Let Xavier's favourite fraction be a/b, Yessie's favourite fraction = b/a and Zorro's favourite fraction = c/d,

c/d x a/b = 12/35 . . . . . . . . (1)
c/d x b/a = 15/7 . . . . . . . . (2)

(1) x (2) = c/d x a/b x c/d x b/a = 12/35 x 15/7
c^2 / d^2 = 36/49
c^2 = 36
c = 6
d^2 = 49
d = 7

Xaviers favourite fraction = 12/35 / 6/7 = 2/5
Yessies favourite fraction = 5/2
Zorro favourite fraction = 6/7

8 0

3 years ago

Given right triangle A w/a hypotenuse length of x+4 and a leg of x, and right triangle B, w/ a hypotense length of 3y and a leg

alex41 [277]

Answer:

Value is x=8 and y=4

Step-by-step explanation:

Given : A right triangle 'A' hypotenuse length of x+4 and a leg of x,

and right triangle 'B' hypotenuse length of 3y and a leg length of y+4

To find : For what values of x and y are the triangles congruent by HL?

Solution :

Triangle A,

Hypotenuse: x+4

Leg: x

Triangle B,

Hypotenuse: 3y

Leg: y+4

Since the triangle A and B are congruent so, the sides of triangles are equal.

$x+4=3y$ (Hypotenuse are equal) ..........[1]

and $x=y+4$ (Legs are equal) ..........[2]

Solving the equation of system,

Put the value of x from [2] in [1],

$x+4=3y$

$y+4+4=3y$

$8=2y$

$y=4$

Substitute y in [2],

$x=y+4$

$x=4+4$

$x=8$

Therefore, The value of x=8 and y=4

Verifying for the values of x and y:

Triangle A,

Hypotenuse: x+4=8+4=12

Leg: x=8

Triangle B,

Hypotenuse: 3y=3(4)=12

Leg: y+4=4+4=8

Both hypotenuses and both legs are equal hence they are congruent.

6 0

3 years ago

In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$

Hence, the required number of ways =27434

5 0

3 years ago