(4x^2-y^2)^2 2simplify fully

The diameter of a circle is 15 ft. Find its circumference in terms

Luda [366]

Answer:

C≈47.25

Step-by-step explanation:

Use the formula C= πd, where C is circumference, π is pi, and d is diameter.

Input the numbers and we'll shorten π to 3.15

C=3.15(15)

C≈47.25

5 0

2 years ago

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If f(x) = 2x + 3 for all values of x, what is the value of f(-3)?

IrinaK [193]

Answer:

f(-3)= -3

Step-by-step explanation:

We are given the function:

f(x) = 2x+3

and asked to find f(-3). Essentially, we want to find f(x) when x is equal to -3.

Therefore, we can substitute -3 for each x in the function.

f(x)= 2x+3 at x= -3

f(-3)= 2(-3) +3

Solve according to PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition and Subtraction

Multiply 2 and -3.

f(-3) = (2*-3) +3

f(-3)= (-6)+3

Add -6 and 3.

f(-3)= (-6+3)

f(-3)= -3

If f(x)= 2x+3, then<em> f(-3)= -3</em>

7 0

3 years ago

Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so tha

Pavel [41]

Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

(Expand to obtain an equivalent expression for the sphere: $x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ )

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between $\left(x_1, \, y_1, \, z_1\right)$ and $\left(x_2, \, y_2, \, z_2\right)$ would be:

$\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right)$ .

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .