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iogann1982 [59]
3 years ago
11

Check all the statement(s) that are true about the polynomial function graphed.

Mathematics
2 answers:
slava [35]3 years ago
7 0
The true statements are:

2 - we can tell this by looking at the far right of the graph, as the slope is going downwards, therefore the leading coefficient must be negative

3 - this is a cubic, meaning its degree is 3

6 - by looking at the graph, we can see that there are 3 points where it cuts the x axis, hence 3 real zeros

7 - even multiplicity is where the curve "bounces off" the x axis and does not cross it. This curve have no zeros with even multiplicity

Hope this helped!
Dennis_Churaev [7]3 years ago
4 0

Answer:

The answer for ed is 2,3,6,7

Step-by-step explanation:


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Thats hard i have never seen any thing like that
5 0
2 years ago
A path 1 m wide is build around a garden that is a 30-meter square. What is the area taken by the path?
maria [59]

Length of outer boundary =30m

So area of outer boundary =30×30

= 900m²

Length of inner boundary of path=30-1= 29m

Inner area =29×29 =841m²

So area of path is= outer area - inner area

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6 0
1 year ago
Root 3 cosec140° - sec140°=4<br>prove that<br><br>​
Lorico [155]

Answer:

Step-by-step explanation:

We are to show that \sqrt{3} cosec140^{0} - sec140^{0} = 4\\

<u>Proof:</u>

From trigonometry identity;

cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}

\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\

From trigonometry, 2sinAcosA = Sin2A

= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\=  \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\=  \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2  \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

\frac{4(sin(420-140)) }{sin280}

= \frac{4sin280}{sin280}\\ = 4 \ Proved!

3 0
2 years ago
the endpoints of the diameter of a circle are (−6, 6) and (6, −2), what is the standard form equation of the circle?
miss Akunina [59]
The equation of a circle in standard form is

(x - h)^2 + (y - k)^2 = r^2

where (h, k) is the center of the circle, and r is the radius if the circle.

We need to find the radius and center of the circle.
We are given a diameter, so to find the center, we need the midpoint of the diameter.

M = ((-6 + 6)/2, (6 + (-2))/2) = (0, 2)
The center is (0, 2).

To find the radius, we find the length of the given diameter and divided by 2.

d = \sqrt{(-6 - 6)^2 + (6 - (-2))^2)}

d = \sqrt{144 + 64}

d = \sqrt{208}

r = \dfrac{d}{2} = \dfrac{\sqrt{208}}{2} = \dfrac{\sqrt{208}}{\sqrt{4}} = \sqrt{52}

(x - 0)^2 + (y - 2)^2 = (\sqrt{52})^2

x^2 + (y - 2)^2 = 52
7 0
3 years ago
The answer should look like this. (___,___) plzzz fast. ​
BartSMP [9]

Answer:

( -1, 15/2) or ( -1, 7 1/2)

Step-by-step explanation:

Formula: ( (x1 + x2)/2, (y1 + y2)/2 )

1. ( (-4 +6)/2, (8 + 7)/2 ) Plug in

2. ( -2/2, 15/2) Add/Subtract

3. ( -1, 15/2) or ( -1, 7 1/2) Divide

5 0
3 years ago
Read 2 more answers
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