All categories

2 years ago

Carmen has 4 baby bunnies. Each bunny weighs 1/5 of a pound. How much do the bunnies weigh all together?

Mathematics

1 answer:

Anestetic [448]2 years ago

4 0

Step-by-step explanation:

The answer is 0.8 as a fraction is 4/5

You might be interested in

Does the following table show a proportional relationship between the variables xxx and yyy?

Eva8 [605]

Answer:

Yes

Step-by-step explanation:

It's right

7 0

3 years ago

Read 2 more answers

40 + 56 ÷ p - q<br><br> i need the answer asap

Vadim26 [7]

Answer:

8 0

3 years ago

HELPPPPP PLEASEEEEE!!!

Pepsi [2]

Answer:

The height of right circular cone is h = 15.416 cm

Step-by-step explanation:

The formula used to calculate lateral surface area of right circular cone is: $s=\pi r\sqrt{r^2+h^2}$

where r is radius and h is height.

We are given:

Lateral surface area s = 236.64 cm²

Radius r = 4.75 cm

We need to find height of right circular cone.

Putting values in the formula and finding height:

$s=\pi r\sqrt{r^2+h^2}\\236.64=3.14(4.75)\sqrt{(3.75)^2+h^2} \\236.64=14.915\sqrt{(3.75)^2+h^2} \\\frac{236.64}{14.915}=\sqrt{14.0625+h^2} \\15.866=\sqrt{14.0625+h^2} \\Switching\:sides\:\\\sqrt{14.0625+h^2} =15.866\\Taking\:square\:on\:both\:sides\\(\sqrt{14.0625+h^2})^2 =(15.866)^2\\14.0625+h^2=251.729\\h^2=251.729-14.0625\\h^2=237.6665\\Taking\:square\:root\:on\:both\:sides\\\sqrt{h^2}=\sqrt{237.6665} \\h=15.416$

So, the height of right circular cone is h = 15.416 cm

4 0

2 years ago

Complete the table and draw the graph

Neko [114]

Answer:

So the vaule of y is 7, 10 & 13

Step-by-step explanation:

Given that y = 3x+1 . So we can just substitute x-value into the equation :

Let x=2,

y = 3(2)+1

= 7

Let x=3,

y = 3(3)+1

= 10

Let x=4,

y = 3(4)+1

= 13

5 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago