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3 years ago

If you know that 7 - square root 5 is it root of a polynomial function

Mathematics

1 answer:

Kitty [74]3 years ago

4 0

Answer:

As you must know, If one root of the polynomial 7-√5, the other will be 7+√5 i.e irrational root occur in pairs.

A polynomial function cannot have single unreal i.e irrational root. It always occur in pairs.

So , consider a polynomial function of any degree, if it has a root 7-√5, then it must have another root as 7+√5.

A polynomial can't have 7-√5 as a single root.

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PLEASE HELP!!

SIZIF [17.4K]

(x, y)

Domain = x

Range = y

It is a function when none of the x values are repeating.

Answer - C) Both A and B

Hope this helps and makes sense!!

8 0

3 years ago

There are 16 teams in a tournament. if during the first round, each team plays every other team exactly once, how many games wil

jasenka [17]

16x17=272/2=136. That is answer.

5 0

3 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2

Read more about the plane

brainly.com/question/1962726

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5 0

1 year ago

A company produces a cardboard box in the shape of a rectangular prism. The surface area of the box is represented by S=10x2, wh

OLEGan [10]

S = 10x^2

S/10 = x^2

+or-sqrt(S/10) = x

+or- sqrt(S10)/10 = x

_inverse = +or-sqrt(10x)/10

5 0

3 years ago

Sandra bought 4 bars of chocolate and a packet of nuts for $11.50. If a bar of chocolate costs $0.50 less than a packet of nuts,

Luden [163]

Answer:

A packet of nuts costs $2.70

Step-by-step explanation:

Let C represent the cost of a bar of chocolate and P represent the cost of a packet of nuts

4 bars of chocolate will cost $4C

1 packet of nuts costs $P

Total price Sandra paid is $11.50

This can be represented using

4C + P = 11.50 (1)

A bar of chocolate costs $0.50 less than a packet of nuts. So

P - C =0.50

-C + P = 0.50 (2)

Multiply eq (2) by 4 to get

-4C + 4P = 4 x 0.50

-4C + 4P = 2 (3)

Add equations (1) and (3) to eliminate the 4C term

4C + P - 4C + 4P = 11.50 + 2

P+4P = 13.50

5P = 13.50

Dividing by 5 on both sides

5P/5 = 13.50/5

P = 2.70

So a packet of nuts costs $2.70 (Answer)

To verify this is correct,

Use equation (2) to get

-C + P = 0.50

-C + 2.70 = 0.50

Add C to both sides

-C + 2.70 + C = 0.50 + C

2.70 = 0.50 + C

Subtract 0.50 from both sides

2.70-0.50 = 0.50 + C - 0.50

2.20 = C which is C= 2.20

Plug this value into equation (1)

4(2.20) + 2.70 = 8.80 + 2.70 = 11.50

So verified that answers are correct

8 0

1 year ago