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2 years ago

−125+|−136|+(−323) plzzzz need it asap

Mathematics

2 answers:

OLEGan [10]2 years ago

3 0

Answer:

-312

Step-by-step explanation:

nasty-shy [4]2 years ago

3 0

−125+|−136|+(−323)

Answer= −312

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Expressions that are equal to 4 times the square root of 5

Morgarella [4.7K]

Answer:

Simplify 4/ (square root of 5) 4 √5 4 5 Multiply 4 √5 4 5 by √5 √5 5 5. 4 √5 ⋅ √5 √5 4 5 ⋅ 5 5

Step-by-step explanation:

4 0

2 years ago

Jamal makes $12.00/hour and worked 9 hours today. What is his total pay? Don't forget to calculate overtime pay.

Elanso [62]

Answer:

He made $108 in 9 hours

Step-by-step explanation:

I first multiplied 12 by 9 and got 108

5 0

3 years ago

Read 2 more answers

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

1 year ago

Read 2 more answers

-2/3*3/5+5/2-3/5*1/6

Kipish [7]

Answer:

2and19/90

Step-by-step explanation:

-2/3*3/5= _2/5

-2/5+5/2-(2/3×1/6)

-2/5+5/2-1/9

21/10-1/9

=199/90

=2/19/90

3 0

3 years ago

Please help with this question, I need the help urgently

grin007 [14]

Answer:

P in terms of V is:

P = 432/V

Step-by-step explanation:

We know that y varies inversely as x, we get the equation

y ∝ 1/x

y = k/x

k = yx

where k is called the constant of proportionality.

In our case,

P is inversely proportional to V

Given

P = 18

V = 24

P = k/V

k = PV

substituting P = 18 and V = 24 to determine k

k = 18 × 24

k = 432

now substituting k = 432 in P = k/V

P = 432/V

Therefore, P in terms of V is:

P = 432/V

5 0

3 years ago