C and D are correct because 25 > 35 so that's C.
Its also D because they are not equal.
Hope this helps :)
your answer should be 1/3 if I did my math right
Answer:
See the step-by-step explanation
Step-by-step explanation:
Let c be any element of C. (I'm not sure wether you have to assume that C is non-empt or not)
C is a subset of B. That means that as c is in C, it is also in B. (
)
Now, B is a subset of A. It follows that as
.
That means c is an element of A. The predicate Q is true for all elements of A, including c.
Because we let c be any element of C, we have proven that the predicate Q is true for all elements in C.
Answer:
191
Step-by-step explanation:
17.70---- 177
19.10-----191
First just take 177/17.70= 10
all you have to do is multiply by 10
so
19.10= 191
Let;
A(-8,6) B(6,6) C(6, -4) D(-8, -4)
Let's find the length AB
x₁= -8 y₁=6 x₂=6 y₂=6
We will use the distance formula;
![d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
![=\sqrt[]{(6+8)^2+(6-6)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%286%2B8%29%5E2%2B%286-6%29%5E2%7D)
![=\sqrt[]{14^2+0}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B14%5E2%2B0%7D)

Next, we will find the width BC
B(6,6) C(6, -4)
x₁= 6 y₁=6 x₂=6 y₂=-4
substitute into the distance formula;
![d=\sqrt[]{(6-6)^2+(-4-6)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%286-6%29%5E2%2B%28-4-6%29%5E2%7D)
![=\sqrt[]{(-10)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%28-10%29%5E2%7D)
![=\sqrt[]{100}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B100%7D)

Area = l x w
= 14 x 10
= 140 square units