What is the factored form of the expression? d^2 - 14d + 49

d^2 - 14d + 49 = (d-7)(d-7).

Step-by-step explanation:

Given the expression is d^2 - 14d + 49.

d^2 - 14d + 49 = d^2 - 2*7*d + 7^2

We know the formula of square of difference as given below:-

x^2 - 2*x*y + y^2 = (x-y)^2

Using the above formula in the problem, here x = d and y = 7.

d^2 - 2*7*d + 7^2 = (d-7)^2

d^2 - 2*7*d + 7^2 = (d-7)(d-7)

So, d^2 - 14d + 49 = (d-7)(d-7).

(d-7)(d-7)

Step-by-step explanation:

We have to make factored form of the expression.

Given expression is :

d²-14d+49

Split the middle term to make the equation into two groups such that the sum of 2 numbers should be -14 and their product be 49.

d²-7d-7d+49

Making two groups and taking the common term out:

d(d-7)-7(d-7)

Take (d-7) common

(d-7)(d-7) which is factored form of d²-14d+49.

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.