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prohojiy [21]
4 years ago
12

What is the product of 523 and 48

Mathematics
1 answer:
Aleksandr [31]4 years ago
6 0

Answer:

25104

Step-by-step explanation:

PRODUCT=MULTIPLICATION, SO

423 X 48= 25104

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Which is the approximate mass of a button?
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4 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

Also, from the attachment

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

Apply Pythagoras Theorem:

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

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

289 - 36 = 12w + 2w^2

253 = 12w + 2w^2

Reorder

2w^2 + 12w - 253 = 0

Solve using quadratic equation:

w = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}

Where

a = 2

b = 12

c = -253

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>

5 0
3 years ago
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