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3 years ago

Write an expression for 35 minus b.

Mathematics

1 answer:

aalyn [17]3 years ago

8 0

Answer:

35-b

Step-by-step explanation:

the answer is 35-b

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denpristay [2]

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2 years ago

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Vedmedyk [2.9K]

Answer:

ST = 8 m

Step-by-step explanation:

Since ∆PQR is congruent to ∆STU, therefore, the three sides of ∆PQR would be congruent or equal to the corresponding three sides of ∆STU.

PQ corresponds to ST.

Therefore, PQ is congruent to corresponding side ST, which is 8 m

ST = 8 m

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3 years ago

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matrenka [14]

$n > \frac{81}{2}$

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3 years ago

The tennis team is selling tickets to wash for six dollars. And they don't do not sell very much tickets, the team decreases the

schepotkina [342]

Answer:

The new cost of tickets after decrease of price at 23 % rate is $ 4.62

Step-by-step explanation:

Given as :

The initial cost of the selling tickets = $ 6

Let The final cost of selling tickets = $ x

The rate of decrease of ticket price = 23 %

Now,

The final cost of tickets = initial cost of tickets × ( 1 - $\dfrac{\textrm rate}{100}$ )

or, $ x = $ 6 × ( 1 - $\dfrac{\textrm 23}{100}$ )

or, x = $ 6 × $\frac{77}{100}$

∴ x = $ 4.62

So decrease price = x = $ 4.62

Hence The new cost of tickets after decrease of price at 23 % rate is $ 4.62 Answer

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4 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