All categories

3 years ago

Which of the graphs above is the graph of the equation below? Y=x^3-6x^2+11x-6=(x-3)(x-2)(x-1)

Mathematics

1 answer:

____ [38]3 years ago

7 0

Answer:

answer Z

Step-by-step explanation:

Look for a graph that contains the following zeros: x = 1, x = 2 , x= 3, following the info derived by the binomial factors that the function contains. Also look ate the fact that the function in question has for leading term positive $x^3$ , then this function must go towards plus infinity when x becomes large. This is the case for the graph option Z (the last graph of the group)

You might be interested in

• The shortest side of a triangle is 8 centimeters long. The longest side is 8 cm longer than the

gizmo_the_mogwai [7]

The shortest side is 9, longest is 17 of the triangle IS NOT a 90 degree triangle

5 0

3 years ago

A manufacturer of new light bulbs claims the average lifetime of its long-life bulb is more than 4000 hours. To test this claim,

natka813 [3]

Answer: C. 12.5

Step-by-step explanation:

Given : A manufacturer of new light bulbs claims the average lifetime of its long-life bulb is more than 4000 hours.

Population mean : $\mu=4000$

Sample size : n= 100

Sample mean : $\overline[x}=4500$

Standard deviation: $s=400$

The value of test-statistic is given by :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}\\\\\Rightarrow\ z=\dfrac{4500-4000}{\dfrac{400}{\sqrt{100}}}\\\\\Rightarrow\ z= 12.5$

Hence, the value of the test statistic for this problem is 12.5.

5 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

6 lb 4 oz − 5 lb 10 oz =

FinnZ [79.3K]

The answer to your problem is 10 oz

5 0

3 years ago

Read 2 more answers

Three wires are 6.5 m, 8.19 m, and 4.457 m long. What is the total length of the wires? Give your answer with the appropriate pr

LenaWriter [7]

The answer for the problem above is 19.147 meters is the total length of the wires.

6 0

3 years ago

Read 2 more answers