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3 years ago

I don’t get this lol...really need help with this!

Mathematics

1 answer:

Hunter-Best [27]3 years ago

4 0

Look up the equation for it

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Solve |x|+7<4 A. 0 B. {x|x<-11 or x>-3} C. {x|-3

mars1129 [50]

Answer:

{3xl42}

Step-by-step explanation:

3 0

2 years ago

A square has sides with lengths of (x - 2) units. A rectangle has a length of x units and a width of

vagabundo [1.1K]

Answer:

(A)The area of the square is greater than the area of the rectangle.

(C)The value of x must be greater than 4

(E)The area of the rectangle is $(x^2-4x) \:Square \:Units$

Step-by-step explanation:

The Square has side lengths of (x - 2) units.

Area of the Square

$(x-2)^2=(x-2)(x-2)=x^2-2x-2x+4\\=(x^2-4x+4) \:Square \:Units$

The rectangle has a length of x units and a width of (x - 4) units.

Area of the Rectangle = $x(x-4)=(x^2-4x) \:Square \:Units$

The following statements are true:

(A)The area of the square is greater than the area of the rectangle.

This is because the area of the square is an addition of 4 to the area of the rectangle.

(C)The value of x must be greater than 4

If x is less than or equal to 4, the area of the rectangle will be negative or zero.

(E)The area of the rectangle is $(x^2-4x) \:Square \:Units$

3 0

3 years ago

Ann can swim 20 laps in 8 min. how long will it take to swim 9 laps? show your equations(s).

Volgvan

Take 20 laps divide by 8 then take the number times 9 laps

6 0

2 years ago

Plz.. Help me.. True or false?

zimovet [89]

Answer:

$\boxed{\mathrm{False}}$

Step-by-step explanation:

$(p-q)^2$

$(p-q)(p-q)$

Use FOIL method.

$p^2-pq-pq +q^2$

$p^2-2pq +q^2$

6 0

3 years ago

ohan found that the equation –2|8 – x| – 6 = –12 had two possible solutions: x = 5 and x = –11. which explains whether his solut

klio [65]

-2 | 8-x| - 6 = -12...add 6 to both sides
-2 |8-x| = -12 + 6
-2 |8-x| = - 6...divide both sides by -2
|8-x| = -6/-2
|8-x| = 3

8 - x = -3 -(8 - x) = -3
-x = -3-8 -8 + x = -3
-x = -11 x = -3 + 8
x = 11 x = 5

not correct.....sign error

5 0

3 years ago