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3 years ago

James has 16 balloons.1/4 of his balloons are red.How many red balloons does he have?

1 answer:

4 0

Hello!

First you have to convert the fraction to a decimal

1/4 = 0.25

Multiply the amount of balloons by the amount that are red

16 * 0.25 = 4

The answer is 4

Hope this helps!

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Frank and his family went on vacation one summer, and drove 3,000 miles in 10 days. If they drove an equal distance every day, h

kiruha [24]

Answer:

Distance cover by Frank and family each day = 300 miles per day

Step-by-step explanation:

Given:

Total distance cover by Frank and family = 3,000 miles

Number of days = 10

Find:

Distance cover by Frank and family each day

Computation:

Distance cover by Frank and family each day = Total distance cover by Frank and family / Number of days

Distance cover by Frank and family each day = 3,000 / 10

Distance cover by Frank and family each day = 300 miles per day

3 0

3 years ago

Read 2 more answers

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

In a function, y varies inversely with x. The consistent of variation is 4. Which table could represent the function?

lianna [129]

Answer:

none of the tables shown

Step-by-step explanation:

Inverse variation is

xy = k where k is the constant

xy = 4

in the top table

-2*2 = -4 so it cannot be the first table

In the middle table

-2 * -8 = 16 so it is not the middle table

In the bottom table

-2 *4 = -8 so it is not the bottom table

There must be a table not shown

5 0

3 years ago

Teach me how to add this problem

Arada [10]

start with the ones and every time it goes over ten take the first number and add it to the tens for example 1+8+4+8=21 take the one and put it in the one spot below. Then take the 2 and add it to the six in the tens spot so it would be Instead of 6+8+3+2 it would be 8+8+3+2 which would equal 21 then put the 1 u=in the ten spot below add the 2 to the hundreds spot making is 5+5+7=4=14 put the 4 Below and add the 1 the the thousands making it 3+4+5+1=13 put the one on the ten thousands and put the three below the whole problem would be 13,411.

7 0

3 years ago

Please help!

lions [1.4K]

Michelle's result shows f(2) = -13, selection C.

3 0

3 years ago

Read 2 more answers