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GREYUIT [131]
3 years ago
10

A 4th degree polynomial function has zeros at 3 and 5-i. Can 4+i also be a zero of the function?

Mathematics
1 answer:
kicyunya [14]3 years ago
8 0
Sure, provided that the coefficients of the polynomial are complex, and not strictly real.

If you require that the coefficients of the polynomial be real, then the answer would be no. For such a polynomial, any complex roots occur in conjugate pairs, so if 5-i is a root, then so must be 5+i.

A degree-4 polynomial can have up to four distinct roots, so if the coefficients are all real, then the remaining fourth root must be real.
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∠ABC and​ ∠DEF ​are supplementary.
MatroZZZ [7]
As the angles are supplementary, their sum would be equal to 180
∠ABC + <span>∠DEF = 180
138 + </span><span>∠DEF = 180
</span><span>∠DEF = 180 - 138
</span><span>∠DEF = 42

In short, Your Answer would be Option A

Hope this helps!</span>
8 0
3 years ago
plz help... Emily wants to make 1000 cranes. she can make 80 cranes a day. if she has already made 304 cranes,how many dayz are
forsale [732]
She would take about 9 days to make 1000 cranes
1000-304=696
696/80= 8.7
8.7 rounded is 9
3 0
4 years ago
Use the figure for Questions 3 and 4.
harina [27]

Answer:

Use the figure for Questions 3 and 4.

(700

3. What is m2?

B 70

A 50

60° 3/4

C 110

D 120

40

4. What is m24?

A 10 B 60

C 100

D 120

Step-by-step explanation:

Use the figure for Questions 3 and 4.

(700

3. What is m2?

B 70

A 50

60° 3/4

C 110

D 120

40

4. What is m24?

A 10 B 60

C 100

D 120

7 0
3 years ago
Someone help me with this
puteri [66]

Answer:

A) t = 1 second to reach the highest point

B) h = 96 feet

C) t = 3.449 seconds to hit the ground

Step-by-step explanation:

We notice that the equation that describes the position of the object in terms of the time, is a parabola (quadratic) with negative leading coefficient. therefore this is a parabola with arms pointing down, and with vertex at the top. It is at that top vertex that the maximum altitude of the object is reached.

We need therefore to find the position of the vertex (time "t" is the horizontal coordinate and height "h" is the vertical coordinate).

We know that the vertex (maximum of a parabola of this type - normally described by y=ax^2+bx+c) is given by x_{vertex} = -\frac{b}{2*a}. Therefore in our case, understanding that time "t" is equivalent to the variable "x", and eight "h" is equivalent to the variable "y", that "-16" is a, and "32" is b, we have that the maximum occurs for:

t_{max}=-\frac{32}{2*(-16)} = \frac{-32}{-32} = 1

which means at one second from the launching.

We then can find the answer to part B by replacing t with "1" second in the formula for the height "h":

h(1)=-16*(1)^2+32(1)+80=-16+32+80=96

That is : a height of 96 feet.

To find the answer for part C: the time to hit the ground, we solve for the variable "t" in the given expression for the height, by setting the height to zero (object touching the ground) and using the quadratic formula:

0=-16t^2+32t+80\\t=\frac{-32+/-\sqrt{32^2-4*(-16)*80} }{2*(-16)} \\ t=\frac{-32+/-\sqrt{6144} }{-32}

which gives us two solutions: t = -1.449 seconds, and t = 3.449 seconds

Since the negative time does not have physical meaning in our case (time before the object was launched), we adopt the second answer: t = 3.449 seconds

5 0
3 years ago
Answer asap please dont get this either
iren2701 [21]

Answer:

(0,-1)

Step-by-step explanation:

The y intercept is where it crosses the y axis

It crosses at -1

The y intercept is (0,-1)

3 0
4 years ago
Read 2 more answers
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