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3 years ago

What number should be added to both sides of the equation to complete the square? x2 - 6x = 5

Mathematics

2 answers:

Arlecino [84]3 years ago

7 0

Answer:

Step-by-step explanation:

Ymorist [56]3 years ago

4 0

Answer:

Step-by-step explanation:

Given

x² - 6x = 5

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 3)x + 9 = 5 + 9, that is

(x - 3)² = 14

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Find the value of the variable that results in congruent triangles

sdas [7]

Answer:

y = 5

Step-by-step explanation:

The triangles will be congruent when corresponding sides are congruent. __

We already have ...

KL ≅ NP . . . = 15 mm

JL ≅ MP . . . = 22 mm

So, we need ...

JK ≅ MN

4y +12 = 32 . . . . substitute given values

This will be the case when ...

4y = 20 . . . . . subtract 12

y = 5 . . . . . . . divide by 4

The value of y that makes the triangles congruent is 5.

8 0

2 years ago

What is the range of y=−x^2−2x+3? x≤−4 x≤4 x≥−4 x≥4

saveliy_v [14]

Answer:

Domain: (−∞,∞),{x|x∈R}

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4 0

3 years ago

Write an expression to represent the following:

nevsk [136]

9u+5

9u represents 9 multiplied by a number.

+5 means 5 more than this product.

Hope this helps.

4 0

3 years ago

Write a word problem that can be solved by finding the numbers that have 4 as a factor

Ksenya-84 [330]

8 10 12 14 16 I think

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3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago