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LUCKY_DIMON [66]
3 years ago
10

I need help ,,, and tahnks☺

Mathematics
2 answers:
Mashcka [7]3 years ago
6 0
Pythagora: 20^2=(4x)^2+(3x^2)
400=16x^2+9x^2
x^2=400/25
x^2=16
x=4
Lubov Fominskaja [6]3 years ago
4 0
We can use the Pythagoras theorem:

leg 1² + leg 2² = hypotenuse²
(4x)² + (3x)² = 20²
16x² + 9x² = 400
combine like terms
25x² = 400

divide by 25 on both sides to isolate x²
\frac{25x^2}{25} =  \frac{400}{25}
25 and 25 cancels out

x² = 16

Take square root on either sides tot find the value
\sqrt{x^2} = \sqrt{16}
x = +/- 4

Cancels -4 because lengths cannot be in negative.

x = 4

check:
(4 × 4)² + (3 × 4)² = 20²
(16)² + (12)² = 400
256 + 144 = 400
400 = 400
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3 years ago
The weights of bags of ready-to-eat salad are normally distributed with a mean of 290 grams and a standard deviation of 10 grams
pentagon [3]
290-10=280

280 is one standard deviation below.
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2 years ago
A, b, c and d are positive integers, such that a+b+ ab = 76, c+d+ cd = 54. Find (a+b+c+d)·a·b·c·d.
lutik1710 [3]

Notice that

(1 + <em>x</em>)(1 + <em>y</em>) = 1 + <em>x</em> + <em>y</em> + <em>x y</em>

So we can add 1 to both sides of both equations, and we use the property above to get

<em>a</em> + <em>b</em> + <em>a b</em> = 76  ==>  (1 + <em>a</em>)(1 + <em>b</em>) = 77

and

<em>c</em> + <em>d</em> + <em>c d</em> = 54  ==>  (1 + <em>c</em>)(1 + <em>d</em>) = 55

Now, 77 = 7*11 and 55 = 5*11, so we get

<em>a</em> + 1 = 7  ==>  <em>a</em> = 6

<em>b</em> + 1 = 11  ==>  <em>b</em> = 10

(or the other way around, since the given relations are symmetric)

and

<em>c</em> + 1 = 5  ==>  <em>c</em> = 4

<em>d</em> + 1 = 11  ==>  <em>d</em> = 10

Now substitute these values into the desired quantity:

(<em>a</em> + <em>b</em> + <em>c</em> + <em>d</em>) <em>a</em> <em>b</em> <em>c</em> <em>d</em> = 72,000

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