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3 years ago

Name: Michael Gutierrez

Mathematics

1 answer:

serious [3.7K]3 years ago

5 0

Answer:

Step-by-step explanation:

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Please help!!!!!

drek231 [11]

Answer:

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4 0

2 years ago

Find the HCF and LCM of 4x^2 - 36 ,2x^2 - 12x + 18 and 2x^2 + x - 21

olga_2 [115]

Answer:

(x-3), 4 (x - 3)^2 (x + 3) (2 x + 7)

Step-by-step explanation:

Factor all the expressions,

1st expression= 4x^2 - 36=4(x^2-9)=4(x+3)(x-3)

2nd expression=2x^2 - 12x + 18 =2(x^2-6x+9)=2 (x - 3)^2=2(x-3)(x-3)

3rd expression=2x^2 + x - 21=(x - 3) (2 x + 7)

HCF=Commo factor=(x-3)

LCF=Common factor*Remaining factor=4(x+3)(x-3)(x-3) (2 x + 7)=4 (x - 3)^2 (x + 3) (2 x + 7)

4 0

2 years ago

What is the Square root of -5

lilavasa [31]

The square root of -5

2.23606798

or

-2.23606798

8 0

2 years ago

Read 2 more answers

Write the equation of a circle for which the endpoints of a diameter are (-2,-2) and (4,-10)

Maurinko [17]

Answer:

Therefore the required Equation of Circle

$x^{2}+y^{2}-2x+12y+12=0$

Step-by-step explanation:

Given:

End point of Diameter be

point A( x₁ , y₁) ≡ ( -2 ,-2 )

point B( x₂ , y₂) ≡ ( 4 , -10 )

To Find:

Equation of a circle =?

Solution:

When end points of the Diameter are A( x₁ , y₁) , B( x₂ , y₂). then the Equation of Circle is given as

$(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0$

Substituting the end point are

$(x-(-2))(x-4)+(y-(-2))(y-(-10))=0\\(x+2))(x-4)+(y+2))(y+10))=0\\$

Applying Distributive Property we get

$x^{2} -2x-8+y^{2}+12y+20 =0\\\\x^{2}+y^{2}-2x+12y+12=0$

Therefore the required Equation of Circle

$x^{2}+y^{2}-2x+12y+12=0$

6 0

2 years ago

What is 99 divided bye 100

WINSTONCH [101]

Answer:

= 0.99

Step-by-step explanation:

6 0

3 years ago