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Basile [38]
3 years ago
15

Two brothers decide that a 21-hour task is to be divided so thst the older brother works twice as long as the younger brother. H

ow long must the older brother work?​
Mathematics
2 answers:
Mice21 [21]3 years ago
7 0

Answer: the older brother must work for 14 hours.

Step-by-step explanation:

Let x represent the number of hours that the younger brother must work.

Two brothers decide that a 21-hour task is to be divided so that the older brother works twice as long as the younger brother. This means that the number of hours that the older brother must work is expressed as 2x.

Since the total number of hours is 21, then

x + 2x = 21

3x = 21

x = 21/3

x = 7

The number of hours that the older brother must work is

2x = 2 × 7 = 14 hours

lyudmila [28]3 years ago
3 0

Answer:

14 hours

Step-by-step explanation:

Let's call the task 3x

The older bro does twice as younger so it would be 2x long and younger brother work x long

Since the totsl of work is 3x = 21 x woul be 21÷3 = 7

the older brother must work 2 × 7 = 14 hours long

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Answer:

36 cm^2

Step-by-step explanation:

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S.A. = 2 ((4)(2) + (2)(2) + (4)(2))

S.A. = 2 (8 + 4 + 6)

S.A. = 2 (18)

S.A. = 36

4 0
3 years ago
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4 years ago
Choose the abbreviation of the postulate or theorem that supports the conclusion that the triangles are congruent. Given: C, F a
agasfer [191]

Answer:

The abbreviation of the postulate or theorem that supports the conclusion that the triangles are congruent.

HL.

Step-by-step explanation:

The Figure is attached below

Given:

AB = DE ; BC = EF

∠C = ∠F = 90°

To Prove:

ΔACB ≅ ΔDFE

Proof:

Hypotenuse Leg Theorem:

The hypotenuse leg theorem states that any two right triangles that have a congruent hypotenuse and a corresponding, congruent leg are congruent triangles.

The abbreviation is HL

In  ΔACB  and ΔDFE  

AB ≅ DE      ……….{Hypotenuse congruent Given}

BC  ≅ EF     ……….{Given}

∠C ≅ ∠F = 90°     …………..{Measure of each angle is 90° given}

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7 0
3 years ago
Use a Maclaurin series to obtain the Maclaurin series for the given function.
Rama09 [41]

Answer:

14x cos(\frac{1}{15}x^{2})=14 \sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}

Step-by-step explanation:

In order to find this Maclaurin series, we can start by using a known Maclaurin series and modify it according to our function. A pretty regular Maclaurin series is the cos series, where:

cos(x)=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}

So all we need to do is include the additional modifications to the series, for example, the angle of our current function is: \frac{1}{15}x^{2} so for

cos(\frac{1}{15}x^{2})

the modified series will look like this:

cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15}x^{2})^{2k}}{(2k)!}

So we can use some algebra to simplify the series:

cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15^{2k}}x^{4k})}{(2k)!}

which can be rewritten like this:

cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}

So finally, we can multiply a 14x to the series so we get:

14xcos(\frac{1}{15}x^{2})=14x\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}

We can input the x into the series by using power rules so we get:

14xcos(\frac{1}{15}x^{2})=14\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}

And that will be our answer.

3 0
3 years ago
find three consecutive integers for which 3 times the sum of the first and third integers is -342? show step by step
Montano1993 [528]
X + (X+1)+ (x+2)=-342
3x+3 =-342
3X=-339
X=-113, -114, -115
4 0
3 years ago
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