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2 years ago

12

The radius of a circle is (2x-3). What is a function for the area of the circle?

1 answer:

Oduvanchick [21]2 years ago

6 0

Answer

A = pi * r^2

= pi * (2x^-3)^2

= pi * 2^2 * (x^-3)^2

= pi * 4 * (x^-6)

= 4*pi * x^-6

Step-by-step explanation:

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you measure 2 cups of peanut butter into mixing bowl. How many 1/4 cups will you need to complete the recipe?

tatuchka [14]

Answer:

I believe the answer would be 8 1/4 cups.

Hope this helps!

5 0

2 years ago

Which of the following best describes how to use of the addition property of equality to isolate the variable x below?. . 5 = x

sergiy2304 [10]

<span>The best answer is C.add 2 to both sides of the equation.
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2 years ago

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 $(1,-1)$ .

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2 years ago

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Sauron [17]

Answer:

Step-by-step explanation:

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What is the equation of a line that is parallel to x−2y=−4 and passes through the point (0, 0) ?

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2 years ago

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