Answer:
x=5
Step-by-step explanation:
3x^2=15x
Subtract 15x from both sides
3x^2-15x=0
Factor out 3x
3x(x-5)=0
Since 0 divided by any number is zero, we get that:
x-5=0
x=5
Answer: 25%
Step-by-step explanation:
Given : Previous baby's weight = 18 pounds
New baby's weight = 24 pounds.
Increase in weight = New weight -Previews weight
=24 pounds - 18 pounds =6 pounds
Percentage increase in baby's weight =
Hence, the percentage increase in baby's weight = 25%
https://www.slader.com/discussion/question/if-a-truck-traveled-248-miles-in-4-hours-then-the-truck-9c45464f/
-x - y = 8
2x - y = -1
Ok, we are going to solve this in 2 parts. First we have to solve for one of the variables in one of the equation in terms of the other variable. I like to take the easiest equation first and try to avoid fractions, so let's use the first equation and solve for x.
-x - y = 8 add y to each side
-x = 8 + y divide by -1
x = -8 - y
So now we have a value for x in terms of y that we can use to substitute into the other equation. In the other equation we are going to put -8 - y in place of the x.
2x - y = -1
2(-8 - y) - y = -1 multiply the 2 through the parentheses
-16 - 2y - y = -1 combine like terms
-16 - 3y = -1 add 16 to both sides
-3y = 15 divide each side by -3
y = -5
Now we have a value for y. We need to plug it into either of the original equations then solve for x. I usually choose the most simple equation.
-x - y = 8
-x - (-5) = 8 multiply -1 through the parentheses
-x + 5 = 8 subtract 5 from each side
-x = 3 divide each side by -1
x = -3
So our solution set is
(-3, -5)
That is the point on the grid where the 2 equations are equal, so that is the place where they intersect.
Answer:
ΔV = 0.36π in³
Step-by-step explanation:
Given that:
The radius of a sphere = 3.0
If the measurement is correct within 0.01 inches
i.e the change in the radius Δr = 0.01
The objective is to use differentials to estimate the error in the volume of sphere.
We all know that the volume of a sphere
The differential of V with respect to r is:
dV = 4 πr² dr
which can be re-written as:
ΔV = 4 πr² Δr
ΔV = 4 × π × (3)² × 0.01
ΔV = 0.36π in³
