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2 years ago

11

Consider two unique parallel lines. What aspects of these two lines are the same? What aspects of these two lines would have to

be different?

1 answer:

nalin [4]2 years ago

5 0

Their slopes would be the same but their positions or points ( if the lines were on a graph) would be different.

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Factor the following<br> polynomial:<br> X^3 + 125

ra1l [238]

X^3 + 5^3
(x+5)x(x^2-X x 5+5^2)

(x+5)x(x^2 -5x+5^2)

(x+5)x(x^2 -5x+25)

(x+5) x (x^2 -5x+25)

6 0

2 years ago

Given: the function defined by f(x)=3x^2-4 Which statement is true? 1. f(0) = 0 2. f(-2) = f(2) 3. f(5)+ f(2) = f(7) 4. f(5) x f

sergeinik [125]

$f(x)=3x^2-4 \\ \Downarrow \\ f(0)=3 \times 0^2-4=0-4=-4 \\ -4 \not= 0 \\ f(0) \not= 0 \\
\hbox{statement 1 is false} \\ \\
f(-2)=3 \times (-2)^2-4 =3 \times 4-4=12-4=8 \\
f(2)=3 \times 2^2-4=3 \times 4-4=12-4=8 \\ 8=8
\\ f(-2)=f(2) \\
\hbox{statement 2 is true}$

$f(5)=3 \times 5^2-4=3 \times 25-4=75-4=71 \\ f(2)=8 \\
f(7)=3 \times 7^2-4=3 \times 49-4=147-4=143 \\
71+8 \not= 143 \\
f(5)+f(2) \not= f(7) \\ \hbox{statement 3 is false} \\ \\
f(5)=71 \\ f(2)=8 \\ f(10)=3 \times 10^2-4=3 \times 100-4=300-4=296 \\
71 \times 8 \not= 296 \\
f(5) \times f(2) \not= f(10) \\
\hbox{statement 4 is false}$

Stamement 2. f(-2)=f(2) is true.

3 0

2 years ago

!!please, please help me!!

grigory [225]

Answer:

Length: 12

Width: 2 $\sqrt{10\\}$

Area: 24 $\sqrt{10\\}$

Step-by-step explanation:

We first find the points of C and D to find the length.

C: (-4,-5)

D: (2,-3)

The distance is 2 $\sqrt{10\\$ .

Now for the height.

A: (2,9)

D: (2,-3)

The distance is 12.

Area: 2 $\sqrt{10\\$ ×12 = 24 $\sqrt{10\\}$

6 0

2 years ago

Yo someone answer this because if I get a bad grade my xbox is getting taken away

blsea [12.9K]

You just have to substitute the x for the given number and complete the equation. Leave the x-box and laziness and study

8 0

2 years ago

masha68 [24]

Answer:

Step-by-step explanation:

Volume of the figure = Base area * height

Base area:

Base area = area of trpezium + area of triangle

<u />

<u>Trapezium:</u>

a, b area the length of parallel sides and h is the height of trapezium.

a = 8 m ; b = 5 m & h = 3 m

$\boxed{ \text{ Area of trapezium = $\dfrac{(a +b)*h}{2} $}}$

$=\dfrac{(8+5)*3}{2}\\\\\\=\dfrac{13*3}{2}\\\\\\= \dfrac{39}{2}\\\\\\= 19.5 \ m^{2}$

<u>Triangle:</u>

height = 4 m & base = 8 m

$\boxed{\text{Area of triangle =$\dfrac{1}{2}*base*height$}}$

$=\dfrac{1}{2}*8*4\\\\\\= 4*4\\\\= 16 \ m^{2}$

Base area = 19.5 + 16

= 35.5 m²

Volume of the figure = Base area * height

= 35.5 * 14

= 497 m³

8 0

2 years ago