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1 year ago

19. Since April 1, 1986 was a Tuesday, March,

Mathematics

1 answer:

Elena-2011 [213]1 year ago

6 0

Answer:

Since April 1, 1986 was a Tuesday, March,

1986 had? Mondays.

the answer A

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All points from which of the following patterns would be contained on the given graph?

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The next number in the linear pattern would be 18

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In how many ways can 10 people form couples of two?

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It should be 10 raised to power 2 which is a hundred.

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Evaluate the expression and enter your answer in the box below |98|

olganol [36]

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to evaluate $\pmb{|98|}$ .

$\triangle~\fbox{\bf{KEY:}}$

The notation |x| denotes absolute value.

The absolute value of a number is its distance from zero.

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So we need to find the absolute value of 98, which is:

$\star~\star\mathrm{98}~\star~\star$

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$\fbox{The\;absolute\;value\;is\;a\;number's\;distance\;from\;zero}$

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1 year ago

Work out the area of this quarter circle take pie to be 3.142 radius in image attached thanks (:

iVinArrow [24]

Answer:

Solution,

Radius(R)= 7 cm

Area of quarter circle:

$\frac{\pi \: {r}^{2} }{4}$

$= \frac{3.142 \times 49}{4}$

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3 years ago

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Solve pls, ans should be 0, add working

Bond [772]

Step-by-step explanation:

Given: {x+(1/x)}³ = 3

Asked: x³ + (1/x³) = ?

Solution:

Method 1:

We have, {x+(1/x)}³ = 3

Comparing the expression with (a+b)³, we get

a = x

b = (1/x)

Using identity (a+b)³ = a³+b³+3ab(a+b), we get

⇛{x+(1/x)}³ = 3

⇛(x)³ + (1/x)³ + 3(x)(1/x){x + (1/x)} = 3

⇛(x*x*x) + (1*1*1/3*3*3) + 3(x)(1/x){x + (1/x)} = 3

⇛x³ + (1/x³) + 3(x)(1/x){x + (1/x)} = 3

⇛x³ + (1/x³) + 3{x + (1/x)} = 3

⇛x³ + (1/x³) + 3(x) + 3(1/x) = 3

⇛x³ + (1/x³) + 3x + (3/x) = 3

Our answer came incorrect.

Let's try..

Method 2:

We have,

[x+(1/x)]³ = 3

On taking cube root both sides then

⇛³√[{ x+(1/x)}³ ] = ³√3

⇛x+(1/x) = ³√3 -----(1)

We know that

a³+b³ = (a+b)³-3ab(a+b)

⇛x³+(1/x)³ = [x+(1/x)]³ - 3(x)(1/x)[x+(1/x)]

⇛x³+(1/x³) = (3)-3(1)(³√3)

[since, {x + (1/x)} = ³√3 from equation (1)]

⇛x³+(1/x)³ = 3-3 ×³√3

⇛x³ + (1/x³) = 3- ³√81 (or )

⇛x³ + (1/x³) = 3(1-³√3)

Therefore, x³ + (1/x³) = 3(1 - cube root of 3)

It is impossible to get zero

6 0

2 years ago