Answer:
cos(θ) = 3/5
Step-by-step explanation:
We can think of this situation as a triangle rectangle (you can see it in the image below).
Here, we have a triangle rectangle with an angle θ, such that the adjacent cathetus to θ is 3 units long, and the cathetus opposite to θ is 4 units long.
Here we want to find cos(θ).
You should remember:
cos(θ) = (adjacent cathetus)/(hypotenuse)
We already know that the adjacent cathetus is equal to 3.
And for the hypotenuse, we can use the Pythagorean's theorem, which says that the sum of the squares of the cathetus is equal to the square of the hypotenuse, this is:
3^2 + 4^2 = H^2
We can solve this for H, to get:
H = √( 3^2 + 4^2) = √(9 + 16) = √25 = 5
The hypotenuse is 5 units long.
Then we have:
cos(θ) = (adjacent cathetus)/(hypotenuse)
cos(θ) = 3/5
Answer:∠1=55° ∠2=55° ∠3=55° ∠4=70° ∠5=55°
Step-by-step explanation:
We know that opposite angles of rhombus are congruent and the diagonals of rhombus cut the vertex angles into half. So ∠4=70° and ∠2=∠3=∠1=∠5.
Let ∠2+∠3=x
the sum of four angles of rhombus= 360°
70+70+ x +x = 360°
140+2x = 360°
2x=220°
x=110°
∠2=110°/2=55°=∠3=∠1=∠5
Answer:
1950
Step-by-step explanation:
......i hope it help
Based on the wording of your question, I believe the answer is C because from any parent function that is similar to A, B, or D, which can be interchangeable, you cannot make an Absolute | | from these types of equations, and you cannot remove an absolute by squaring it, adding to it, and so on. Therefore, C is the only function that differs from the other options and cannot have a relationship to the parent function. If you meant something else by this question please clarify.
Answer:
Hope this helps
Step-by-step explanation:
When new substances are produced in a reaction we say a chemical change has occurred. These changes are very difficult, sometimes impossible, to reverse.
Atoms are not created or destroyed in a chemical reaction, they are just rearranged from the reactants to form the products. This is the reason why symbol equations must be balanced.
If a chemical reaction is completed in a closed system (when nothing extra can get in and nothing can escape), then the mass will remain constant. But, if the chemical reaction happens in an open system (where air can get in and out), then mass may appear to change.
In this simulation we will explore the mass changes in chemical reactions and endeavour to explain why they have occurred.
