Answer:
A dress originally costs $120
Step-by-step explanation:
To find the original cost of the dress, we will follow the steps below;
let x represent the original cost of the dress
75% of x = $90
× x = $90
= $90
MULTIPLY both-side of the equation by 100
× 100 = $90×100
At the right-hand side of the equation, 100 will cancel out 100, leaving us with just 75x
75x = $9000
DIVIDE both-side of the equation by 75
75x/75 = $9000/75
At the left-hand side of the equation 75 will cancel-out 75 leaving us with just x
x= $9000/75
x = $120
A dress originally costs $120
9514 1404 393
Answer:
14.01, 493, 87
Step-by-step explanation:
Subtracting 28 from both sides tells you the range of values you need to be looking at.
28 + x > 42
x > 14
Any values more than 14 will make the inequality true. Three of them are ...
14.01, 493, 87
Answer:
None of these.
Step-by-step explanation:
Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.
Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.
In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.
If we look at the first term x^3 and we divide it by 13 we get 
we cannot reduce it so it is not a fraction so 13 is not a factor of P
None of these is the right option.
The expression "x is less than or equal to zero" can be written in several forms; some examples, however, could be:
x ≤ 0
0 ≥ x
(You could substitute any number in place of 3 for the below)
3 - 3 ≥ x
When x=-1:
Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?
The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.
From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4
and solve for x'.
Hope that helps!
