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3 years ago

15

Physical science is ________

1 answer:

lorasvet [3.4K]3 years ago

7 0

The answer is going to be b

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what is the center of a circle whose equation is x2 + y2 – 12x – 2y + 12 = 0? o (-12,-2) o (-6, -1) o (6, 1) (12, 2)

Leno4ka [110]

Answer:

(6,1)

Step-by-step explanation:

1.) group the x and y terms (and move the constant to the other side)

x²-12x+y²-2y= -12

divide the coeficcents for the terms with just 1 or 1 x variable by 2 and then square it (and add that to both sides)

x²-12x+36+y²-2y+1= -12+36+1

factor

(x-6)²+(y-1)²= 25

the center is therefore (6,1)

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3 years ago

Find the area of the shape shown below.

nevsk [136]

Answer:

828

Step-by-step explanation:

12

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4 years ago

Read 2 more answers

What is the domain of f(x)=(1/4)^x?

atroni [7]

$\huge \bf༆ Answer ༄$

The domain of given function is <u>All real numbers</u>

$\sf \: f(x) = { \bigg( \dfrac{1}{4} \bigg) }^{x}$

For each value of x the function is defined.

Example -

For x = 1

The value is $\sf\frac{1}{4}$

For x = -1

The value is 4

$꧁ \: \large \frak{Eternal \: Being } \: ꧂$

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2 years ago

What’s the answer to this? 4x-4=4x is it infinitive solution?

emmasim [6.3K]

Answer:

16 ?

Step-by-step explanation:

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3 years ago

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X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

2 years ago