Answer: The cost for each subscription is $19.5
Step-by-step explanation:
32m=624
Divide 32 on both sides:
32m/32=624/32
m=19.5
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:
10
Step-by-step explanation:
Answer:
a) The discriminant of the equation = - 44
b)The nature of the roots will be imaginary.
c) 
Step-by-step explanation:
Here, the given expression is 
or, 
Now the discriminant (D) of a quadratic equation 
D = 
Hence, the discriminant of the equation = - 44
As D< 0, so the roots will be imaginary.
Now,by quadratic formula : 
So, here 
So, either 
or,
The correct answer is A.
You have direct variation if x and y are modeled by the equation
In this case, m is the constant of proportionality. So, if the constant has to be 2, the equation becomes
A side note: Actually, option C has a constant of proportionality of two as well, except the roles of x and y are interchanged. I chose option A because usually you want the y = mx form, but the names of the variables are obviously meaningless.
