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Aleksandr [31]
3 years ago
14

When added, angles B, C, and D should all equal ___ degrees.

Mathematics
1 answer:
jenyasd209 [6]3 years ago
3 0

Answer:

Maybe show a picture?

Step-by-step explanation:

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HELP PLEASEEEEEEEEEEEEEE
vazorg [7]

Answer:

-2/4,  8/40, 21/30, 9, 9 20/25

Step-by-step explanation:

9/20/25 is a bit greater than 9.

21/30 is 9 away from a whole 8/40 is 32 away from 40 -2/4 is way less because its on the negative side which is like so far away from the other numbers.

7 0
3 years ago
What are the coordinates of the vertices of the polygon in the graph that are in Quadrant II? A) (4,–2) B) (4,3), (0,5), (0,1) C
Vinil7 [7]

Answer:

C) (–5,2), (–3,2), (–3,4)

Step-by-step explanation:

A) (4,–2)

B) (4,3), (0,5), (0,1)

<em><u>C) (–5,2), (–3,2), (–3,4)</u></em>

D) (–1,0), (–5,2), (–3,2), (–3,4), (0,5), (0,1)

For quadrant two the points are always (-x,y) and x is always negative.

Image shows quadrant places.

7 0
4 years ago
Read 2 more answers
Let $m$ be the number of integers $n$, $1 \le n \le 2005$, such that the polynomial $x^{2n} + 1 + (x + 1)^{2n}$ is divisible by
Kisachek [45]

x^2+x+1=\dfrac{x^3-1}{x-1}

so we know x^2+x+1 has roots equal to the cube roots of 1, not including x=1 itself, which are

\omega=e^{2i\pi/3}\text{ and }\omega^2=e^{4i\pi/3}

Any polynomial of the form p_n(x)=x^{2n}+1+(x+1)^{2n} is divisible by x^2+x+1 if both p(\omega)=0 and p(\omega^2)=0 (this is the polynomial remainder theorem).

This means

p(\omega)=\omega^{2n}+1+(1+\omega)^{2n}=0

But since \omega is a root to x^2+x+1, it follows that

\omega^2+\omega+1=0\implies1+\omega=-\omega^2

\implies\omega^{2n}+1+(-\omega^2)^{2n}=0

\implies\omega^{4n}+\omega^{2n}+1=0

and since \omega^3=1, we have \omega^{4n}=\omega^{3n}\omega^n=\omega^n so that

\implies\omega^{2n}+\omega^n+1=0

From here, notice that if n=3k for some integer k, then

\omega^{2(3k)}+\omega^{3k}+1=1+1+1=3\neq0

\omega^{4(3k)}+\omega^{2(3k)}+1=1+1+1=3\neq0

which is to say, p(x) is divisible by x^2+x+1 for all n in the given range that are *not* multiples of 3, i.e. the integers 3k-2 and 3k-1 for k\ge1.

Since 2005 = 668*3 + 1, it follows that there are m=668 + 669 = 1337 integers n such that x^2+x+1\mid p(x).

Finally, m\equiv\boxed{337}\pmod{1000}.

8 0
4 years ago
Does anyone know the answer of all these questions if so I will be Very glad if you help ​
USPshnik [31]

Answer:

6) 22,408

7) 8,993

8) 68,701

Step-by-step explanation:

6) 122408 - 100000 = 22,408

7) 8993/999 = 9 2/999

8) 98,700 - 29,999 = 68,701

7 0
3 years ago
Can someone help me with this?
Alex73 [517]

Answer:

A. 6 hours

B. No students studied for 6 hours

C. 4 students

(I’m just using a educated guess for B)

5 0
3 years ago
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