Answer: Choice C) 10.5
The distance from A to C is 7 units (count out the spaces between the two points, or subtract y coordinates 4-(-3) = 4+3 = 7)
Let AC = 7 be the base of the triangle. You might want to rotate the image so that AC is laying horizontally rather than being vertical.
Now move to point P. Walk 3 spaces to the right until you land on segment AC. This shows that the height of the triangle is 3 when the base is AC = 7.
base = 7, height = 3
area of triangle = (1/2)*base*height
area of triangle = 0.5*7*3
area of triangle = 10.5 square units
Answer:
22 and 7
Step-by-step explanation:
Let x and y be the two numbers we are looking for:
The statement “their sum is 29” means x + y = 29
The statement “their difference is 15” means x - y = 15
we get this system:
x + y = 29 (1)
x - y = 15 (2)
Let’s solve it :
(1) + (2) ⇒ x + y + x - y = 29 + 15 ⇒ 2x = 44 ⇒ x = 44/2 = 22
Now we replace x by 22 in equation (1) :
22 + y = 29 ⇒ y = 29 - 22 = 7
Answer:
The equation does not have a real root in the interval ![\rm [0,1]](https://tex.z-dn.net/?f=%5Crm%20%5B0%2C1%5D)
Step-by-step explanation:
We can make use of the intermediate value theorem.
The theorem states that if 
is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:
- If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
- The image of a continuous function over an interval is itself an interval.
Of course, in our case, we will make use of the first one.
First, we need to proof that our function is continues in , which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval , which means to evaluate the equation in 0 and 1:
Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval . I attached a plot of the equation in the interval ![\rm [-2,2]](https://tex.z-dn.net/?f=%5Crm%20%5B-2%2C2%5D)
where you can clearly observe how the graph does not cross the x-axis in the interval.
Three possible answers are:
y = 2x-1
y = 5x-1
y = -3x-1
All of them have the same y intercept of -1. The slopes are the only thing that differ.
Any linear equation can be written in the form y = mx+b where
m = slope
b = y intercept
Answer:
25
Step-by-step explanation:
