Use the discriminant to determine the number of real solutions to the equation. −4x^2=−8−10x
2 answers:
Answer:
Step-by-step explanation:
You don't get a free answer. I am, after all, a high school math teacher, so there has to be a lesson in with this.
The discriminant is part of the quadratic formula. It is:
. If this value is found to be > 0 and a perfect square, there are 2 real roots; if this value is found to >0 and not a perfect square, there are 2 complex roots; if this value is found to be = 0, then there is 1 real root with a multiplicity of 2; and finally, if this value is found to be < 0, then there are 2 imaginary roots. Also, it would help to know that, because we are dealing with the discriminant, which comes from the quadratic formula, and quadratics, by definition, have 2 solutions, you must have 2 solutions listed as the possible roots for the equation. Our equation is:
but in order to determine the a, b, and c for the discriminant, that equation has to be in standard form, set equal to 0:
From this we can see that a = -4, b = 10, and c = 8. Filling in the discriminant:
which gives us a value of
100 - [4(-4)(8)] (don't forget orders of operation here!)
100 - (-128) = 100 + 128 = 228
This value is greater than 0 but is not a perfect square, so there are 2 complex roots. That means that there will be radicals in the solutions.
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Answer: 8
Step-by-step explanation:
The triangle shown in the image attached is a right triangle.
Therefore, to calculate the missing lenght of the triangle you can apply the Pythagorean Theorem, which is shown below:
Where <em>a</em> is the hypotenuse and <em>b</em> and <em>c</em> are the legs.
The problem gives you the value of the hypotenuse and the value of one leg. Therefore, you must solve for the other leg from , as following:
Therefore, the lenght of the missing side is: 8
PART A
The barn is constructed by the following 2D shapes
Two triangles of height = 4' and base = 20'
Area = 2 × (0.5×4×20) = 80
Two rectangles of length = 20' and width = 15'
Area = 2 × 20 × 15 = 600
Two rectangles of length = 45' and width = 15'
Area = 2 × 45' × 15' = 1350
One base of length = 45' and width = 15'
Area = 45 × 15 = 675
Two rectangles on each on one side of the roof. We have the length = 45' but not the width. We can work out the width by using Pythagoras theorem
w² = 4² + 10²
w² = 16 + 100
w² = 116
w = √116 = 10.77
Area of the two rectangles on the roof part is = 2 ×10.77 × 45 = 969.33
Total area to paint = 969.33+675+1350+600+80 = 3674.33 ≈ 3700 (to the nearest hundreth)
PART B
Numbers of paint cans needed = 3700 ÷ 57 = 64.9 ≈ 65 cans
PART C
Total cost of paint = 65 × 23.50 = $1527.50
PART D
The barn is constructed by a cuboid and a rectangular prism
V of cuboid = length × width × height
V of cuboid = 20 × 45 × 15
V of cuboid = 13500
V of triangular prism = Area of cross section × depth
V = [0.5×4×20] × 45
V = 1800
Total volume = 1800 + 13500 = 15300
The barn is constructed by the following 2D shapes
Two triangles of height = 4' and base = 20'
Area = 2 × (0.5×4×20) = 80
Two rectangles of length = 20' and width = 15'
Area = 2 × 20 × 15 = 600
Two rectangles of length = 45' and width = 15'
Area = 2 × 45' × 15' = 1350
One base of length = 45' and width = 15'
Area = 45 × 15 = 675
Two rectangles on each on one side of the roof. We have the length = 45' but not the width. We can work out the width by using Pythagoras theorem
w² = 4² + 10²
w² = 16 + 100
w² = 116
w = √116 = 10.77
Area of the two rectangles on the roof part is = 2 ×10.77 × 45 = 969.33
Total area to paint = 969.33+675+1350+600+80 = 3674.33 ≈ 3700 (to the nearest hundreth)
PART B
Numbers of paint cans needed = 3700 ÷ 57 = 64.9 ≈ 65 cans
PART C
Total cost of paint = 65 × 23.50 = $1527.50
PART D
The barn is constructed by a cuboid and a rectangular prism
V of cuboid = length × width × height
V of cuboid = 20 × 45 × 15
V of cuboid = 13500
V of triangular prism = Area of cross section × depth
V = [0.5×4×20] × 45
V = 1800
Total volume = 1800 + 13500 = 15300