All categories

3 years ago

Please help .......

Mathematics

1 answer:

WITCHER [35]3 years ago

7 0

f(n) = n² + 20

f(1) = 1² + 20 = 21

f(2) = 2² + 20 = 24

f(3) = 3² + 20 = 29

f(4) = 4² + 20 = 36

f(5) = 5² + 20 = 45

f(6) = 6² + 20 = 56

a) first three terms

Answer: 21, 24, 29

b) how many < 50

f(5)=45, f(6)=56 so

Answer: 5

You might be interested in

4. (01.02 MC)

sleet_krkn [62]

Answer:

It is A [1st option]

Step-by-step explanation:

Use process of elimination.

The last 2 would not work since 3^0 is 1 not 0.

Then you can eliminate the 2nd option as it says 2^3^2 is 2^5 which is wrong... it would be 2^6

Therefore, it is A

5 0

3 years ago

The variables A, B, and C represent polynomials where A = x + 1, B = x2 + 2x − 1, and C = 2x. What is AB + C in simplest form?

topjm [15]

AB + C
= (x + 1)(x^2 + 2x - 1) + 2x
= x^3 + 2x^2 - x + x^2 + 2x - 1 + 2x
= x^3 + 3x^2 + 3x - 1

3 0

3 years ago

Read 2 more answers

What is 1,360/16 reduced to smallest form? (please hurry this assignment is due soon)

Keith_Richards [23]

Answer:

85/1

Step-by-step explanation:

because both numbers are divisible by 16

5 0

2 years ago

Read 2 more answers

Lisa draws several right triangles and measures the angles. Her measurements are shown in the diagram. Each right triangle has o

Ray Of Light [21]

Answer:

a and b

Step-by-step explanation:

7 0

3 years ago

Suppose a, b denotes of the quadratic polynomial x² + 20x - 2022 & c, d are roots of x² - 20x + 2022 then the value of ac(a

Alja [10]

<h3><u>Correct Question :- </u></h3>

$\sf\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0 \: and \: \\ \sf \: c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0 \: then \:$

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) =$

(a) 0

(b) 8000

(d) 16000

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0}$

We know

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:ab = \dfrac{ - 2020}{1} = - 2020$

And

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:a + b = - \dfrac{20}{1} = - 20$

Also, given that

$\red{\rm :\longmapsto\:c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0}$

$\rm \implies\:c + d = - \dfrac{( - 20)}{1} = 20$

and

$\rm \implies\:cd = \dfrac{2020}{1} = 2020$

Now, Consider

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)$

$\sf \: = {ca}^{2} - {ac}^{2} + {da}^{2} - {ad}^{2} + {cb}^{2} - {bc}^{2} + {db}^{2} - {bd}^{2}$

$\sf \: = {a}^{2}(c + d) + {b}^{2}(c + d) - {c}^{2}(a + b) - {d}^{2}(a + b)$

$\sf \: = (c + d)( {a}^{2} + {b}^{2}) - (a + b)( {c}^{2} + {d}^{2})$

$\sf \: = 20( {a}^{2} + {b}^{2}) + 20( {c}^{2} + {d}^{2})$

$\sf \: = 20\bigg[ {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\bigg]$

We know,

$\boxed{\tt{ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta \: }}$

So, using this, we get

$\sf \: = 20\bigg[ {(a + b)}^{2} - 2ab + {(c + d)}^{2} - 2cd\bigg]$

$\sf \: = 20\bigg[ {( - 20)}^{2} + 2(2020) + {(20)}^{2} - 2(2020)\bigg]$

$\sf \: = 20\bigg[ 400 + 400\bigg]$

$\sf \: = 20\bigg[ 800\bigg]$

$\sf \: = 16000$

Hence,

$\boxed{\tt{ \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = 16000}}$

<em>So, option (d) is correct.</em>

4 0

2 years ago

Please help .......​

Please help .......