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3 years ago

(6.03) Which of the following best describes the expression 6(y + 3)?

Mathematics

2 answers:

ivanzaharov [21]3 years ago

6 0

Answer:

what are the followings..........

Lesechka [4]3 years ago

5 0

Answer:

i think

6y+18

...........

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Activity 4: Performance Task

Nookie1986 [14]

An arithmetic progression is simply a progression with a common difference among consecutive terms.

The sum of multiplies of 6 between 8 and 70 is 390
The sum of multiplies of 5 between 12 and 92 is 840
The sum of multiplies of 3 between 1 and 50 is 408
The sum of multiplies of 11 between 10 and 122 is 726
The sum of multiplies of 9 between 25 and 100 is 567
The sum of the first 20 terms is 630
The sum of the first 15 terms is 480
The sum of the first 32 terms is 3136
The sum of the first 27 terms is -486
The sum of the first 51 terms is 2193

(a) Sum of multiples of 6, between 8 and 70

There are 10 multiples of 6 between 8 and 70, and the first of them is 12.

This means that:

$\mathbf{a = 12}$

$\mathbf{n = 10}$

$\mathbf{d = 6}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{10} = \frac{10}2(2*12 + (10 - 1)6)}$

$\mathbf{S_{10} = 390}$

(b) Multiples of 5 between 12 and 92

There are 16 multiples of 5 between 12 and 92, and the first of them is 15.

This means that:

$\mathbf{a = 15}$

$\mathbf{n = 16}$

$\mathbf{d = 5}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*15 + (16 - 1)5)}$

$\mathbf{S_{16} = 840}$

(c) Multiples of 3 between 1 and 50

There are 16 multiples of 3 between 1 and 50, and the first of them is 3.

This means that:

$\mathbf{a = 3}$

$\mathbf{n = 16}$

$\mathbf{d = 3}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*3 + (16 - 1)3)}$

$\mathbf{S_{16} = 408}$

(d) Multiples of 11 between 10 and 122

There are 11 multiples of 11 between 10 and 122, and the first of them is 11.

This means that:

$\mathbf{a = 11}$

$\mathbf{n = 11}$

$\mathbf{d = 11}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{11}2(2*11 + (11 - 1)11)}$

$\mathbf{S_{11} = 726}$

(e) Multiples of 9 between 25 and 100

There are 9 multiples of 9 between 25 and 100, and the first of them is 27.

This means that:

$\mathbf{a = 27}$

$\mathbf{n = 9}$

$\mathbf{d = 9}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{9} = \frac{9}2(2*27 + (9 - 1)9)}$

$\mathbf{S_{9} = 567}$

(f) Sum of first 20 terms

The given parameters are:

$\mathbf{a = 3}$

$\mathbf{d = 3}$

$\mathbf{n = 20}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{20} = \frac{20}2(2*3 + (20 - 1)3)}$

$\mathbf{S_{20} = 630}$

(f) Sum of first 15 terms

The given parameters are:

$\mathbf{a = 4}$

$\mathbf{d = 4}$

$\mathbf{n = 15}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{15} = \frac{15}2(2*4 + (15 - 1)4)}$

$\mathbf{S_{15} = 480}$

(g) Sum of first 32 terms

The given parameters are:

$\mathbf{a = 5}$

$\mathbf{d = 6}$

$\mathbf{n = 32}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{32} = \frac{32}2(2*5 + (32 - 1)6)}$

$\mathbf{S_{32} = 3136}$

(g) Sum of first 27 terms

The given parameters are:

$\mathbf{a = 8}$

$\mathbf{d = -2}$

$\mathbf{n = 27}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{27} = \frac{27}2(2*8 + (27 - 1)*-2)}$

$\mathbf{S_{27} = -486}$

(h) Sum of first 51 terms

The given parameters are:

$\mathbf{a = -7}$

$\mathbf{d = 2}$

$\mathbf{n = 51}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{51} = \frac{51}2(2*-7 + (51 - 1)*2)}$

$\mathbf{S_{51} = 2193}$

Read more about arithmetic progressions at:

brainly.com/question/13989292

4 0

2 years ago

Read 2 more answers

Monique has 2 less than half as many pairs of shoes as Dalia . Monique has four pairs of shoes. Which equation can be used to fi

viktelen [127]

Answer:

4m +2m = D

Step-by-step explanation:

i think i havent done this in a long time but idk

ik that dalia has 6 shoes

4 0

3 years ago

A rectangular lawn measures 80 ft by 120 ft. Part of the lawn is torn up to install a sidewalk of uniform width around it. The a

IRISSAK [1]

Answer:

Both answers will give an area of 2400 ft2

but with x=60 we have lawn dimensions -60 ft by -40 ft so this is out

x = 10 ft width for the sidewalk

Check: New lawn dimensions

(80-2x)(60-2x) = 60(40) = 2400 ft^2

Step-by-step explanation:

Draw a diagram:

We have a rectangle inside a rectangle.

The larger outside rectangle is the original lawn: 80ft by 60 ft with area 4800 ft2

The smaller inside rectangle is (80-2x)by(60-2x) where x is width of the new sidewalk.

Area of new lawn is 2400 ft^2

(80-2x)(60-2x) = 2400

4800 - 160x - 120x + 4x2 = 2400

4x2 - 280x + 2400 = 0

Factor out a 4

x2 - 70x + 600 = 0

(x-60)(x-10) = 0

x = 60 ft or x = 10 ft

7 0

3 years ago

Read 2 more answers

what if your making a 68 in a class and then you make a 50 on a test and its worth 50% of your grade what are you making then in

BlackZzzverrR [31]

You'd have a 59 in the class

4 0

3 years ago

A box contains 20 light box of which five or defective it for lightbulbs or pick from the box randomly what's the probability th

Snowcat [4.5K]

Answer:

Step-by-step explanation:

Given:-

- The box has n = 20 light-bulbs

- The number of defective bulbs, d = 5

Find:-

what's the probability that at most two of them are defective

Solution:-

- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.

- We will define random variable X : The number of defective bulbs picked.

Such that, P ( X ≤ 2 ) is required!

- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.

- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:

X = 0 , Number of choices = 15 C r = 15C2 = 105 ways

- The probability of selecting 2 non-defective bulbs:

P ( X = 0 ) = number of choices with no defective / Total choices

= 105 / 20C2 = 105 / 190

= 0.5526

- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

X = 1 , Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways

- The probability of selecting 1 defective bulbs:

P ( X = 1 ) = number of choices with 1 defective / Total choices

= 75 / 20C2 = 75 / 190

= 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

X = 2 , Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

P ( X = 2 ) = number of choices with 2 defective / Total choices

= 10 / 20C2 = 10 / 190

= 0.05263

- Hence,

P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

= 0.5526 + 0.3947 + 0.05263

= 1

7 0

3 years ago