All categories

3 years ago

PLEASE HELP ME .......!!

Mathematics

1 answer:

Kruka [31]3 years ago

7 0

Answer:

wait is that a test if it is cheating

Step-by-step explanation:

well the first one go on the bottom 2 row secound one first row third first row but it is in happpy i am not reall good at math XD third last row forth very happy fifth on unhappy sixth very unhappy seven very unhappy eight on very happy i think i got it i did this before i just don't rembear but i tried

You might be interested in

Which graphic organizer would help you visualize important events in the history of the U.S. space program? timeline point-by-po

Sergeu [11.5K]

Timeline would help your the most<span />

6 0

3 years ago

Read 2 more answers

Roman has a rope that is 25 1/2 feet long. he wants to cut it into 6 pieces that are equal in length. how long will each piece b

eduard

25.5*6=4.25 or 4.25
Your welcome

8 0

3 years ago

2 hundreds 6 tens 12 ones

adell [148]

Answer:

umm that would be 272

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Please help with this Calculus questions

Triss [41]

Answer:

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\frac{193}{100}=1.93$ .

Step-by-step explanation:

To find $\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du$ .

First, calculate the corresponding indefinite integral:

Integrate term by term:

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u}} =\int{2 d u} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant rule $\int c\, du = c u$

$\int{2 d u}} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u} = {\left(2 u\right)} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$

$2 u - {\int{\frac{3 u^{9}}{2} d u}} + \int{\frac{2 u^{4}}{5} d u} = 2 u - {\left(\frac{3}{2} \int{u^{9} d u}\right)} + \left(\frac{2}{5} \int{u^{4} d u}\right)$

Apply the power rule $\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$

$2 u - \frac{3}{2} {\int{u^{9} d u}} + \frac{2}{5} {\int{u^{4} d u}}=2 u - \frac{3}{2} {\frac{u^{1 + 9}}{1 + 9}}+ \frac{2}{5}{\frac{u^{1 + 4}}{1 + 4}}$

Therefore,

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u} = - \frac{3 u^{10}}{20} + \frac{2 u^{5}}{25} + 2 u = \frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)$

According to the Fundamental Theorem of Calculus, $\int_a^b F(x) dx=f(b)-f(a)$ , so just evaluate the integral at the endpoints, and that's the answer.

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}=\frac{193}{100}$

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=0$

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}-\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=\frac{193}{100}$

6 0

3 years ago

Simplify (-3/5) divided by (7/6)

eduard

-18/35 is the answer to (-3/5) divided by (7/6)

3 0

3 years ago