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2 years ago

Answer both questions please!!! I need help....

Mathematics

2 answers:

Temka [501]2 years ago

6 0

Answer

The first one is equal and the second one is the first choice.

Strike441 [17]2 years ago

6 0

Answer:

1.=

2. 1 line of symmetry.

Could I please have BRAINLIEST?

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In 1999 Daniel was 146 cm tall. He grew to be 176 cm by the year 2006. What was Daniel's rate of growth over this period of his

andreyandreev [35.5K]

Answer:

he grows by 5 cm every year between 1999 and 2006

Step-by-step explanation:

This is a arithmetic progression problem with the formula;

T_n = a + (n - 1)d

We are told that In 1999 Daniel was 146 cm tall. He grew to be 176 cm by the year 2006.

Thus;

a = 146

d = 2006 - 1999 = 7

Thus;

176 = 146 + (7 - 1)d

176 - 146 = 6d

30 = 6d

d = 30/6

d = 5 cm

Thus, he grows by 5 cm every year between 1999 and 2006

3 0

3 years ago

PLEASE HELP FIND THE VOLUME OF EACH FIGURE ROUND TO THE NEAREST TENTH! Thank you

swat32

Answer:

Sphere=2144.6

Cylinder=1693.3

Cone=2093.3

Step-by-step explanation:

6 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$