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2 years ago

Rate of change for y= 2/3x-48

Mathematics

1 answer:

tangare [24]2 years ago

4 0

Answer:

2/3

Step-by-step explanation:

The rate of change is the same thing as the slope of a line

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can someone explain it to me, i don't need the answer, i just need an detailed explanation of how they got the answer using the

zubka84 [21]

Answer:

Step-by-step explanation:

Good idea to review quadratic functions and the quadratic formula.

Quadratics have three coefficients: ax² + bx + c, and the "discriminant" is defined as b²-4ac. Please review these rules:

1) if the discriminant is +, the quadratic equation has two real, unequal roots.

2) if the disc. is 0, the equation has two real, equal root.

3) If the disc. is - , the equation has two complex roots.

Here a = 1, b = -3 and c = 4. Therefore the discriminant is (-3)²-4(1)(4), or

-7. Rule 3) applies: the equation has two complex roots, but no real ones. Thus we know that the graph does not cross the x-axis.

Graphing the given quadratic, x² - 3x + 4, using a dashed "line," is helpful. As you can see in the illustration of this graph, the graph neither touches nor crosses the x-axis. Thus, y = x² - 3x + 4 is greater than 0 for all x. The answer: All real numbers.

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2 years ago

(2x^3+6x-1)+(3x^2-7x)

Neporo4naja [7]

Answer:

2x^3+2x-1

Step-by-step explanation:

3 0

2 years ago

Which angles are linear pairs. check all that apply

snow_tiger [21]

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6 0

2 years ago

Read 2 more answers

Determine whether the following lines represented by the vector equations below intersect, are parallel, are skew, or are identi

KiRa [710]

Answer:

r(t) and s(t) are parallel.

Step-by-step explanation:

Given that :

the lines represented by the vector equations are:

r(t)=⟨1−t,3+2t,−3t⟩

s(t)=⟨2t,−3−4t,3+6t⟩

The objective is to determine if the following lines represented by the vector equations below intersect, are parallel, are skew, or are identical.

NOTE:

Two lines will be parallel if $\dfrac{x_1}{x_2}= \dfrac{y_1}{y_2}= \dfrac{z_1}{z_2}$

here;

$d_1 = (-1, \ 2, \ -3)$

Thus;

$r(t) = \dfrac{x-1}{-1} = \dfrac{y-3}{2}=\dfrac{z-0}{-3} = t$

$d_2 =(2, \ -4, \ +6)$

$s(t) = \dfrac{x-0}{2} = \dfrac{y+5}{-4}=\dfrac{z-3}{6} = t$

∴

$\dfrac{d_1}{d_2}= \dfrac{-1}{2} = \dfrac{2}{-4}= \dfrac{-3}{-6}$

Hence, we can conclude that r(t) and s(t) are parallel.

6 0

2 years ago

What’s the answer to this?

Elina [12.6K]

Step-by-step explanation:

Line is passing through the points:

$(-2,\:3)=(x_1,\:y_1) \:\&\: (2,\:5)=(x_2 ,\:y_2)$

Equation of line in two point form is given as:

$\frac{y -y_1 }{y_1 -y_2 } = \frac{x -x_1 }{x_1 -x_2 } \\ \\ \therefore \frac{y -3}{ 3 -5} = \frac{x -(-2) }{-2 -2} \\ \\ \therefore \frac{y -3}{ - 2} = \frac{x -(-2) }{-4} \\ \\ \therefore \frac{y - 3}{1} = \frac{x +2}{2} \\ \\ \therefore \: y-3= \frac{x}{2} + \frac{2}{2} \\ \\\therefore \: y= \frac{x}{2} + 1 +3\\ \\ \huge \red{ \boxed{\therefore \: y= \frac{1}{2} \:x+ 4}}\\ is \: the \: required \: equation \: of \: line.$

4 0

2 years ago