Answer:
D E + E F greater-than D F
5 less-than D F less-than 13
Triangle D E F is a scalene triangle
Step-by-step explanation:
we know that
The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side
we have the triangle EDF
where
<u><em>Applying the triangle inequality theorem</em></u>
1)
2)
so
The length of DF is the interval -----> (5,13)
The triangle DEF is a scalene triangle (the three length sides are different)
therefore
<em>The statements that are true are</em>
D E + E F greater-than D F
5 less-than D F less-than 13
Triangle D E F is a scalene triangle
Answer:
x=-2
y= 8
Step-by-step explanation:
Given data
10x – 2y = -36-----------1
7x – 2y = -30--------------2
-Subtract to eliminate y
3x-0= -6
3x= -6
x= -6/3
x= -2
Put x= -2 in
7(-2)– 2y = -30
-14-2y=-30
-14+30=2y
16=2y
y= 16/2
y= 8
Answer:
Force = - 2.1 Newton
Step-by-step explanation:
According to question ,
The value of Spring constant , K = 7 N/m Newton per meter
And the spring is extended up to 0.3 meter
The Force experience on the spring is calculated from the Hook's Law
<u>The principal state that , the force applied on spring to extend or compress the spring to some distance is proportion to that distance. </u>
Let the force be F
So, F = - K X
where K is spring constant and X is its displacement
Negative sign shows displacement of spring once it is stretched
And K is spring constant
So ,Force = ( - ) 7 × 0.3
I.e Force = - 2.1 Newton Answer
Answer:
The temp is increasing
Step-by-step explanation:
I just did it
Step-by-step explanation:
(a) ∫₋ₒₒ°° f(x) dx
We can split this into three integrals:
= ∫₋ₒₒ⁻¹ f(x) dx + ∫₋₁¹ f(x) dx + ∫₁°° f(x) dx
Since the function is even (symmetrical about the y-axis), we can further simplify this as:
= ∫₋₁¹ f(x) dx + 2 ∫₁°° f(x) dx
The first integral is finite, so it converges.
For the second integral, we can use comparison test.
g(x) = e^(-½ x) is greater than f(x) = e^(-½ x²) for all x greater than 1.
We can show that g(x) converges:
∫₁°° e^(-½ x) dx = -2 e^(-½ x) |₁°° = -2 e^(-∞) − -2 e^(-½) = 0 + 2e^(-½).
Therefore, the smaller function f(x) also converges.
(b) The width of the intervals is:
Δx = (3 − -3) / 6 = 1
Evaluating the function at the beginning and end of each interval:
f(-3) = e^(-9/2)
f(-2) = e^(-2)
f(-1) = e^(-1/2)
f(0) = 1
f(1) = e^(-1/2)
f(2) = e^(-2)
f(3) = e^(-9/2)
Apply Simpson's rule:
S = Δx/3 [f(-3) + 4f(-2) + 2f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)]
S ≈ 2.5103
