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SCORPION-xisa [38]
3 years ago
11

A polynomial that has 8 turning points is multiplied by another polynomial that has 7 zeros. What is the minimum degree of the n

ew polynomial?
Mathematics
1 answer:
Nata [24]3 years ago
6 0
If there are 8 turning points in a polynomial graph, then the degree of the polynomial is n+1 or 9. On the other hand, this also applies to the number of zeros present in the function. If there are 7 zeros, then the degree is 8. Adding both degrees results to the resulting polynomial's degree is 16. 
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When looking at sides AC and CB, what is the included angle?<br> Question 22 options:
Vladimir [108]

Answer:

64 degrees

Step-by-step explanation:

since it is in between the both of them and they both share the 62 degree angle

5 0
3 years ago
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2×3+4-5 <br>please help ​
Andru [333]

Step-by-step explanation:

PEMDAS

parentheses, exponents, multiplication, division, addition, subtraction.

first multiply:

2×3=6

6+4-5

Now add

6+4=10

Now subtract

10-5=5

5 is the answer

<em><u>I</u></em><em><u> </u></em><em><u>hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>helped</u></em><em><u>!</u></em><em><u> </u></em><em><u>:</u></em><em><u>)</u></em>

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3 years ago
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The table above shows the number of hours that Robin worked each week for 5 weeks. What would be the number of hours Robin worke
ELEN [110]
Robin would have worked a total of 10 days!!! Hope this helps :) :) :)
8 0
3 years ago
3RD TIME ASKING PLS HELP
Svet_ta [14]

The angles surrounding the point p are 90 degrees

Then using the angle given, find the missing angles.

JPH = 90

HJP = 49

90+49 = 139

180 - 139 = 41

Therefore JHP = 41

HLP = HJP = 49

HLP = 49

KLP = 72

KLP + HLP = 72 + 49 = 121

Therefore, HLK = 121

LP = JP = 16

JP = 16

HL = HJ = 13

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4 0
3 years ago
Can someone please solve this problem
ruslelena [56]

ANSWER

\boxed { \sqrt{} }30 \degree

\boxed { \sqrt{} }210 \degree

EXPLANATION

We want to solve

\cot( \theta)  =  \sqrt{3}

where

0 \degree \:  \leqslant x \leqslant 360 \degree

We reciprocate both sides of this trigonometric equation to obtain:

\tan( \theta)  =  \frac{1}{ \sqrt{3} }

We take arctangent of both sides to get;

\theta =  \tan ^{ - 1} ( \frac{1}{ \sqrt{3} } )

\theta = 30 \degree

This is the principal solution.

The tangent ratio is also positive in the third quadrant.

The solution in the third quadrant is

180 +  \theta = 180 + 30 = 210 \degree

3 0
3 years ago
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