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JulijaS [17]
3 years ago
12

Simplify the expression –3(x + 3)2 – 3 + 3x. What is the simplified expression in standard form?

Mathematics
2 answers:
andrew11 [14]3 years ago
8 0

For this case we must simplify the following expression:

-3 (x + 3) ^ 2-3 + 3x

We solve the parenthesis:

-3 (x ^ 2 + 2 (x) (3) + 3 ^ 2) -3 + 3x =\\-3 (x ^ 2 + 6x + 9) -3 + 3x =

We apply distributive property to the terms within parentheses:

-3x ^ 2-18x-27-3 + 3x =

We add similar terms:

-3x ^ 2-18x + 3x-27-3 =\\-3x ^ 2-15x-30

Answer:

-3x ^ 2-15x-30

NNADVOKAT [17]3 years ago
7 0

Answer: -3x^2-15x-30

Step-by-step explanation:

We need to remember that (a\±b)^2=a^2\±2ab+b^2

Knowing this, we can simplify  the expression:

-3(x + 3)^2 - 3 + 3x=-3[x^2+2(x)(3)+3^2]-3+3x=-3[x^2+6x+9]-3+3x

Apply Distributive property:

=-3x^2-18x-27-3+3x

Add like like terms:

=-3x^2-15x-30

Since it has the form ax^2+bx+c, it is already expressed in Standad form.

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Determine whether the following lines represented by the vector equations below intersect, are parallel, are skew, or are identi
KiRa [710]

Answer:

r(t) and s(t) are parallel.

Step-by-step explanation:

Given that :

the  lines represented by the vector equations are:

r(t)=⟨1−t,3+2t,−3t⟩

s(t)=⟨2t,−3−4t,3+6t⟩

The objective is to determine if the following lines represented by the vector equations below intersect, are parallel, are skew, or are identical.

NOTE:

Two lines will be parallel if \dfrac{x_1}{x_2}= \dfrac{y_1}{y_2}= \dfrac{z_1}{z_2}

here;

d_1 = (-1, \ 2, \ -3)

Thus;

r(t) = \dfrac{x-1}{-1} = \dfrac{y-3}{2}=\dfrac{z-0}{-3} = t

d_2 =(2, \ -4, \  +6)

s(t) = \dfrac{x-0}{2} = \dfrac{y+5}{-4}=\dfrac{z-3}{6} = t

∴

\dfrac{d_1}{d_2}= \dfrac{-1}{2} = \dfrac{2}{-4}= \dfrac{-3}{-6}

Hence, we can conclude that r(t) and s(t) are parallel.

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