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3 years ago

Find the volume of a cube where the length of each side is 6 centimeters.

1 answer:

4 0

Well if each side is 6cm each then u Have to multiply 6x how many sides to the cube there are. So if there are 6 sides to a cube and you have 6cm each side then you have to multiply 6cm X 6 witch = 36 then multiply 36X6=216. So the final answer is 216.

Hope this helps!

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Which is the equation of a hyperbola with directrices at x = ±3 and foci at (4, 0) and (−4, 0)?

Alex Ar [27]

Answer:

$\frac{x^2}{12}-\frac{y^2}{4}=1$

Step-by-step explanation:

Since our foci are located on the x-axis, then our major axis is going to be the horizontal transverse axis of the hyperbola:

Formula for hyperbola with horizontal transverse axis centered at origin

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Directrices -> $x=\pm\frac{a^2}{c}$
Foci -> $(\pm c,0)$ where $a^2+b^2=c^2$
$a>b$

Since we are given our directrices of $x=\pm3$ and foci of $(\pm4,0)$ , then we can set up the directrices equation to solve for $a^2$ :

$x=\pm\frac{a^2}{c}\\ \\\pm3=\pm\frac{a^2}{4}\\ \\12=a^2$

Now we can determine $b^2$ to complete our equation for the hyperbola:

$a^2+b^2=c^2\\\\12+b^2=4^2\\\\12+b^2=16\\\\b^2=4$

Therefore, our equation for our hyperbola is $\frac{x^2}{12}-\frac{y^2}{4}=1$

5 0

2 years ago

1 nnnnnnnneeeeeeeeeeeed help

zavuch27 [327]

Answer is 84.86 cm^2

8 0

3 years ago

Which fraction is equal to ⅛? ⁴⁄₆ ⁴⁄₃₀ ²⁄₁₆

aliya0001 [1]

Answer:

2/16

Step-by-step explanation:

1/8 * 2/2 = 2/16

4 0

3 years ago

Read 2 more answers

Which statement describes the polygon? A. The polygon is equilateral but not equiangular. B. The polygon is equiangular but not

matrenka [14]

B.the polygon is equiangular but not equilateral. this should be correct

5 0

3 years ago

Read 2 more answers

Without graphing is the system independent, dependent, or inconsistent?

rewona [7]

Y = -2 x - 9
3 x - 4(-2 x - 9 ) = -8
3 x + 8 x + 36 = - 8
11 x = - 44, x = - 4
y = 8 - 9 , y = -1
This is a unique solution, the system is independent.

3 0

3 years ago