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3 years ago

6

What is the smallest of 3 consecutive odd numbers if the product of the smaller two integers is 3 less than 6 times the largest

integer?

1 answer:

pochemuha3 years ago

3 0

Answer:

7,9,11

Step-by-step explanation:

Let first number is x, second number is (x+2) and third number is (x+4).

ATQ,

The product of the smaller two integers is 3 less than 6 times the largest integer.

x(x+2) = 6(x+4)-3

x² + 2x = 6x+24-3

x² + 2x = 6x+21

x² + 2x -6x -21=0

x²-4x -21=0

It is a quadratic equation whose solution is :

x = -3 and x = 7

First number = 7

Second number = 9

Third number = 11

Hence, this is the required ssolution.

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Whats the slope and y-intercept of y=-3x+4?

Helga [31]

Answer:

slope= -3. y intercept is 4

Step-by-step explanation:

the x value is slope and the y value is y intercept.

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3 years ago

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Which equation matches the given points? (0, 7.6), (1, 6.2), (2, 4.8), (3, 3.4), (4,2)

Cerrena [4.2K]

The equation that matches the given points is g(x) = -1.4x + 7.6

The standard form of a linear equation is expressed as $g(x)=mx + b$

m is the slope of the line

b is the y-intercept:

Using the coordinate points (3, 3.4), (4,2)

$m=\frac{2-3.4}{4-3}\\m=\frac{-1.4}{1}\\m=-1.4$

Substitute m = -1.4 and the coordinate (4, 2) into the formula:

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Get the required equation:

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3 years ago

Segment FG begins at point F(-2, 4) and ends at point G (-2, -3). Segment FG is translated by (x, y) → (x – 3, y + 2) and then r

Mariana [72]

Answer:

The length of the segment F'G' is 7.

Step-by-step explanation:

From Linear Algebra we define reflection across the y-axis as follows:

$(x',y')=(-x, y)$ , $\forall\, x, y\in \mathbb{R}$ (Eq. 1)

In addition, we get this translation formula from the statement of the problem:

$(x',y') =(x-3,y+2)$ , $\forall \,x,y\in \mathbb{R}$ (Eq. 2)

Where:

$(x, y)$ - Original point, dimensionless.

$(x', y')$ - Transformed point, dimensionless.

If we know that $F(x,y) = (-2, 4)$ and $G(x,y) = (-2,-3)$ , then we proceed to make all needed operations:

Translation

$F''(x,y) = (-2-3,4+2)$

$F''(x,y) = (-5,6)$

$G''(x,y) = (-2-3,-3+2)$

$G''(x,y) = (-5,-1)$

Reflection

$F'(x,y) = (5, 6)$

$G'(x,y) = (5,-1)$

Lastly, we calculate the length of the segment F'G' by Pythagorean Theorem:

$F'G' = \sqrt{(5-5)^{2}+[(-1)-6]^{2}}$

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3 years ago

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