Answer:
D. (7, 0)
Step-by-step explanation:
The rule for a reflection over the y-axis is (x, y) → (x, -y)
This means that the x-values stay the same while the y-values change.
Q(x, y) → (x, -y)
Q(3, 0) → (3, 0)
Q'(3, 0)
P(x, y) → (x, -y)
P(5, 6) → (5, -6)
P'(5, -6)
R(x, y) → (x, -y)
R(7, 0) → (7, 0)
R'(7, 0)
Therefore, the correct answer is D.
Hope this helps!
To do this, we're going to use the order of operations (PEMDAS):
P - Parentheses
E - Exponents
M - Multiplication
D - Division
A - Addition
S - Subtraction
First let's do parentheses, there isn't anythig in parentheses we need to simplify, so we can skip this step.
Next let's look for exponents. I see we have a
so let's replace that with
:
![-3(25)-(-2)(-8)](https://tex.z-dn.net/?f=-3%2825%29-%28-2%29%28-8%29)
Now let's look for multiplcation. We know that things that are right next to eachother in parentheses represent multiplcation, so let's simply this more:
![-3(25)=-75](https://tex.z-dn.net/?f=-3%2825%29%3D-75)
![(-2)(-8)=16](https://tex.z-dn.net/?f=%28-2%29%28-8%29%3D16)
![-75-16](https://tex.z-dn.net/?f=-75-16)
And now we're left with a simple problem we know how to solve.
Answer: ![-91](https://tex.z-dn.net/?f=-91)
Hope this helps!
Hi there.
A triangle's interior angles must always add up to 180 degrees. Since we already have one measurement, 56, we can set up an equation to solve for the missing angles.
(2x + 4) + 56 + x= 180; solve for x.
Subtract 56 from both sides.
(2x + 4) + x = 124;
Combine like-terms (x).
3x + 4 = 124;
Subtract 4 from both sides.
3x = 120
Divide both sides by 3 to solve for x.
x = 40.
Now, we need to substitute x with 40 in each of our angles to determine their measurements.
2x + 4; x = 40.
2(40) + 4 = 80 + 4 = 84;
One measurement is 84 degrees.
x = 40 is another measurement on its own.
Our measurements are:
56, 84, and 40.
Your corresponding answer choice is H.) 56, 84, 40.
I hope this helps!
Answer:
V'(t) = ![-250(1 - \frac{1}{40}t)](https://tex.z-dn.net/?f=-250%281%20-%20%5Cfrac%7B1%7D%7B40%7Dt%29)
If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.
Step-by-step explanation:
Given:
V =
, where 0≤t≤40.
Here we have to find the derivative with respect to "t"
We have to use the chain rule to find the derivative.
V'(t) = ![2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )](https://tex.z-dn.net/?f=2%285000%29%281%20-%20%5Cfrac%7B1%7D%7B40%7D%20t%29d%2Fdt%20%281%20-%20%5Cfrac%7B1%7D%7B40%7Dt%20%29)
V'(t) = ![2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )](https://tex.z-dn.net/?f=2%285000%29%281%20-%20%5Cfrac%7B1%7D%7B40%7D%20t%29%28-%5Cfrac%7B1%7D%7B40%7D%20%29)
When we simplify the above, we get
V'(t) = ![-250(1 - \frac{1}{40}t)](https://tex.z-dn.net/?f=-250%281%20-%20%5Cfrac%7B1%7D%7B40%7Dt%29)
If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.