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3 years ago

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Mathematics

1 answer:

blsea [12.9K]3 years ago

6 0

Answer:

Dop

Step-by-step explanation:

Calculus

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Please answer look at pitot

klio [65]

Answer:

z = 7

Step-by-step explanation:

Based on the midsegment theorem of a triangle, the length of the midsegment of the triangle (z) = ½ of the length of the 3rd side of the triangle (14)

Thus:

z = ½(14)

z = 7

4 0

3 years ago

jacob just had lunch at a restaurant and leaves a 18% tip. Choose all the expressions that will give his total cost in terms of

Leokris [45]

Answer:

1.18b

Step-by-step explanation:

Jacob had lunch at a restaurant and his bill was b.

Now, Jacob leaves 18% of his bill amount as a tip.

So, the tip amount at the rate of 18% of the bill amount will be $\frac{18 \times b}{100} =0.18 \times b$

Therefore, the total cost of Jacob for the lunch will be b + 0.18b = 1.18b. (Answer)

8 0

3 years ago

Mary ran 4 miles in 37 minutes. If she continued at the same pace, how long would it take her to run 11 miles?

Pavlova-9 [17]

Solution:

we are given that

Mary ran 4 miles in 37 minutes.

we have been asked to find

how long would it take her to run 11 miles?

Since Mary ran 4 miles in 37 minutes, it mean time taken to run 1 miles can be given by

$t_{1 \ mile}=\frac{37}{4} \ minutes.$

Hence we can write , time taken in running 11 miles as

$t_{11 \ mile}=\frac{37}{4}*11 \ minutes.\\ \\ t_{11 \ mile}=\frac{407}{4} \ minutes\\ \\ t_{11 \ mile}=101.75 \ minutes\\$

6 0

3 years ago

A store is giving out cards labeled 1 through 10 when customers enter the store. If the card is an even number, you get a % di

m_a_m_a [10]

Answer:

yrbffbfzbc

Step-by-step explanation:

brfzfbfz

8 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