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Mnenie [13.5K]
3 years ago
11

The table shows some values of a function of the form y = ax2 + bx + c.

Mathematics
1 answer:
natali 33 [55]3 years ago
5 0

Answer:

Value of constant term c is (-4)

Step-by-step explanation:

The given table represents a function which is in the form of a quadratic equation,

y = ax² + bx + c

We choose three points (3, -10), (4, -16) and (5, -24) from the table and satisfy the equation to get the values of a, b, and c.

For point (3, -10)

-10 = a(3)² + 3b + c

9a + 3b + c = -10 -------(1)

For point (4, -16)

-16 = a(4)² + 4b + c

16a + 4b + c = -16 ------(2)

For point (5, -24)

-24 = a(5)² + 5b + c

25a + 5b + c = -24 -----(3)

Equation (1) - equation (2)

(9a + 3b + c) - (16a + 4b + c) = -10 + 16

-7a - b = 6

7a + b = -6 ------(4)

Equation (2) - equation (3)

(16a + 4b + c) - (25a + 5b + c) = -16 + 24

-9a - b = 8

9a + b = -8 -------(5)

Equation (4) - Equation (5)

(7a + b) - (9a + b) = -6 + 8

-2a = 2

a = -1

From equation (4),

-7(1) + b = -6

b = -6 + 7

b = 1

From equation (1)

9(-1) + 3(1) + c = -10

-9 + 3 + c = -10

c = -10 + 6

c = -4

Therefore, the value of constant term c is (-4).

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a pyramid whose altitude is 5ft weighs 800lbs. at what distance from its vertex must it be cut by a plane parallel to its base s
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A structure which has a square base and four triangular sides meeting at a point is called pyramid.

At distance of 3.97 feet from its vertex , pyramid is cut by plane  so that the two solids of equal weight will be formed.

<u>It is assumed that weight of pyramid is proportional to its volume.</u>

So,   w  = k V , where V is volume of original pyramid and v is volume of small pyramid and k is constant.

Let us consider that at h  distance,  pyramid is cut from its vertex. So a small pyramid is also formed.

Assume that base area of original pyramid is A and base of small pyramid is a.

Volume of original pyramid is,   V= \frac{1}{3}  *A* 5

So, weight of original pyramid,  W = k *(\frac{1}{3}  *A* 5) = 800

Volume of small pyramid is,  v = \frac{1}{3}* a* h

So, weight of small pyramid, w = k*(\frac{1}{3}* a* h)=400

<u>Since, base and height of small pyramid and original pyramid are in proportion.</u>

So,  \frac{a}{A}  = (\frac{h}{5}) ^{2}

       a = (\frac{h}{5} )^{2}A

Substituting value of a in  equation k*(\frac{1}{3}* a* h)=400

So, k*(\frac{1}{3}* \frac{h^{2} }{25}A * h)=400

       (k*\frac{1}{3}* A*5)*\frac{1}{5} *\frac{h^{2} }{25} * h)=400

Since, (k *\frac{1}{3}  *A* 5) = 800, substitute in above equation.

    So, 800*\frac{h^{3} }{125}=400\\\\h^{3}=\frac{125}{2}\\\\h=\sqrt[\frac{1}{3} ]{62.5}\\\\h=3.97 feet

Learn more:

brainly.com/question/17950304

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