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3 years ago

7

3 12 27 48. what is the rule?

2 answers:

Shkiper50 [21]3 years ago

8 0

Answer:

Step-by-step explanation:

3(1)^2=3, 3(2)^2=12, 3(3)^2=27, 3(4)^2=48

f(n)=3n^2, n≥1

pshichka [43]3 years ago

3 0

Answer:

this rule follow the airthmatics progression.

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The volume of a rectangular prism is (x4 + 4x3 + 3x2 + 8x + 4), and the area of its base is (x3 + 3x2 + 8). If the volume of a r

muminat

x + 1 minus StartFraction 4 Over x cubed + 3 x squared + 8

Step-by-step explanation:

Volume of the prism is given by the product of its base area and height. To get the height, divide the volume by the base area

Perform long division of the volume polynomial by the polynomial for base area to get the polynomial for height

x³+3x²+8 √x⁴+4x³+3x²+8x+4

-x⁴+3x³+8x

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1x³+3x²+0+4

- 1x³+3x²+8

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

-4

The answer will be : x+1 + -4/(x³+3x²+8)

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Dividing polynomials : brainly.com/question/11792208

Keywords : volume, rectangular prism, area,height

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3 0

2 years ago

Read 2 more answers

The equation 2x2 − 12x + 1 = 0 is being rewritten in vertex form. Fill in the missing step. Given 2x2 − 12x + 1 = 0 Step 1 2(x2

Bezzdna [24]

Answer:

Part 1) $2(x-3)^{2}-17=0$ (the missing steps in the explanation)

Part 3) (8, 4); The vertex represents the maximum profit

Part 4) x = 3.58, 0.42

Part 5) x = 6, x = 44; The zeros represent the number of monthly memberships where no profit is made

Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

Part 8) (x + 3)^2 − 4

Part 9) 2(x − 1)^2 = 4

Part 10) 8(x − 4)^2 + 592

Step-by-step explanation:

Part 1) we have

$2x^{2} -12x+1=0$

Convert to vertex form

step 1

Factor the leading coefficient and complete the square

$2(x^{2} -6x)+1=0$

$2(x^{2} -6x+9)+1-18=0$

step 2

$2(x^{2} -6x+9)+1-18=0$

$2(x^{2} -6x+9)-17=0$

step 3

Rewrite as perfect squares

$2(x-3)^{2}-17=0$

Part 3) we have

$f(x)=-x^{2}+16x-60$

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

$f(x)+60=-x^{2}+16x$

Factor the leading coefficient

$f(x)+60=-(x^{2}-16x)$

Complete the squares

$f(x)+60-64=-(x^{2}-16x+64)$

$f(x)-4=-(x^{2}-16x+64)$

Rewrite as perfect squares

$f(x)-4=-(x-8)^{2}$

$f(x)=-(x-8)^{2}+4$

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

$-2(x-2)^{2}+5=0$

$-2(x-2)^{2}=-5$

$(x-2)^{2}=2.5$

square root both sides

$(x-2)=(+/-)1.58$

$x=2(+/-)1.58$

$x=2(+)1.58=3.58$

$x=2(-)1.58=0.42$

Part 5) we have

$f(x)=-x^{2}+50x-264$

we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

so

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

$-x^{2}+50x-264=0$

$-x^{2}+50x=264$

Factor -1 of the leading coefficient

$-(x^{2}-50x)=264$

Complete the squares

$-(x^{2}-50x+625)=264-625$

$-(x^{2}-50x+625)=-361$

$(x^{2}-50x+625)=361$

Rewrite as perfect squares

$(x-25)^{2}=361$

square root both sides

$(x-25)=(+/-)19$

$x=25(+/-)19$

$x=25(+)19=44$

$x=25(-)19=6$

Part 6) we have

$-2x^{2}+28x+20$

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

$-2(x^{2}-14x)+20$

Complete the square

$-2(x^{2}-14x+49)+20+98$

$-2(x^{2}-14x+49)+118$

Rewrite as perfect square

$-2(x-7)^{2}+118$

The vertex is the point (7,118)

therefore

The video game price that produces the highest weekly profit is x=$7

Part 7) we have

$f(x)=-x^{2}+8x-10$

Convert to vertex form

$f(x)+10=-x^{2}+8x$

Factor -1 the leading coefficient

$f(x)+10=-(x^{2}-8x)$

Complete the square

$f(x)+10-16=-(x^{2}-8x+16)$

$f(x)-6=-(x^{2}-8x+16)$

Rewrite as perfect square

$f(x)-6=-(x-4)^{2}$

$f(x)=-(x-4)^{2}+6$

The vertex is the point (4,6)

therefore

The maximum height of the puck is 4 feet.

Part 8) we have

$x^{2}+6x+5$

Convert to vertex form

Group terms

$(x^{2}+6x)+5$

Complete the square

$(x^{2}+6x+9)+5-9$

$(x^{2}+6x+9)-4$

Rewrite as perfect squares

$(x+3)^{2}-4$

Part 9) we have

$2x^{2}-4x-2=0$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 2 the leading coefficient

$2(x^{2}-2x)-2=0$

Complete the square

$2(x^{2}-2x+1)-2-2=0$

$2(x^{2}-2x+1)-4=0$

Rewrite as perfect squares

$2(x-1)^{2}-4=0$

$2(x-1)^{2}=4$

The vertex is the point (1,-4)

Part 10) we have

$8x^{2}-64x+720$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

$8(x^{2}-8x)+720$

Complete the square

$8(x^{2}-8x+16)+720-128$

$8(x^{2}-8x+16)+592$

Rewrite as perfect squares

$8(x-4)^{2}+592$

the vertex is the point (4,592)

The population has a minimum at x=4 years ( that is after 4 years since 1998 )

6 0

3 years ago

How are observations different from inferences? (choose)

sergejj [24]

Answer:

b

Step-by-step explanation:

4 0

3 years ago

Use the following diagram and information to complete the proof.

Rashid [163]

Answer:

Given: ABCD is a rectangle.

Prove: The diagonals AC¯¯¯¯¯¯¯¯ and BD¯¯¯¯¯¯¯¯ are congruent.

Match each numbered statement to the correct reason to complete the proof.

PS : i will mark brainliest if they answer the question fully..

Step-by-step explanation:

Given: ABCD is a rectangle.

Prove: The diagonals AC¯¯¯¯¯¯¯¯ and BD¯¯¯¯¯¯¯¯ are congruent.

Match each numbered statement to the correct reason to complete the proof.

PS : i will mark brainliest if they answer the question fully..

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3 years ago

Find two consecutive even integers such that twice the smaller is 16 more than the larger.

qwelly [4]

Answer: smaller: 18 ; larger: 20

Step-by-step explanation:

Let n be the smaller integer. Then the larger integer is n+2. So:

2n=n+2+16

n=18

The smaller integer is 18; the larger one is 20.

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