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Jobisdone [24]
3 years ago
13

What is the slope for y=2

Mathematics
1 answer:
Arada [10]3 years ago
7 0
The slope would be zero
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How do you solve this question
Wewaii [24]

<h2>Solution:.</h2>

Let the ceilings be <em>a</em><em> </em><em>&</em><em> </em><em>b</em>

and the distance from one corner of the ceiling to the opposite be <em>c</em>

<em>then </em><em>using</em><em> </em><em>Pythagoras</em><em> theorem</em>

c =  \sqrt{a {}^{2} +  b  {}^{2} } \\  c =    \sqrt{16 {}^{2} +  12  {}^{2} } \\ c =  \sqrt{256 + 144 }  \\ c =  \sqrt{400}  \\ c = 20

hence ,c .°. the distance from one corner of the ceiling to the opposite is<em> </em><em>2</em><em>0</em>

3 0
2 years ago
Help brainliest and 10 points
evablogger [386]
A= 1
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4 0
3 years ago
Read 2 more answers
Write a polynomial function of least degree with the given zero. -1+ √ 2, √ 3
kirill115 [55]

Answer:

f(x) = x^{4} + 2x³ - 4x² - 6x + 3

Step-by-step explanation:

Note that radical zeros occur in conjugate pairs, thus

- 1 + \sqrt{2} is a zero then - 1 - \sqrt{2} is also a zero

\sqrt{3} is a zero then - \sqrt{3} is also a zero

Thus the corresponding factors are

(x - (- 1 + \sqrt{2}) ), (x - (- 1 - \sqrt{2}) ), (x - \sqrt{3}), (x - (- \sqrt{3})), that is

(x + 1 - \sqrt{2}), (x + 1 + \sqrt{2}), (x - \sqrt{3}), (x + \sqrt{3})

The polynomial is then the product of the roots

f(x) = (x + 1 - \sqrt{2})(x + 1 + \sqrt{2})(x - \sqrt{3})(x + \sqrt{3})

     = ((x + 1)² - (\sqrt{2})²)((x² - (\sqrt{3})²)

     = (x² + 2x + 1 - 2)(x² - 3)

     = (x² + 2x - 1)(x² - 3) ← distribute

     = x^{4} - 3x² + 2x³ - 6x - x² + 3

     = x^{4} + 2x³ - 4x² - 6x + 3

6 0
3 years ago
What is the value 49,254
erastova [34]
40,000
9,000
200
50
4

I'm guessing that's what you were looking for?
8 0
3 years ago
The point (x, y) is the top of a polygon. Which point would it map to if the polygon were reflected over the y-axis and then tra
Pachacha [2.7K]
It’s either b or C because your going down.... but I think it’s B
7 0
3 years ago
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