Answer:
(5/2)√13 ≈ 9.0139
Step-by-step explanation:
The longer hypotenuse is found using the Pythagorean theorem as ...
The right triangles are AA similar, and vertical sides correspond. Their ratio, smaller/larger is 5/8. That means the hypotenuse of the smaller triangle, A, is 5/8 of the hypotenuse of the larger one.
Answer:
Either they converse s little more like if Kate asks "You came just at the right time! When did you get here?" or Alyiah says something like "Now come with me, I'm taking you to somewhere safe"
Step-by-step explanation:
This story sounds super cool, keep it up !!!
Step-by-step explanation:
In this case we have:
Δx = 3/n
b − a = 3
a = 1
b = 4
So the integral is:
∫₁⁴ √x dx
To evaluate the integral, we write the radical as an exponent.
∫₁⁴ x^½ dx
= ⅔ x^³/₂ + C |₁⁴
= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)
= ⅔ (8) + C − ⅔ − C
= 14/3
If ∫₁⁴ f(x) dx = e⁴ − e, then:
∫₁⁴ (2f(x) − 1) dx
= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx
= 2 (e⁴ − e) − (x + C) |₁⁴
= 2e⁴ − 2e − 3
∫ sec²(x/k) dx
k ∫ 1/k sec²(x/k) dx
k tan(x/k) + C
Evaluating between x=0 and x=π/2:
k tan(π/(2k)) + C − (k tan(0) + C)
k tan(π/(2k))
Setting this equal to k:
k tan(π/(2k)) = k
tan(π/(2k)) = 1
π/(2k) = π/4
1/(2k) = 1/4
2k = 4
k = 2
Answer: The dimensions are 7cm and 3cm.
Step-by-step explanation:
The following equation can be deduced from the question
L × W = 21 ....... i
2L + 2W = 20 .......... ii
Divide equation ii through by 2
(2L + 2W = 20) / 2
L + W = 10 ....... iii
From equation iii
L = 10 - W ......... iv
Put equation iv into I
L × W = 21
(10 - W) × W = 21
10W - W² = 21
W² - 10W + 21 = 0
W² - 7W - 3W + 21 = 0
W(W - 7) - 3(W - 7) = 0
W - 3 = 0
W = 0+3
W = 3 centimeter
Since,
Length × Width = 21
Length = 21/3
Length = 7 centimeters
The dimensions are 7cm and 3cm.
Answer:
y = -4x + 36
Step-by-step explanation:
First, find the slope by calculation (y2-y1)/(x2-x1), you should get -8/2, or -4.
Next, plug the points into the equation y-y1 = slope(x-x1), getting y-28 = -4(x-2).
Simplify the equation and you get y =-4x + 36.
Hopefully this helps- let me know if you have any questions!
