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3 years ago

Please help will give brainliest thank you.

Mathematics

1 answer:

tester [92]3 years ago

5 0

Two events are dependent if the outcome of the first event affects the outcome of the second.

The last answer is the correct one.

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Can anyone help me with this geometry test please like I will do anything!! I will give brainlist!!

sammy [17]

Answer:

It c

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Mr. Yi buys vegetables at a market. He purchases 6 pounds of potatoes, p, and 3 pounds of onions, n, for $18. Onions cost twice

Alex17521 [72]

6p + 3n = 18
2p = n

6p + 3(2p) = 18
6p + 6p = 18
12p = 18
p = 18/12 = 1.50

2p = n
2(1.50) = n
3 = n

true statements : The equation 2n = p should be 2p = n
The actual cost of onions is $ 3 per lb
Potatoes cost $ 1.50 per lb

9 0

3 years ago

Which set of three angles could represent the interior angles of a triangle?

irina [24]

Answer:

26°,51°,103°

Step-by-step explanation:

sum of angles=180°

8 0

3 years ago

An eight-sided game piece is shaped like

DiKsa [7]

Answer:

27.5 mm

Step-by-step explanation:

Let the side length of each of the side of the base of the pyramid be a, hence:

perimeter of square = 4a

80 = 4a

a = 20 mm.

The distance from the middle of the square pyramid to the side = r = half of the side length = a/2 = 20 / 2 = 10 mm.

r = 10 mm

The slant height (s) = 17 mm, Let h be the height of each of the pyramid. Using Pythagoras theorem:

r² + h² = s²

17² = 10² + h²

h² = 17² - 10² = 189

h = √189

h = 13.748 mm

The length of the game piece = 2 * h = 2 * 13.748 = 27.5 mm to nearest tenth of a mm.

6 0

3 years ago

Compute the differential of surface area for the surface S described by the given parametrization.

AysviL [449]

With $S$ parameterized by

$\vec r(u,v)=\langle e^u\cos v,e^u\sin v,uv\rangle$

the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

We have

$\dfrac{\partial\vec r}{\partial u}=\langle e^u\cos v,e^u\sin v,v\rangle$

$\dfrac{\partial\vec r}{\partial v}=\langle -e^u\sin v,e^u\cos v,u\rangle$

with cross product

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle ue^u\sin v-ve^u\cos v,-ve^u\sin v-ue^u\cos v,e^{2u}\cos^2v+e^{2u}\sin^2v\rangle$

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle e^u(u\sin v-v\cos v),-e^u(v\sin v+u\cos v),e^{2u}\rangle$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{e^{2u}(u\sin v-v\cos v)^2+e^{2u}(v\sin v+u\cos v)^2+e^{4u}}$

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=e^u\sqrt{u^2+v^2+e^{2u}}$

So we have

$\mathrm dS=\boxed{e^u\sqrt{u^2+v^2+e^{2u}}\,\mathrm du\,\mathrm dv}$

8 0

3 years ago