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3 years ago

5

Expand and simplify 2(a+3)+3(2a-1)

1 answer:

AlexFokin [52]3 years ago

7 0

2(a+3)+3(2a-1)
=2a+6+6a-3
=8a+3

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Addi places the letters from the word DECEMBER into a bag, a letter will be randomly selected and not replaced. Thnen another le

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Answer:

Step-by-step explanation:

for c it would be 1\8

for E it would be 3\8

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2 years ago

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Okay sorry about the other one

Neporo4naja [7]

Number ten is 60 73+36=108

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2 years ago

The equation of the circle whose center is at (3, 7) and whose radius is 6 is a. (x + 3)² + (y + 7)² = 6 b. (x + 3)² + (y + 7)²

kicyunya [14]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
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3 years ago

26 divided by 5 1/2 I dont know!!

zavuch27 [327]

Answer:

4.727

Step-by-step explanation:

An easy way to think of something being divided by a fraction is by turning the fraction into a decimal. In this case, it's 5 1/2, which is the same as 5.5. So, you can just calculate 26 / 5.5, by hand, or on a calculator, and the answer is 4.727.

<em>Hope this helps!</em>

5 0

3 years ago

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Find the zeros of y = x2 – 6x – 4 by completing the square.

katrin [286]

Answer:

The solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$

Step-by-step explanation:

Given the function

$y\:=\:x^2\:-\:6x\:-\:4$

substitute y = 0 in the equation to determine the zeros

$0\:=\:x^2\:-\:6x\:-\:4$

Switch sides

$x^2-6x-4=0$

Add 4 to both sides

$x^2-6x-4+4=0+4$

Simplify

$x^2-6x=4$

Rewrite in the form (x+a)² = b

But, in order to rewrite in the form x²+2ax+a²

Solve for 'a'

2ax = -6x

a = -3

so add a² = (-3)² to both sides

$x^2-6x+\left(-3\right)^2=4+\left(-3\right)^2$

$x^2-6x+\left(-3\right)^2=13$

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

$\left(x-3\right)^2=13$

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

solve

$x-3=\sqrt{13}$

Add 3 to both sides

$x-3+3=\sqrt{13}+3$

Simplify

$x=\sqrt{13}+3$

now solving

$x-3=-\sqrt{13}$

Add 3 to both sides

$x-3+3=-\sqrt{13}+3$

Simplify

$x=-\sqrt{13}+3$

Thus, the solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$

4 0

2 years ago