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3 years ago

Find the volume of a cone that has radius = 2 1/2 and height = 5.

Mathematics

2 answers:

Andreas93 [3]3 years ago

7 0

Answer:

98.1 is your anwser I know this kinds of stuff

muminat3 years ago

6 0

Step-by-step explanation:

v=22/7×r^2×h/3

=22/7×21/2×5/3

=33/5

=11cm^3

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A bouqet had 10 flower and sold it for 90$ which is the rate of $_ per flower

lana [24]

Answer:

$9 per flower

Step-by-step explanation:

divide cost ($90) by the number of items (10 flowers)

6 0

3 years ago

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Alec's password consists of 1 letter and 4 digits. What is the probability of correctly guessing Alec's password?

valentinak56 [21]

Answer:

11A22

Step-by-step explanation:

7 0

3 years ago

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Marlene went shopping and bought a bunch of candy for her swim team. She bought packs of Skittles for $1.50 each and packs of M&

lord [1]

Answer:

18 Skittles

6 M&Ms

Step-by-step explanation:

Set up an equation:

Variable x = number of skittles

Variable y = number of M&Ms

1.50x + 2y = 39

x + y = 24

In the second equation, isolate a variable:

x = 24 - y

Substitute the value of x for 24 - y in the first equation:

1.50(24 - y) + 2y = 39

Use distributive property

36 - 1.5y + 2y = 39

Combine like terms

36 + 0.5y = 39

Isolate variable y:

0.5y = 3

y = 6

Substitute the value of y for 6 in the second equation:

x + 6 = 24

Isolate variable x:

x = 18

Plug these values into any equation of your choice to see if these values are correct (I'll do both equations just to prove it):

1.50(18) + 2(6) = 39

27 + 12 = 39

39 = 39

Correct

x + y = 24

18 + 6 = 24

24 = 24

Correct

8 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,

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

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

using the same approach as above.

5 0

2 years ago

Name three decimals with a product that is about 40

yulyashka [42]

4.0
.40
.040
Sorry if im wrong but im sure the second one is correct

6 0

3 years ago