Find x, y, and z please!!

A.

x= sqrt(40)  ( =6.325 to 3 dec. places)

y = sqrt(56)  (= 7.483 to 3 dec. places)

z = sqrt(140)  (= 11.832 to 3 dec. places)

B.

x = sqrt(24)   (= 4.899 to 3 dec. places)

y = sqrt(28)    (=5.292 to 3 dec. places)

z = sqrt(168)    (= 12.961 to 3 dec. places)

Step-by-step explanation:

Assuming x is an altitude of the triangles.

Using metric relations,

A.

x^2=4*10=40 => x= sqrt(40) = 2sqrt(10) (= 6.325 to 3 dec. places)

y^2=4*(10+4) => y = sqrt(56) (= 7.483 to 3 dec. places)

z^2=10*(10+4) => z = sqrt(140) (= 11.832 to 3 dec. places)

check: y^2 + z^2 = (10+4)^2  ok.

similarly

B.

x^2= 2*12=24  => x = sqrt(24) = 2 sqrt(6)    (= 4.899 to 3 dec. places)

y^2 = 2*(12+2) = 28  => y = sqrt(28) = 2 sqrt(7)  (= 5.292 to 3 dec. places)

z^2 = 12*(12+2) = 168  =>   z = sqrt(168) = 4sqrt(42)   (= 12.961 to 3 dec. places)

check: y^2 + z^2 = 28+168 = 196  = (2+12)^2   ok.

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2