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3 years ago

1. Choose two 2. Count the on the line then the 3. Write the slope as _

Mathematics

1 answer:

NikAS [45]3 years ago

6 0

Answer:

y=2×+1

Step-by-step explanation:

for the slope of the question that was given

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$1500 at 9% for 2 years

kotykmax [81]

Answer:$270

Step-by-step explanation:

I=PRT

Interest=principal times rate times time

Principal=1500

Rate=9%=0.09

Time=2

I=1500 times 0.09 times 2

I=1500 times 0.18

I=270

Interest=270

Total is 1500+270=1770 total paid

8 0

3 years ago

Which ordered pair is a solution of the equation?

Oksi-84 [34.3K]

Answer:

Step-by-step explanation:

Only (4, -1)

5 0

3 years ago

The dot plot shows the number of miles Jamal biked per week for ten weeks. Which measure of central tendency best represents the

vitfil [10]

Answer:The answer is Mean or Median

Step-by-step explanation:

Mean represents the average of a set of values which is gotten by adding the miles then diving by the total number .

Mean generally gives you the average ,so it is the best means to get the number of miles and it is also the most frequently used

Median is the mid number (s) in a data set .

Median is useful in this data because there are some missing values in the data .

8 0

3 years ago

terracE baked 15 cookies he gave 6 to his sister what percentage of the cookies did he give his sister

torisob [31]

Answer:

40%

Step-by-step explanation:

6 is 40% of 15

3 0

3 years ago

\int (x+1)\sqrt(2x-1)dx

Nezavi [6.7K]

Answer:

$\int (x+ 1) \sqrt{2x-1} dx = \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15}(2x-1)^{\frac{5}{2}} + C$

Step-by-step explanation:

$\int (x+1)\sqrt {(2x-1)} dx\\Integrate \ using \ integration \ by\ parts \\\\u = x + 1, v'= \sqrt{2x - 1}\\\\v'= \sqrt{2x - 1}\\\\integrate \ both \ sides \\\\\int v'= \int \sqrt{2x- 1}dx\\\\v = \int ( 2x - 1)^{\frac{1}{2} } \ dx\\\\v = \frac{(2x - 1)^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}} \times \frac{1}{2}\\\\v= \frac{(2x - 1)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{2}\\\\v = \frac{2 \times (2x - 1)^{\frac{3}{2}}}{3} \times \frac{1}{2}\\\\v = \frac{(2x - 1)^{\frac{3}{2}}}{3}$

$\int (x+1)\sqrt(2x-1)dx\\\\ = uv - \int v du$

$= (x +1 ) \cdot \frac{(2x - 1)^{\frac{3}{2}}}{3} - \int \frac{(2x - 1)^{\frac{3}{2}}}{3} dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ u = x + 1 => du = dx \ ]$

$= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \int (2x - 1)^{\frac{3}{2}}} dx\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{5}{2}}}{\frac{5}{2} }) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15} \times (2x-1)^{\frac{5}{2}} + C\\\\$

6 0

3 years ago