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3 years ago

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Marta was pregnant 270 days and 260 days and 272 days for her first three pregnancies in order for Marta average pregnancy to eq

ual the worldwide average of 266 days for how long was her fourth pregnancy last

2 answers:

aivan3 [116]3 years ago

6 0

Answer:262 days

Step-by-step explanation:

Pregnancy 1=270 days

Pregnancy 2= 260days

Pregnancy 3=272 days

To find cumulative days for all four pregnancies to equal an average of 266days

For the average to be 266 days

266×4=1064

Sum of all 3 pregnancies = 270+260+272=802

Sum of all cumulative days for all four pregnancies-sum of the three pregnancies

1064-802=262

So for the average to be 266days

The forth pregnancy has to

262 days

wlad13 [49]3 years ago

5 0

266 * 4 = 1064 days. The total days she’s been pregnant is the sum of those 3 pregnancies which is 802 days. Therefore, she needs her pregnancy to be 262 days to equal the average.

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Ahmad travelled 1/4 of a journey by Train X, 3/4 of the remainder by Train Y and the remaining 10km by Train Z. a) How far did A

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3 years ago

In the right ∆ABC, the hypotenuse AB = 17 cm. M is the midpoint of the hypotenuse. Find the legs if PAMC=32 cm and PBMC=25 cm

jeyben [28]

Answer:

The length of the legs is 8.64cm and 14.64cm respectively

Step-by-step explanation:

I've added an attachment to aid my explanation.

At different intervals, I'll be making reference to it.

Given

$AB = 17$

$PAMC = 32$

$PBMC = 25$

From the attachment, we have:

$y + z = AB$

Since, M is the Midpoint

$y = z = \½AB$

Substitute 17 for AB

$y = z = \½ * 17$

$y = z = 8.5$

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$v + x + z = PAMC$

$v + x + y = 32$

Substitute 8.5 for y

$v + x + 8.5 = 32$

$v + x = 32 - 8.5$

$v + x = 23.5$ --------- (1)

Also, from the attachment

$v + w + z = 25$

Substitute 8.5 for z

$v + w + 8.5 = 25$

$v + w = 25 - 8.5$

$v + w = 17.5$ ----------- (2)

Subtract (2) from (1)

$v - v + x - w = 23.5 - 17.5$

$x - w = 6$

Make x the subject

$x = 6 + w$

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We have that:

$AB^2 = AC^2 + BC^2$

The above can be replaced with

$17^2 = x^2 + w^2$ (see attachment)

$289 = x^2 + w^2$

Substitute 6 + w for x

$289 = (6 + w)^2 + w^2$

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$289 - 36 = 12w + 2w^2$

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$2w^2 + 12w - 253 = 0$

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Where

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$w = \frac{-12 \± \sqrt{12^2 - 4 * 2 * -253}}{2 * 2}$

$w = \frac{-12 \± \sqrt{144 + 2024}}{4}$

$w = \frac{-12 \± \sqrt{2168}}{4}$

$w = \frac{-12 \± 46.56}{4}$

Split:

$w = \frac{-12 + 46.56}{4}$ or $w = \frac{-12 - 46.56}{4}$

$w = \frac{34.56}{4}$ or $w = \frac{-58.56}{4}$

$w = 8.64$ or $w = -14.64$

But length can't be negative

So:

$w = 8.64$

Recall that: $x = 6 + w$

$x = 6 + 8.64$

$x = 14.64$

<em>Hence, the length of the legs is 8.64cm and 14.64cm respectively</em>

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