X varies inversely as y^2, and x = 4 when y = 10. Find x when y = 2.

What's the area of this triangle

Luda [366]

Answer:

D A = (1/2) (y₂ - y₁)(x₃ - x₁)

Step-by-step explanation:

b = (y₂ - y₁)

h = (x₃ - x₁)

A = (1/2)bh

A = (1/2) (y₂ - y₁)(x₃ - x₁)

8 0

3 years ago

X^2-18x=-72<br><br> I need two solutions. Thanks.

eimsori [14]

Hey Astute!

Solve: x^2 - 18x = -72

Solution:

x^2 − 18x − (−72) = −72 − (−72) (Subtract -72 from both sides of the equation)

x^2 − 18x + 72 = 0

(x − 6) (x − 12) = 0 (Factor left side)

x − 6 = 0 or x − 12 = 0 (factors have to equal 0)

6 - 6 = 0 or 12 - 12 = 0

Answer:

x = 6 or x = 12

Hope This Helps, Have a Great Day!

7 0

3 years ago

Read 2 more answers

PLS HELP ME!!! The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of

Pachacha [2.7K]

Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: $\pi \approx 3.14$

V (volume) = ?

Solving: (Cone volume)

$V = \dfrac{ \pi *r^2*h}{3}$

$V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}$

$V = 3.14*16*4$

$\boxed{V = 200.96\:cm^3}$

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: $\pi \approx 3.14$

If: We know that the volume of a sphere is $V = 4* \pi * \dfrac{r^3}{3}$ , but we have a hemisphere, so the formula will be half the volume of the hemisphere $V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}$