All categories

3 years ago

Need more help on this question

Mathematics

2 answers:

Bond [772]3 years ago

8 0

79
hope this helped

Alisiya [41]3 years ago

7 0

Answer: 79

Step-by-step explanation:

6 ÷ 3 + 9 x² - 4=79

You might be interested in

How do I solve this using long division?

Natalija [7]

I know this solution but I don't know how I send you the answer....

7 0

3 years ago

Read 2 more answers

Maria is covering an entire box with 1-in. mosaic tiles and wants to find out how many square inches of mosaic files will cover

natka813 [3]

The amount of mosaic tiles covering the box is the surface area, which is 157 in.

4 0

3 years ago

Read 2 more answers

It takes Matt 3 minutes to read one page in a book. If he continues to read at the same pace, he can read 15 similar pages in mi

elena-14-01-66 [18.8K]

Answer:

a)45min

b)900min

Step-by-step explanation:

We were told that it takes Matt 3 minutes to read one page in a book.

It implies that 3minutes = 1page of the book

Then he continues to read at the same pace, he can read 15 similar pages in minutes.

a)Then, he can read 15 similar pages in

15x3 minutes.which is 45minutes.

b)Then if the book has 300 pages, it will take him. 300×3 minutes to read it.which is 900mins

5 0

3 years ago

Help picture below problem 9

Agata [3.3K]

Answer:

i think the answer might be E.

7 0

2 years ago

Read 2 more answers

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