What value of b will cause 16x ^ 2 + bx + 4 = 0 to have one real solution ?

b  =  16

Step-by-step explanation:

16*x  + b*x  + 4  =  0

Solving for x we get:

x₁,₂  =   - b  ±  √  b² - 4*a*c   ] / 2*a         (1)

If we take a look to the square root

√ b² - 4*4*16      =  √ b²  - 256

if we make √ b²  - 256    = 0

b²  =  256       ⇒      b  =  16

Then equation (1) becomes:

x₁,₂  = ( - 16  ±  0  ) / 32

x  = -  1/2

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)