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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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Maslowich

Answer:

Step-by-step explanation:

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Long division is often considered one of the most challenging topics to teach. Luckily, there are strategies that we can teach to make multi-digit division easier to understand and perform.

The Box Method, or  the Area Model, is one of these strategies. It is a mental math based approach that will enhance number sense understanding. Students solve the equation by subtracting multiples until they get down to 0, or as close to 0 as possible.

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Check the picture below.

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\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&P&(~ -9 &,& 7~) &#10;%  (c,d)&#10;&Q&(~ -3 &,& 7~)&#10;\end{array}\quad&#10;%   coordinates of midpoint &#10;\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)&#10;\\\\\\&#10;S=\left( \cfrac{-3-9}{2}~,~\cfrac{7+7}{2} \right)\implies S=(-6,7)\\\\&#10;-------------------------------

\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&Q&(~ -3 &,& 7~) &#10;%  (c,d)&#10;&R&(~ -3 &,& 1~)&#10;\end{array}&#10;\\\\\\&#10;T=\left(\cfrac{-3-3}{2}~,~\cfrac{1+7}{2}  \right)\implies T=(-3,4)

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\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&S&(~ -6 &,& 7~) &#10;%  (c,d)&#10;&T&(~ -3 &,& 4~)&#10;\end{array}\\\\\\ &#10;%  distance value&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;ST=\sqrt{[-3-(-6)]^2+[4-7]^2}\implies ST=\sqrt{(-3+6)^2+(4-7)^2}&#10;\\\\\\&#10;ST=\sqrt{3^2+(-3)^2}\implies ST=\sqrt{18}\implies ST=\sqrt{9\cdot 2}&#10;\\\\\\&#10;ST=\sqrt{3^2\cdot 2}\implies ST=3\sqrt{2}

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