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3 years ago

Hi:) anyone able to explain how to get the quadratic equation into the form

Mathematics

2 answers:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

y = a(x-p)(x-q)

p and q are roots,

0 and 10

y = a(x-0)(x-10)

y = ax(x-10)

To find a use (5,5)

5 = a(5)(5-10)

5 = -25a

a = -⅕

y = -⅕(x)(x-10)

Nina [5.8K]3 years ago

3 0

Answer:

Step-by-step explanation:

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What is the sum of 1/9, 2/3, and 5/18? A. 121 B.4/15 C. 19/18 D. 8/30

sergey [27]

Answer is C...Use calculator

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3 years ago

In Jeremiah class

Irina18 [472]

Answer:

2/5 = 4/10

4/10 = 40%

Explanation:

2/5 is a simplified fraction, and one of the bigger fractions it’s made by 4/10. Do you can use 4/10, but can also use 8/20, 20/50, 40/100, so on. Since we know that any fraction that of ten can be turned into a percentage easily (2/10=20%, 5/10=50%,9/10=%90), then 4/10 is equal to 40%. So 40% of the people in Jeremiah’s class are boys.

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2 years ago

Suppose a railroad rail is 5 kilometers and it expands on a hot day by 12 centimeters in length. Approximately how many meters w

jolli1 [7]

Answer:

about 17 meters

Step-by-step explanation:

We can use the Pythagorean theorem to put an upper bound on the height of the bump in the rail. This assumes half the expanded rail length (d+e) is the hypotenuse of a right triangle whose legs are the bump height (b) and the 2500 meter distance (d) from the center of the rail to its end.

The Pythagorean theorem relates these distances this way:

b^2 + d^2 = (d+e)^2

Expanding the square on the right, we can simplify the expression to find b.

b^2 = (d^2 +2de +e^2) -d^2

b^2 = e(2d +e)

b = √(e(2d +e))

Using lengths in meters, we can fill this in to calculate b.

b = √(.06(2·2500 +.06)) = √300.0036

b ≈ 17.32 . . . . meters

_____

Comment on this solution

We don't expect rails to tear loose from the rail bed and rise up to a height matching that of a 3-story building. That is why there are typically expansion joints and shorter rail lengths used in the construction of railways.

The height is a little lower if we take physics into account and distribute the stress in the rail along its length. No doubt the final curve is somewhat more complicated than the triangle we have assumed.

If it were an ellipse, the height might only be 9.4 meters, with the steepest rise occurring near the ends of the rail. The math for this model is beyond the scope of this answer.

7 0

3 years ago

Express 5.1146• 10^3 in standard notation

charle [14.2K]

Answer:

5,114.6

Step-by-step explanation:

5.1146 x 10^3 means move decimal point 3 to the right to get the answer

4 0

3 years ago

In a local ice sculpture contest, one group sculpted a block into a rectangular based pyramid. The dimensions of the base were 3

m_a_m_a [10]

Answer:

1. The amount of ice needed = 18 m²

2. The amount of fabric needed to manufacture the umbrella is 0.76 m²

3. The height of the cone, is 3.75 cm

4. The dimensions of the deck are;

Width = 28/3 m, breadth = 28/3 m

The area be 87.11 m²

5. The dimensions of the optimal design for setting the storage area at the corner, we have;

Width = 10m

Breadth = 10 m

The dimensions of the optimal design for setting the storage area at the back of their building are;

Width = 7·√2 m

Breadth = 7·√2 m

Step-by-step explanation:

1. The amount of ice needed is given by the volume, V, of the pyramid given by V = 1/3 × Base area × Height

The base area = Base width × Base breadth = 3 × 5 = 15 m²

The pyramid height = 3.6 m

The volume of the pyramid = 1/3*15*3.6 = 18 m²

The amount of ice needed = 18 m²

2. The surface area of the umbrella = The surface area of a cone (without the base)

The surface area of a cone (without the base) = π×r×l

Where:

r = The radius of the cone = 0.4 m

l = The slant height = √(h² + r²)

h = The height of the cone = 0.45 m

l = √(0.45² + 0.4²) = 0.6021 m

The surface area = π×0.4×0.6021 = 0.76 m²

The surface area of a cone (without the base) = 0.76 m²

The surface area of the umbrella = 0.76 m²

The amount of fabric needed to manufacture the umbrella = The surface area of the umbrella = 0.76 m²

3. The volume, V, of the cone = 1/3×Base area, A, ×Height, h

The volume of the cone V = 150 cm³

The base area of the cone A = 120 cm²

Therefore we have;

V = 1/3×A×h

The height of the cone, h = 3×V/A = 3*150/120 = 3.75 cm

4. Given that the deck will have railings on three sides, we have;

Maximum dimension = The dimension of a square as it is the product of two equal maximum obtainable numbers

Therefore, since the deck will have only three sides, we have that the length of each side are equal and the fourth side can accommodate any dimension of the other sides giving the maximum dimension of each side as 28/3

The dimensions of the deck are width = 28/3 m, breadth = 28/3 m

The area will then be 28/3×28/3 = 784/9 = $87\frac{1}{9}$ =87.11 m²

5. The optimal design for setting the storage area at the corner of their property with four sides is having the dimensions to be that of of a square with equal sides of 10 m each as follows;

Width = 10m

Breadth = 10 m

The optimal design to have the storage area at the back of their building having a fence on only three sides, is given as follows;

Storage area specified = 98 m²

For optimal use of fencing, we have optimal side size of fencing = s = Side length of a square

s² = 98 m²

Therefore, s = √98 = 7·√2 m

Which gives the width = 7·√2 m and the breadth = 7·√2 m.

8 0

3 years ago