3 years ago

Can someone help me solve this

1 answer:

5 0

Domain: (-infinity,infinity)
Range: (-infinity,infinity)
as x goes to infinity y goes to: infinity
as x goes to -infinity y goes to: -infinity

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The length of the football field is 10x+10 and the width is 4x+20. Write a polynomial that represents the are of the football fi

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The solution is in the attachment

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Simplify the following 3√28-5√63+4√112

alexandr402 [8]

7√7

using the ' rule of radicals '

• √a × √b ⇔ √ab

simplifying the radicals

√28 = √(4 × 7 ) = √4 × √7 = 2√7

√63 = √(9 × 7) =√9 × √7 = 3√7

√112 = √(16 × 7 ) = √16 × √7 = 4√7

substituting into the expression

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3 years ago

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Se considera numerele naturale x=4 si y=6. scrieti raportul numerelor x si y

dimaraw [331]

Answer:

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Step-by-step explanation:

4 0

2 years ago

Diego is solving the equation x^2-12x = 21

uysha [10]

Answer:

The solutions to the quadratic equations will be:

$x=\sqrt{57}+6,\:x=-\sqrt{57}+6$

Step-by-step explanation:

Given the expression

$x^2-12x\:=\:21$

Let us solve the equation by completing the square

$x^2-12x\:=\:21$

Add (-6)² to both sides

$x^2-12x+\left(-6\right)^2=21+\left(-6\right)^2$

simplify

$x^2-12x+\left(-6\right)^2=57$

Apply perfect square formula: (a-b)² = a²-2ab+b²

i.e.

$x^2-12x+\left(-6\right)^2=\left(x-6\right)^2$

so the expression becomes

$\left(x-6\right)^2=57$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x-6=\sqrt{57}$

add 6 to both sides

$x-6+6=\sqrt{57}+6$

Simplify

$x=\sqrt{57}+6$

also solving

$x-6=-\sqrt{57}$

add 6 to both sides

$x-6+6=-\sqrt{57}+6$

Simplify

$x=-\sqrt{57}+6$

Therefore, the solutions to the quadratic equation will be:

$x=\sqrt{57}+6,\:x=-\sqrt{57}+6$

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3 years ago