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3 years ago

What is 10,000 more than 30,000+3,000+600+60+5

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

6 0

I added all the numbers and I got 43,665

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Please helpppP! Please simplify the fractions.

UkoKoshka [18]

Answer:

3/10=#3

2/5=#2

8/9=#1

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

Kaj is flying a kite, holding her hands a distance of 3.5 feet above the ground and letting all the kite’s string play out. She

Savatey [412]

Using trigonometric ratio, the height of the ground above the ground to the nearest hundredth is 44.85 ft

The situation forms a right angle triangle.

<h3>What is a right angle triangle?</h3>

Right angle triangle has one of its angles as 90 degrees. The sides and angles can be found using trigonometric ratios.

Therefore, the length of the string is the hypotenuse side of the triangle formed . The opposite side of the triangle is the height of the kite form the ground.

Therefore, the height of the kite form the ground can be found as follows;

sin 33° = opposite / hypotenuse

sin 33° = h / 75

cross multiply

h = 75 × sin 33

h = 40.8479276261

h = 40.84

The height of kite from ground = 40.847 + 3.5 = 44.3479276261 = 44.85 ft

learn more on right triangles here: brainly.com/question/25799394

4 0

3 years ago

In a batch of 120 manufactured machine parts, 3 are found to be defective. At

arlik [135]

Answer:

300

Step-by-step explanation:

3:120 =1:60 ==300:12000

7 0

4 years ago

Decide whether the rates are equivalent.

atroni [7]

Yes, the rates are equivalent

5 0

4 years ago

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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