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3 years ago

Which pairs of points would result in a line with a rate of change of 2:1? Select all that apply.

Mathematics

1 answer:

Helen [10]3 years ago

7 0

Answer:

Options A, C, D are true.

Step-by-step explanation:

We have to choose the pair of points from that given in the attached file which will result in a line with a rate of change i.e. slope of 2:1 i.e. 2.

A. The points are (-4,-11) and (-2,-7).

Then the slope of the joining line is $\frac{-7-(-11)}{-2-(-4)} =2$

B. The points are (5,6) and (-5,12).

Then the slope of the joining line is $\frac{12-6}{-5-5} =-\frac{3}{5}$

C. The points are (0,-3) and (-4,-11).

Then the slope of the joining line is $\frac{-11-(-3)}{-4-0} =2$

D. The points are (0,-3) and (5,7).

Then the slope of the joining line is $\frac{7-(-3)}{5-0} =2$

Therefore, options A, C, D are true. (Answer)

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Read 2 more answers

the equation is -2x² = 4-3 (x + 1) and the question is justify that it is a 2nd degree equation with the unknown x complete.

Serggg [28]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the equation

-2x² = 4-3 (x + 1)

-2x² = 4-3x-3

-2x² = -3x -7

0 = 2x² -3x -7

We know that the degree of the equation is the highest power of x variable in the given equation.

In the equation 0 = 2x² -3x -7 the highest power of x variable in the given equation is 2.

Thus, the degree of the equation is 2.

Also in the equation 0 = 2x² -3x -7, the unknown variable is 'x'.

Let us determine the value 'x'

2x² -3x -7 = 0

Add 7 to both sides

$2x^2-3x-7+7=0+7$

$2x^2-3x=7$

Divide both sides by 2

$\frac{2x^2-3x}{2}=\frac{7}{2}$

$x^2-\frac{3x}{2}=\frac{7}{2}$

Add (-3/4)² to both sides

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{7}{2}+\left(-\frac{3}{4}\right)^2$

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{65}{16}$

$\left(x-\frac{3}{4}\right)^2=\frac{65}{16}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{3}{4}=\sqrt{\frac{65}{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{\sqrt{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{4}$

Add 3/4 to both sides

$x-\frac{3}{4}+\frac{3}{4}=\frac{\sqrt{65}}{4}+\frac{3}{4}$

$x=\frac{\sqrt{65}+3}{4}$

similarly solving

$x-\frac{3}{4}=-\sqrt{\frac{65}{16}}$

$x=\frac{-\sqrt{65}+3}{4}$

So the solution of the equation will have the values of x such as:

$x=\frac{\sqrt{65}+3}{4},\:x=\frac{-\sqrt{65}+3}{4}$

6 0

3 years ago

this is urgent please help i need this qucik : neetu said that the sum of -8+d is always positive whenthe value of d is greater

hjlf

Answer:

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Step-by-step explanation:

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4 years ago

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