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3 years ago

If the length of the side of a square is 3×-y what is the area of the square in terms of x and y

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

5 0

(3x -y)(3x-y )= 9x^2 - 6xy + y^2

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Can some please help me!!!. I only have 1 hour to complete it.

dedylja [7]

Answer:

The correct option is;

ASA

Step-by-step explanation:

The given information are;

1. ∠A ≅ ∠D, 1. Given

2. $\overline{CD}\left | \right |\overline{AB}$ , 2. Given

3. $\overline{CB}\cong \overline{BC}$ , 3. refl. prop.

4. ∠ABC ≅ ∠ DCB, 4. Alt int. ∠s are ≅

5. ΔABC ≅ ΔDCB, 5. Angle Side Angle ASA rule of congruency

The conditions for two triangles to be congruent includes;

1. All sides equal or SSS rule

2. Side Angle Side or SAS rule

3. Angle Side Angle or ASA rule

4. Angle Angle Side or AAS rule

5. Right angle Hypotenuse Side or RHS also known as Hypotenuse Leg, HL rule

4 0

4 years ago

Sam decides to build a square garden. if the area of the garden is 4x2 - 20x + 25 square feet, what is the length of one side of

Mkey [24]

33-20x heres your answer

3 0

3 years ago

A Fruit Stand sells 6 oranges for $3 and 3 grapefruits for $2.40 Lionel buys 10 oranges and 11 grapefruit how much Lionel spend

lina2011 [118]

$3/6 = .50 per orange
$2.40/3 = .80 per grapefruit

10(.50) + 11(.80) =
$5.00 + $8.80 = $13.80

8 0

4 years ago

Read 2 more answers

5р + 3р + (-9) ? A. 8p+9 B. 3(p+(-3))+5p C. None of the above . . .

kvv77 [185]

Answer:

none of the above

Step-by-step explanation:

because 5p + 3p = 8p and positive integers × negative integers = negative integers.

3 0

3 years ago

Solve x2 − 12x + 5 = 0 using the completing-the-square method. x = six plus or minus the square root of five x = negative six pl

Andru [333]

we are given

$x^2-12x+5=0$

we have to solve it by completing square method

step-1: Move 5 on right side

$x^2-12x+5-5=0-5$

$x^2-12x=-5$

step-2: Break middle term

$x^2-2*6*x=-5$

step-3: Add 6^2 both sides

$x^2-2*6*x+(6)^2=-5+(6)^2$

$(x-6)^2=31$

step-3: Solve for x

$\sqrt{(x-6)^2} =\sqrt{31}$

we wil get two values

First value is

$x-6=\sqrt{31}$

add both sides 6

$x-6+6=6+\sqrt{31}$

$x=6+\sqrt{31}$

Second value is

$x-6=-\sqrt{31}$

add both sides 6

$x-6+6=6-\sqrt{31}$

$x=6-\sqrt{31}$

so, solutions are

$x=6+\sqrt{31}$

$x=6-\sqrt{31}$ .............Answer

5 0

3 years ago