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Scorpion4ik [409]

3 years ago

10

The midpoint of stack A B with bar on top has coordinates of (4, -9). Endpoint A has coordinates (-3, -5). What are the coordina

tes of B?

1 answer:

Radda [10]3 years ago

3 0

I think it's B with the bar on top

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Solve for the base of a triangle where the height is 16cm and area is 104 cm2. What is the base?

Arturiano [62]

Answer:

13cm

Step-by-step explanation:

104 × 2 = 208

208 ÷ 16 = 13cm

(13 × 16) ÷ 2 = 104cm²

6 0

3 years ago

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Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

How to graph 200-25x=y

In-s [12.5K]

Simplifying
y = -25x + 200

Reorder the terms:
y = 200 + -25x

Solving
y = 200 + -25x

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Simplifying
y = 200 + -25x

4 0

4 years ago

Simplify<br> ^3<br> V-27n^27

Sauron [17]

Apply the radical rule to separate terms:

Cubicroot(-27) and cubic root(n^27)

Cubicroot(-27) = -3

Now you have -3 cubicroot(n^27)

Using the exponent rule

N^27 can be rewritten as (n^9)^3

Now you have -3 cubicroot((n^9)^3)

The cubic root and the 3rd power cancel out to get the final answer of

-3n^9

8 0

3 years ago

If F(x) and G(x) are two functions such that F(G(x)) = x then F(x) and G(x) are ..?

lana [24]

They are inverse functions though to be completely thorough your teacher should have also put g(f(x)) = x as well. Though I can see what your teacher is aiming for at least.

The idea is that whatever the output of g(x) is, it's plugged into f(x) and the initial input is the result. So g(x) takes a step forward and f(x) takes a step back undoing everything g(x) did. Which is exactly what an inverse operation does.

4 0

3 years ago

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