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The results are independent:
P(2 heads) = 1/4
P(roll a 4) = 1/6
P(2 heads and a 4) = (1/4)(1/6) = 1/24

6 0

3 years ago

The hypotenuse of a right angled triangle is 2√13 cm . If the smaller side is increased by 2 cm and the larger side is increased

White raven [17]

Statement:

The hypotenuse of a right angled triangle is 2√13 cm. If the smaller side is increased by 2 cm and the larger side is increased by 3 cm, the new hypotenuse will be √117 cm.

To find out:

The length of the larger side of the right angled triangle.

Solution:

Let us consider x as the smaller side and y as the larger side.

Then, in the right angled triangle,

x² + y² = (2√13)² ...(I) [By Pythagoras Theorem]

Now, if the smaller side is increased by 2 cm, then the smaller side will be (x + 2).

And if the larger side is increased by 3 cm, then the larger side will be (y + 3).

Then, in the new right angled triangle,

(x + 2)² + (y + 3)² = (√117)² [By Pythagoras Theorem]

or, x² + 2 × 2 × x + 2² + y² + 2 × 3 × y + 3² = (√117)²

or, x² + 4x + 4 + y² + 6y + 9 = (√117)²

or, x² + y² + 4x + 6y + 13 = (√117)²

Now, put the value of x² + y² from equation (I),

or, (2√13)² + 4x + 6y + 13 = (√117)²

or, (2 × 2 × √13 × √13) + 4x + 6y + 13 = (√117 × √117)

or, 52 + 4x + 6y + 13 = 117

or, 4x + 6y = 117 - 52 - 13

or, 4x + 6y = 52

or, 4x = 52 - 6y

or, x = $\frac {(52 - 6y)}{4}$ ...(II)

Now, put the value of x of equation (II) in (I),

x² + y² = (2√13)²

or, $\frac {(2704-624y +36y²)}{16}$ + y² = 52

or, $\frac {(2704-624y +36y² + 16y²)}{16}$ = 52

or, 52y²-624y + 2704 = 52 × 16

or, 52y² - 624y + 2704 - 832 = 0

or, 52y² - 624y + 1872 = 0

or, 52(y² - 12y + 36) = 0

or, y²-12y +36 = 0 ÷ 52

or, y²-12y +36 = 0

or, (y)² - 2 × 6 × y + (6)² = 0

or, (y - 6)² = 0

or, y - 6=0

or, y = 6

We have taken y as the length of the larger side of the right angled triangle.

So, the length of the larger side is 6 cm.

Answer:

6 cm

Hope you could understand.

If you have any query, feel free to ask.

6 0

3 years ago

A quadrilateral has 4 right angles, 2 pairs of sides of equal length,2 pairs of opposite sides that are parallel. The pairs of o

tekilochka [14]

That is describing a rectangle! The only quadrilaterals with 4 right angles are squares and rectangles. Squares have equal sides but rectangles have one pair of parallel sides that is longer than the other. Hope this helps! If you get more than one answer and see it fit to do so, please mark me Brainliest Answer! Thank you!

6 0

4 years ago

Which sequence matches the recursive formula?

goldenfox [79]

Hello:
an = 2an-1 + 5, where a1 = 5
n =2 a2 = 2a1 +5 = 2(5)+5 = 15
n =3 a3 = 2a2 +5 = 2(15)+5 = 35
n =4 a4 = 2a3 +5 = 235)+5 = 75
answer :
B)
5, 15, 35, 75, ...

4 0

3 years ago

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