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Viktor [21]
3 years ago
15

Suppose a polynomial of degree 4 with rational coefficients has the given numbers as zeros. Find the other zero.

Mathematics
1 answer:
Nady [450]3 years ago
3 0

Answer:

x_4 = -5 + i

Step-by-step explanation:

See comment for complete question.

From the complete question, we have:

Zeros \to -2, -6, -5-i

Required

Determine the other zero

The zeros can be represented as:

x_1 = -2

x_2 = -6

x_3 = -5-i

Considering the 3rd zero, which is a complex number.

If a polynomial has a complex number as one of its zeros, then its conjugate pair is the corresponding zero

A conjugate pair is given as:

a + bi and a - bi

For x_3 = -5-i

The conjugate pair is:

x_4 = -5 + i

Hence, the other zero is: -5 + i

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PLEASE HELP ASAP
Serga [27]

Answer:

A, B, C, and E

Step-by-step explanation:

SSA doesn't work because this leaves room for two possible answers in some situations.

The other postulate, HL (hypotenuse-leg) works for right triangles; it's basically a short cut for SSS postulate because if the hypotenuse and leg of a right triangle are congruent to the hypotenuse and leg on another, the other side also has to be congruent.

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3 years ago
Read 2 more answers
Factor 26r3s + 52r5 – 39r2s4.
algol [13]

Answer:

13r²(2rs + 4r³ - 3s⁴)

Step-by-step explanation:

In equation 26r³s + 52r⁵ - 39r²s⁴;

The GCF of 26, 52, and 39 = 13

The GCF of r³, r⁵ and r² = r²

The GCF of s, (no "s"), and s⁴ = no "s" (Since one of the number doesn't have "s")

Now we can factor out 13r² from all three expressions;

26r³s + 52r⁵ - 39r²s⁴

=> <em>13r²(2rs) + 13r²(4r³) - 13r²(3s⁴)</em>

To factor it all together;

<u>13r²(2rs + 4r³ - 3s⁴)</u>

Hope this helps!

7 0
3 years ago
Examine this system of equations. Which numbers can be multiplied by each equation so that when the two equations are added toge
podryga [215]

Answer:

18 times the first equation and 8 times the second equation

Step-by-step explanation:

we have

\frac{1}{4}x-\frac{1}{6}y=5 ------> first equation

\frac{4}{5}x+\frac{3}{8}y=10 ------> second equation

step 1

Multiply the second equation  by 8 both sides to remove the fraction in the variable y

8(\frac{4}{5}x+\frac{3}{8}y)=10(8)

\frac{32}{5}x+3y=80

step 2

Multiply the first equation  by 18 both sides to obtain the coefficient -3 in the variable y

18(\frac{1}{4}x-\frac{1}{6}y)=5(18)

\frac{18}{4}x-3y=90

step 3

Adds the new equation 1 and equation 2

\frac{18}{4}x-3y=90\\\\ \frac{32}{5}x+3y=80\\ -------

The y-term is eliminated

therefore

The answer is

18 times the first equation and 8 times the second equation

3 0
4 years ago
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