the equation that we can solve using the given system of equations is:
3x^5 - 4x^4 - 11x^3 + 2x^2 - 10x + 15 = 0
<h3>Which equation can be solved using the given system of equations?</h3>
Here we have the system of equations:
y = 3x^5 - 5x^3 + 2x^2 - 10x + 4
y = 4x^4 + 6x^3 - 11
Notice that both x and y should represent the same thing in both equations, then we could write:
3x^5 - 5x^3 + 2x^2 - 10x + 4 = y = 4x^4 + 6x^3 - 11
If we remove the middle part, we get:
3x^5 - 5x^3 + 2x^2 - 10x + 4 = 4x^4 + 6x^3 - 11
Now, this is an equation that only depends on x.
We can simplify it to get:
3x^5 - 4x^4 - 11x^3 + 2x^2 - 10x + 15 = 0
That is the equation that we can solve using the given system of equations.
If you want to learn more about systems of equations:
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38. 6900 grams
39. 19600 grams
40. 27910 grams
41. 32840 grams
42. 610 grams
43. 970 grams
44. 3712 grams
45. 8937 grams
46. 37 grams
47. 69 grams
48. 1510 grams
49. 4700 grams
50. 150 grams
51. 15 grams
52. 15200 grams
53. 460
I'm sorry if these are wrong but I hope this could help.
Answer:
6.1
Step-by-step explanation:
Draw a picture of an equilateral triangle. Cut the triangle in half, so that you get two 30-60-90 triangles. The area of these smaller triangles is 8 square inches.
The short leg of these triangles (the base) is half the side length: ½ s.
According to properties of 30-60-90 triangles, the long leg (the height) is √3 times the short leg: ½ s√3.
Area of a triangle is half the base times the height:
A = ½bh
8 = ½ (½ s) (½ s√3)
8 = ⅛ s²√3
64 = s²√3
s² = 64/√3
s = √(64/√3)
s ≈ 6.1
