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3 years ago

Perform the indicated operation to write given polynomial in standard form: (x+ 2)(2x^2 − 5x+7)

Mathematics

1 answer:

Elan Coil [88]3 years ago

5 0

Answer:

So the expression in standard form will be $2x^3-x^2-3x-14$

Step-by-step explanation:

We have given expression $(x+2)(2x^-5x+7)$

We have to write this expression in standard form

For writing in standard form first we have to multiply the expression

So after multiplication number $(x+2)(2x^2-5x+7)=2x^3-5x^2+7x+4x^2-10x-14=2x^3-x^2-3x-14$

So the expression in standard form will be $2x^3-x^2-3x-14$

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Let a=(1,2,3,4), b=(4,3,2,1) and c=(1,1,1,1) be vectors in R4. Part (a) [4 points]: Find (a⋅2c)b+||−3c||a. Part (b) [6 points]:

love history [14]

Solution :

Given :

a = (1, 2, 3, 4) , b = ( 4, 3, 2, 1), c = (1, 1, 1, 1) ∈ $R^4$

a). (a.2c)b + ||-3c||a

Now,

(a.2c) = (1, 2, 3, 4). 2 (1, 1, 1, 1)

= (2 + 4 + 6 + 6)

= 20

-3c = -3 (1, 1, 1, 1)

= (-3, -3, -3, -3)

||-3c|| = $\sqrt{(-3)^2 + (-3)^2 + (-3)^2 + (-3)^2 }$

$=\sqrt{9+9+9+9}$

$=\sqrt{36}$

= 6

Therefore,

(a.2c)b + ||-3c||a = (20)(4, 3, 2, 1) + 6(1, 2, 3, 4)

= (80, 60, 40, 20) + (6, 12, 18, 24)

= (86, 72, 58, 44)

b). two vectors $\vec A$ and $\vec B$ are parallel to each other if they are scalar multiple of each other.

i.e., $\vec A=r \vec B$ for the same scalar r.

Given $\vec p$ is parallel to $\vec a$ , for the same scalar r, we have

$\vec p = r (1,2,3,4)$

$\vec p = (r,2r,3r,4r)$ ......(1)

Let $\vec q = (q_1,q_2,q_3,q_4)$ ......(2)

Now given $\vec p$ and $\vec q$ are perpendicular vectors, that is dot product of $\vec p$ and $\vec q$ is zero.

$q_1r + 2q_2r + 3q_3r + 4q_4r = 0$

$q_1 + 2q_2 + 3q_3 + 4q_4 = 0$ .......(3)

Also given the sum of $\vec p$ and $\vec q$ is equal to $\vec b$ . So

$\vec p + \vec q = \vec b$

$(r,2r,3r,4r) + (q_1+q_2+q_3+q_4)=(4, 3,2,1)$

∴ $q_1 = 4-r , \ q_2=3-2r, \ q_3 = 2-3r, \ q_4=1-4r$ ....(4)

Putting the values of $q_1,q_2,q_3,q_4$ in (3),we get

$r=\frac{2}{3}$

So putting this value of r in (4), we get

$\vec p =\left( \frac{2}{3}, \frac{4}{3}, 2, \frac{8}{3} \right)$

$\vec q =\left( \frac{10}{3}, \frac{5}{3}, 0, \frac{-5}{3} \right)$

These two vectors are perpendicular and satisfies the given condition.

c). Given terminal point is $\vec a$ is (-1, 1, 2, -2)

We know that,

Position vector = terminal point - initial point

Initial point = terminal point - position point

= (-1, 1, 2, -2) - (1, 2, 3, 4)

= (-2, -1, -1, -6)

d). $\vec b = (4,3,2,1)$

Let us say a vector $\vec d = (d_1, d_2,d_3,d_4)$ is perpendicular to $\vec b.$

Then, $\vec b.\vec d = 0$

$4d_1+3d_2+2d_3+d_4=0$

$d_4=-4d_1-3d_2-2d_3$

There are infinitely many vectors which satisfies this condition.

Let us choose arbitrary $d_1=1, d_2=1, d_3=2$

Therefore, $d_4=-4(-1)-3(1)-2(2)$

= -3

The vector is (-1, 1, 2, -3) perpendicular to given $\vec b.$

6 0

2 years ago

Daniel and his children went into a bakery and where they sell cupcakes for $3.75 each and donuts for $1 each. Daniel has $20 to

grigory [225]

The system of inequalities are x + y ≥ 7 and 3.75x + 1y ≤ 20 The one possible solution is 4 cupcakes and 3 donuts.

Solution:

Given, Daniel and his children went into a bakery and where they sell cupcakes for $3.75 each and donuts for $1 each.

Daniel has $20 to spend and must buy a minimum of 7 cupcakes and donuts altogether.

⇒ "x" represents the number of cupcakes purchased

⇒ "y" represents the number of donuts purchased,

we have to write and solve a system of inequalities graphically and determine one possible solution.

Now, he should buy minimum 7 items, then x + y ≥ 7 ⇒ (1)

And, he has $20, so his maximum purchase is $20 then, 3.75x + 1y ≤ 20 ⇒ (2)

Now, suppose that he took 5 cupcakes, then he must take at least 2 donuts to satisfy (1)

So, substitute x and y values in (2)

⇒ 3.75(5) + 1(2) ≤ 20

⇒ 20.75 ≤ 20 ⇒ condition failed

So he can take maximum of 4 cupcakes only, and subsequently he has to take minimum of 3 donuts.

Hence, the one possible solution is 4 cupcakes and 3 donuts

7 0

2 years ago

Whats the answer to y=3x+9

MakcuM [25]

Answer:

x=-3

Step-by-step explanation:

so you tryna get x by itself, so you are gonna do the opposite. so move nine over by subtracting it.

-9=3x

now you have to divide -9 by 3 to move it over

-2=x

7 0

3 years ago

Use lagrange multipliers to find the maximum volume of a rectangular box that is inscribed in a sphere of radius r

vesna_86 [32]

For the derivative tests method, assume that the sphere is centered at the origin, and consider the circular projection of the sphere onto the xy-plane. An inscribed rectangular box is uniquely determined

by the xy-coordinate of its corner in the first octant, so we can compute the z coordinate of this corner by

x2+y2+z2=r2 =⇒z= r2−(x2+y2). Then the volume of a box with this coordinate for the corner is given by

V = (2x)(2y)(2z) = 8xy r2 − (x2 + y2),

and we need only maximize this on the domain x2 + y2 ≤ r2. Notice that the volume is zero on the boundary of this domain, so we need only consider critical points contained inside the domain in order to carry this optimization out.

For the method of Lagrange multipliers, we optimize V(x,y,z) = 8xyz subject to the constraint x2 + y2 + z2 = r2.

7 0

3 years ago

According to the National Association of Colleges and Employers, the average starting salary for new college graduates in health

katen-ka-za [31]

Answer:

a) The probability that a new college graduate in business will earn a starting salary of at least $65,000 is P=0.22965 or 23%.

b) The probability that a new college graduate in health sciences will earn a starting salary of at least $65,000 is P=0.11123 or 11%.

c) The probability that a new college graduate in health sciences will earn a starting salary of less than $40,000 is P=0.14686 or 15%.

d) A new college graduate in business have to earn at least $77,133 in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences.

Step-by-step explanation:

a. What is the probability that a new college graduate in business will earn a starting salary of at least $65,000?

For college graduates in business, the salary distributes normally with mean salary of $53,901 and standard deviation of $15,000.

To calculate the probability of earning at least $65,000, we can calculate the z-value:

$z=\frac{x-\mu}{\sigma} =\frac{65000-53901}{15000} =0.74$

The probability is then

$P(X>65,000)=P(z>0.74)=0.22965$

The probability that a new college graduate in business will earn a starting salary of at least $65,000 is P=0.22965 or 23%.

b. What is the probability that a new college graduate in health sciences will earn a starting salary of at least $65,000?

For college graduates in health sciences, the salary distributes normally with mean salary of $51,541 and standard deviation of $11,000.

To calculate the probability of earning at least $65,000, we can calculate the z-value:

$z=\frac{x-\mu}{\sigma} =\frac{65000-51541}{11000} =1.22$

The probability is then

$P(X>65,000)=P(z>1.22)=0.11123$

The probability that a new college graduate in health sciences will earn a starting salary of at least $65,000 is P=0.11123 or 11%.

c. What is the probability that a new college graduate in health sciences will earn a starting salary less than $40,000?

To calculate the probability of earning less than $40,000, we can calculate the z-value:

$z=\frac{x-\mu}{\sigma} =\frac{40000-51541}{11000} =-1.05$

The probability is then

$P(X$

The probability that a new college graduate in health sciences will earn a starting salary of less than $40,000 is P=0.14686 or 15%.

d. How much would a new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences?

The z-value for the 1% higher salaries (P>0.99) is z=2.3265.

The cut-off salary for this z-value can be calculated as:

$X=\mu+z*\sigma=51,541+2.3265*11,000=51,541+25,592=77,133$

A new college graduate in business have to earn at least $77,133 in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences.

8 0

3 years ago