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2 years ago

Construct a linear equation for the linear data presented in the table

Mathematics

2 answers:

alina1380 [7]2 years ago

8 0

(-2,14)
(0,0)

Slope

m=y/x
m=14/-2
m=-7

Equation of line

y=mx+0
y=-7x

yaroslaw [1]2 years ago

5 0

Answer:

y = -7x

Explanation:

Take two points from table: (-4, 28), (-3, 21)

Find slope:

$\sf slope: \dfrac{y_2 - y_1}{x_2- x_1} \ \ where \ (x_1 , \ y_1), ( x_2 , \ y_2) \ are \ points$

$\rightarrow \sf slope\ (m): \dfrac{21-28}{-3-(-4)} = -7$

Now find equation:

$\sf y - y_1 = m (x - x_1)$

$\rightarrow \sf y - 21 = -7(x - (-3))$

$\rightarrow \sf y = -7(x +3) + 21$

$\rightarrow \sf y = -7x -21 + 21$

$\rightarrow \sf y = -7x$

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3 years ago

The value 5 is an upper bound for the zeros of the function shown below.

Mice21 [21]

Answer:

The given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

Step-by-step explanation:

Given

$f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$

We know the rational zeros theorem such as:

if $x=c$ is a zero of the function $f(x)$ ,

then $f(c) = 0$ .

As the $f\left(x\right)\:=\:x^2\:+\:x^3\:-\:11x^2\:-\:9x\:+\:18$ is a polynomial of degree $4$ , hence it can not have more than $4$ real zeros.

Let us put certain values in the function,

$f(5) = 448$ , $f(4) = 126$ , $f(3) = 0$ , $f(2) = -20$ ,

$f(1) = 0$ , $f(0) = 18$ , $f(-1) = 16$ , $f(-2) = 0$ , $f(-3) = 0$

From the above calculation results, we determined that $4$ zeros as

$x = -3, -2, 1,$ and $3$ .

Hence, we can check that

$f(x) = (x+3)(x+2)(x-1)(x-3)$

Observe that,

$for x > 3$ , $f(x)$ increases rapidly, so there will be no zeros for $x>3$ .

Therefore, the given statement that value 5 is an upper bound for the zeros of the function f(x) = x⁴ + x³ - 11x² - 9x + 18 will be true.

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strojnjashka [21]

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