅1.2=18.0=18.
Yes Explanation: 120
Answer: The answer is (C) Exponential.
Step-by-step explanation: We are to select out of the given options the type of graph that a savings account with compounded interest be modelled.
We know that compounding gives more interest because we are earning interest on interest, and not just on the principal.
The formula foe compound interest is given by
where, 'P' is the principal, r is the rate of interest and 'n' is the number of years.
Therefore, we can see that the function is of exponential type.
If we draw the graph of compound interest earned every year with a particular rate of interest is of exponential type.
So, the correct option is (C) Exponential.
