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Triss [41]
3 years ago
9

Someone help it’s lateral and surface area for cones

Mathematics
1 answer:
-BARSIC- [3]3 years ago
4 0

2946482947564757575748

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[Inequalities]<br> Please help and thanks. <br> ONLY NEED TO ANSWER 15 AND 17!
ValentinkaMS [17]

Answer:

15. b < 16

17. c (is greater than or equal to) 25

Step-by-step explanation:

i hope this helps :)

3 0
3 years ago
Read 2 more answers
Write an expression that shows how to usethe halving and doubling strategy to find 8 x 50
faust18 [17]
Double of 50 is 100, so 28*100=2800, now half of the answer is 2800/2=1400
3 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
Two angles form a linear pair. The measure of one angle is x and the measure of the other angle is 1.4 times x plus 12∘ . Find t
nekit [7.7K]

Answer:

70° and 110°

Step-by-step explanation:

If two angles forms a linear pair, this means that the sum of the angles is 180°. If the measure of one angle is x and the measure of the other angle is 1.4 times x plus 12∘

Let A be the first angle = x°

Let B be the second angle = (1.4x+12)°

Since they form a linear pair, then

A+B = 180°

x + 1.4x+12 = 180°

2.4x = 180-12

2.4x = 168

x = 168/2.4

x = 70°

The measure of angle A = 70°

The measure if angle B = 1.4x+12

B = 1.4(70)+12

B = 98+12

B = 110°

The measure of both angles are 70° and 110°

6 0
4 years ago
समचतुर्भज का। परिमाप ​
Alexeev081 [22]
45 / 9 = 5 x 100 = 500
3 0
3 years ago
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