Answer:
<h2>x=52</h2>
Basic fact:
<em>angles in a point add up to 360</em>
Step-by-step explanation:
20+165+33+90(that is the right angle)+x=360
20+165+33+90=308
308+x=360
-308
x=52
Step-by-step explanation:
ok so as u know the angles of a triangle add up to 180.
So we will be using that for help.
S+R+T=180
12x+13x+12x-4+2-3=180
37x-5=180
37x=185
x=5
Now we will plug in the value for x to find each angle
S= 13(5)-4
=61
T=12(5)-3
=57
R=12(5)+2
=62
Hope this helped:)
Here here here here here here here
Answer:
The correct option is A
Step-by-step explanation:
Since (1, 3) are the solutions to the function g(x), then
g(x) = (x – 1)(x – 3) = x² –3x – x + 3 = x² – 4x + 3
We know that f(x) is on the same plane and f(1) = 1.
Let's take g(1) = (1 – 1)(1 – 3) = 0
Therefore, when x = 1; f = 1 and g = 0. Hence
g(1) = f(1) – 1
In order words, g(x) could be f(x) – 1.
The correct option is A
<h2>
Hello!</h2>
The answer is:
The first option, the amount dumped after 5 days is 0.166 tons.
<h2>Why?</h2>
To solve the problem, we need to integrate the given expression and evaluate using the given time.
So, integrating we have:


Hence, we have that the amount dumped after 5 days is 0.166 tons.
Have a nice day!