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2 years ago

The equation y=ax2+bx+c goes through the coordinates (-6, 15), (0, -3), and (2, 7). What are the values of a, b, and c?

Mathematics

1 answer:

Romashka-Z-Leto [24]2 years ago

7 0

The typical method of solving this question... (School method)

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What is the 4th term of the linear sequence below?<br> -9,-5, -1,3,7,...

Anton [14]

Answer:

The 4th term of the linear sequence = 3

Step-by-step explanation:

First term, a = -9

Common difference, d = -5 - (-9)

= -5 + 9

= 4

d = 4

The 4th term of the linear sequence = a + (n - 1)d

where,

n = 4

a + (n - 1)d

= -9 + (4 - 1)4

= -9 + (3)4

= -9 + 12

= 3

The 4th term of the linear sequence = 3

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2 years ago

Solve the simultaneous equation y = x^2 and y = 3x - 2

gayaneshka [121]

Answer:

${x}^{2} = 3x - 2 \\ {x}^{2} - 3x + 2 = 0 \\ (x - 1)(x - 2) = 0 \\ x = 1 \: or \: x = 2 \\ for \: x \: = 1 \\ \: y = {(1)}^{2} = 1 \\ for \: x = 2 \\ y = {(2)}^{2} = 4$

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2 years ago

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coldgirl [10]

Answer:

D. A = πr^2

Step-by-step explanation:

The formula of the area of a circle is <u>A = πr^2</u>.

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2 years ago

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mafiozo [28]

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Step-by-step explanation:

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2 years ago

Without actual division prove that x4 +2x3 -2x2 +2x -3 is exactly divisible by<br> x2 +2x-3

Brut [27]

Answer:

Step-by-step explanation:

$\displaystyle\Large\boldsymbol{} x^4+2x^3-2x^2+2x-3= \\\\\\x^4+2x^2(x-1)+2x-\underbrace{2-1}_{-3}= \\\\\\x^4-1+2x^2(x-1)+2x-2 = \\\\\\(x^2-1)(x^2+1)+2x^2(x-1)+2(x-1) = \\\\\\\underline{(x-1)}(x+1)(x^2+1)+2x^2\underline{(x-1)} +2\underline{(x-1)} =\\\\\\(x-1)((x+1)(x^2+1)+2x^2+2)=\\\\\\(x-1)(x^3+x^2+2x^2+x+2+1)=\\\\\\(x-1)(x^3+3x^2+x+3) =\\\\\\(x-1)( \underline{(x+3)}x^2+\underline{(x+3)})=(x^2+1)\underbrace{(x-1)(x+3)}_{x^2+2x-3}= \\\\\\(x^2+1)(x^2+2x-3) : (x^2+2x-3)=x^2+1$

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2 years ago