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nalin [4]
3 years ago
8

Suppose that a polynomial function of degree 5 with real coefficients has 4, -3i, and 9- 2i as zeros. Find the remaining zeros.

Mathematics
1 answer:
Naya [18.7K]3 years ago
6 0

9514 1404 393

Answer:

  3i, 9+2i

Step-by-step explanation:

Polynomials with real coefficients have complex zeros in conjugate pairs. The conjugates of the given complex zeros are the ones that are remaining:

  3i and 9+2i

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If a pyramid is cut with a plane parallel to the base, the intersection of the pyramid and the plane is ----- a cross section
Stella [2.4K]

Answer:

Square cross section (Red)

Step-by-step explanation:

If the pyramid is cut with a plane (green) parallel to the base, the intersection of the pyramid and the plane is a square cross section (red). If the pyramid is cut with a plane (green) passing through the top vertex and perpendicular to the base, the intersection of the pyramid and the plane is a triangular cross section (red)

4 0
4 years ago
In their first year, Josiah will be paid
Alinara [238K]

Answer:

their pay rate is not proportional because to be proportional the chart needs an equivalent ratio for each next layer, this one is not

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
There is a total of 25 bicycles. If the ratio of gray bicycles to black bicycles is 6 to 9, how many of them are black?
sveta [45]

the ratio is 6 to 9, 9 out of (6+9) 15 bicycles will be black.

25 * 9/15 = 15

there will be 15 black bicycles

8 0
3 years ago
If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

6 0
3 years ago
Pls helpppppppppppppppp
vredina [299]

Answer: D

Each time a new row and a new column is added. The last pattern had 3 rows and 4 columns, so the next pattern will have 4 rows and 5 column.

:)

5 0
3 years ago
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