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3 years ago

What is the equation in point-slope form of a line that passes through the points (−4,−1) and (5,7) ?

Mathematics

1 answer:

stepan [7]3 years ago

4 0

Answer:

y+1=8/9(x+4)

Step-by-step explanation:

y-y1=m(x-x1)

m=(y2-y1)/(x2-x1)

m=(7-(-1))/(5-(-4))

m=(7+1)/(5+4)

m=8/9

y-(-1)=8/9(x-(-4))

y+1=8/9(x+4)

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Helpp plz..............

solong [7]

Answer:

Ivory is not correct.

Step-by-step explanation:

Ivory is not correct because of the way the first equation is treated. Ivory thinks 2x+2 can be double of what x+1 is. However, x can be anything, even 0. If we replace x with 0 in 2x+2, that would leave 2x = 0, and adding 2 would equal 2. If we look at x+1, we can add 1 to x and would equal 2. So Ivory is not correct, there is many equations other than this one that can equal.

6 0

3 years ago

Convert 0.4 into a fraction :) thanks! hugs!<3 show ALL work!!

Juliette [100K]

$0.4 =\frac{0.4}{1} = \frac{0.4\times 10}{1\times 10} = \frac{4}{10} = \frac{2}{5}$

When you multiply the numerator and denominator both by the same thing, it is called an equivalent fraction.

3 0

3 years ago

Read 2 more answers

Bonjour pourriez vous m'aider avec cette équation du second degré? j'ai du mal à comprendre le concept. merci

olga_2 [115]

If a quadratic equation has solutions, the standard procedure will always work: given an equation like

$ax^2+bx+c=0$

The solutions (if any) are given by

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

In your case, you have

$a=-9,\ b=0,\ c=1$

So the formula becomes

$x_{1,2} = \dfrac{\pm\sqrt{36}}{18}= \pm\dfrac{6}{18}=\pm\dfrac{1}{3}$

Anyway, this is quite a special case, because you're missing the linear term (since $b=0$ )

This means that you can solve equations like these more easily: rearrange the equation to make it look like $ax^2=c$ . In your case, it becomes

$9x^2=1$

Divide both sides by 9:

$x^2=\dfrac{1}{9}$

Now, you know that the square of a number equals 1/9. By definition, it means that this number is the square root of 1/9. Nevertheless, both the square root and its opposite are solutions of the equation, because the minus sign will cancel out when squaring.

So, in general, you have

$ax^2=c \iff x^2 = \dfrac{c}{a} \iff x=\pm\sqrt{\dfrac{c}{a}$