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4 years ago

Tell me the linear equation for point-slope form? SOMEONE HELP ME

Mathematics

1 answer:

kolezko [41]4 years ago

4 0

I believe that it’s this? y-y₁=m(x-x₁)

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Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

3 years ago

A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.63 and a sample standard deviatio

ss7ja [257]

To solve for the confidence interval for the true average percentage elongation, we use the z statistic. The formula for confidence interval is given as:

Confidence interval = x ± z σ / sqrt (n)

where,

x = the sample mean = 8.63

σ = sample standard deviation = 0.79

n = number of samples = 56

From the standard distribution tables, the value of z at 95% confidence interval is:

z = 1.96

Therefore substituting the known values into the equation:

Confidence interval = 8.63 ± (1.96) (0.79) / sqrt (56)

Confidence interval = 8.63 ± 0.207

Confidence interval = 8.42, 8.84

5 0

3 years ago

WILL GIVE THE BRAINIEST, THANKS, AND A 5 STAR RATING WITH 15 POINTS HELP ME PLZZ.

Alona [7]

Ermm, Don't you think it's 60?

Cause in my caculatons it's 60

5 0

4 years ago

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An 8 pack of light bulbs cost $22.40 what’s the unit price

alekssr [168]

Answer:

$179.20

Step-by-step explanation:

Hope this helps

3 0

2 years ago

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Combine like terms. 2ac+4b+6ac-7a-b=

wolverine [178]

Answer:

8ac+3b-7a

Step-by-step explanation:

2ac+6ac+4b-b-7a=

3 0

4 years ago

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