Ask question

Ask question

All categories

3 years ago

5

What is the product? 3x^5 (2x^2+4x+1)

2 answers:

s2008m [1.1K]3 years ago

8 0

Answer:

The product of the given polynomial $3x^5(2 x^2+4x+1)$ is $6x^7+12x^6+3x^5$

Step-by-step explanation:

Given: Polynomial $3x^5(2 x^2+4x+1)$

We have to find the product of the given polynomial $3x^5(2 x^2+4x+1)$

Consider the given polynomial $3x^5(2 x^2+4x+1)$

Apply distributive rule, $a(b+c)=ab+ac$

Multiply $3x^5$ with each term in brackets, we have,

$=3x^5\cdot \:2x^2+3x^5\cdot \:4x+3x^5\cdot \:1$

$=3\cdot \:2x^5x^2+3\cdot \:4x^5x+3\cdot \:1\cdot \:x^5$

Apply exponent rule, $a^b\cdot \:a^c=a^{b+c}$

Simplify, we have,

$=6x^7+12x^6+3x^5$

Thus, The product of the given polynomial $3x^5(2 x^2+4x+1)$ is $6x^7+12x^6+3x^5$

almond37 [142]3 years ago

5 0

<span><span>the answer is C.) 6x^7...
</span></span>

You might be interested in

Find the amplitude y = -5 sin one half x

Alex_Xolod [135]

Answer:

amplitude=5

Step-by-step explanation:

y=a sin x

amplitude =|a|

here a=-5

|a|=5

8 0

3 years ago

Could someone please help me out? I've tried to solve this but I'm really having some trouble with it. Please try to explain it

guajiro [1.7K]

To find the area of the shaded region you need find the area of the shaded region and subtract the area of the unshaded region.

Area of a rectangle = width x length

A = (x + 10) x (2x + 5)

Next apply FOIL or First Outer Inner Last

A = (x * 2x) (x * 5) (10 * 2x) (10 * 5)

A= 2x2 + 5x + 20x + 50

A= 2x2 +25x +50

Area of a square= sides2

A= (x + 1)2

A= (x+1) (x+1)

Next apply FOIL or First Outer Inner Last

A = (x *x) (1*x) (1*x) (1*1)

A = x2 + 1x + 1x +1

A= x2 + 2x +1

A= 2x2 +25x +50 - 2x2 +25x +50

A= 50x + 100

4 0

3 years ago

Jobisdone [24]

Answer:19683x

Step-by-step explanation:

Evaluate the exponent

27.93x

27.729x

i am pretty sure sorry if I am wrong

7 0

3 years ago

What is the scale factor from ABC to DEF?

ollegr [7]

Answer:

C

Step-by-step explanation:

The scale factor is the ratio of corresponding sides, image to original, so

scale factor = $\frac{EF}{BC}$ = $\frac{11}{66}$ = $\frac{1}{6}$ → C

8 0

3 years ago

A point H is 20m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the po

astra-53 [7]

Answer:

a. See Attachment 1

b. $PT = 12.3\ m$

c. $HT = 31.1\ m$

d. $OH = 28.4\ m$

Step-by-step explanation:

Calculating PT

To calculate PT, we need to get distance OT and OP

Calculating OT;

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

$tan50 = \frac{OT}{20}$

Multiply both sides by 20

$20 * tan50 = \frac{OT}{20} * 20$

$20 * tan50 = OT$

$20 * 1.1918 = OT$

$23.836 = OT$

$OT = 23.836$

Calculating OP;

We have to consider angle 30, distance OH and distance OP

The relationship between these parameters is;

$tan30 = \frac{OP}{20}$

Multiply both sides by 20

$20 * tan30 = \frac{OP}{20} * 20$

$20 * tan30 = OP$

$20 * 0.5774= OP$

$11.548 = OP$

$OP = 11.548$

$PT = OT - OP$

$PT = 23.836 - 11.548$

$PT = 12.288$

$PT = 12.3\ m$ (Approximated)

--------------------------------------------------------

Calculating the distance between H and the top of the tower

This is represented by HT

HT can be calculated using Pythagoras theorem

$HT^2 = OT^2 + OH^2$

Substitute 20 for OH and $OT = 23.836$

$HT^2 = 20^2 + 23.836^2$

$HT^2 = 400 + 568.154896$

$HT^2 = 968.154896$

Take Square Root of both sides

$HT = \sqrt{968.154896}$

$HT = 31.1\ m$ <em>(Approximated)</em>

--------------------------------------------------------

Calculating the position of H

This is represented by OH

See Attachment 2

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

$tan50 = \frac{OH}{OT}$

Multiply both sides by OT

$OT * tan50 = \frac{OH}{OT} * OT$

$OT * tan50 = {OH$

$OT * 1.1918 = OH$

Substitute $OT = 23.836$

$23.836 * 1.1918 = OH$

$28.4= OH$

$OH = 28.4\ m$ <em> (Approximated)</em>

5 0

3 years ago