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Naily [24]
2 years ago
10

a line passes through the point (-9,1) and has a slope of 4/3. Write an equation in slope-intercept form for this line.

Mathematics
2 answers:
Strike441 [17]2 years ago
6 0

Answer:  y=4/3x+1

Step-by-step explanation:

The formula is Y=mx+b

b is the y-intercept and MX is slope

belka [17]2 years ago
5 0
Y=4/3x+1. This will be your answer.
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Simplify, 99/121 whats the GCF?
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To find the GCF:

-- List the factors of the first number.
-- List the factors of the second number.
-- Make a short list of the factors that show up for BOTH numbers.
-- Find the biggest number on the short list.

Factors of the first number (99):  <u>1</u>, 3, 9, <u>11</u>, 33, 99 .

Factors of the second number (121):  <u>1</u>, <u>11</u>, 121 .

Short list (factors that show up for both numbers):  1, 11

Biggest number on the short list:  <em>11</em>


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Line segment GH is congruent to line segment IJ. Which of the following is an equivalent statement?
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2 years ago
Read 2 more answers
Help needed plez<br><br>Ans in (ii) is 1.5
Deffense [45]
OK first let's check the x=1.5. 

y = \frac x 6 (x^2 - 10)

y = 1-2x

1-2x = \frac x 6 (x^2 - 10)

6-12x = x^3 - 10 x

x^3 + 2x-6 = 0

Oh my, that's called a depressed cubic, no x^2 term. There's a formula for these very much like the quadratic formula but you're probably not quite old enough for that.  Anyway, x\approx 1.45616 \approx 1.5 is a solution, but that's not what they're asking.  They are asking us to compare 

x^3 + 2x-6 = 0

with

x^3 + bx^2 + cx + d = 0

and conclude b=0, c= 2, d=-6

It turns out we did need all the rest of it.  Save those brain cells, there's lots more math coming.


~~~~~~~~~~~~~~

I love it when the student asks for more.  Here's the formula for a depressed cubic. I won't derive it here (though I did earlier today, coincidentally, but I'm probably not allowed to link to my Quora answer "what led to the discovery of complex numbers" from here).  We use the trick of putting coefficients on the coefficients to avoid fractions.

x^3 + 3 p x = 2 q

has solutions

x = \sqrt[3] { q - \sqrt{p^3 + q^2} } + \sqrt[3] {q + \sqrt{p^3 + q^2} } &#10;&#10;


That's pretty simple, though sometimes we end up having to take the cube roots of complex numbers, which isn't that helpful.  Let's try it out on


x^3 + 2x=6


That's p=2/3, q=3 so

x = \sqrt[3] { 3 - \sqrt{(2/3)^3+9} } + \sqrt[3] {3 + \sqrt{(2/3)^3+9} }

x = \sqrt[3] { 3 - \sqrt{753}/9 } +\sqrt[3]{3 + \sqrt{753}/9 }

x \approx 1.4561642461359084609748069666






6 0
3 years ago
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