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3 years ago

What is 20kg divide by 1120

Mathematics

1 answer:

Stels [109]3 years ago

5 0

Answer:

56kg

Step-by-step explanation:

This question is pretty simple. First you want to divide 1120 by 20.

This will give you the answer of 56.

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Put these questions in order from least to greatest 3/8 19/24 2/3

aleksandr82 [10.1K]

First 3/8, second 2/3, and third 19/24

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3 years ago

Solve linear equations Solve for x -2(4x + 3) = 3(x + 1)

inna [77]

Answer:

x = - $\frac{9}{11}$

Step-by-step explanation:

Given

- 2(4x + 3) = 3(x + 1) ← distribute parenthesis on both sides

- 8x - 6 = 3x + 3 ( subtract 3x from both sides )

- 11x - 6 = 3 ( add 6 to both sides )

- 11x = 9 ( divide both sides by - 11 )

x = - $\frac{9}{11}$

7 0

3 years ago

Read 2 more answers

A pizza parlor offers 3 sizes of pizzas, 2 types of crust, and one of 4 different toppings. How many different pizzas can be mad

KonstantinChe [14]

Answer: The total number of pizzas that can be made from the given choices is 24.

Step-by-step explanation: Given that a pizza parlor offers 3 sizes of pizzas, 2 types of crust, and one of 4 different toppings.

We are to find the number of different pizzas that can be made from the given choices.

We have the COUNTING PRINCIPLE :

If we have m ways of doing one task and n ways of doing the second task, then the number of ways in which we can do both the tasks together is m×n.

Therefore, the number of different pizzas that can be made from the given choices is

$N=3\times2\times4=24.$

Thus, the total number of pizzas that can be made from the given choices is 24.

5 0

4 years ago

Which of the following equations matches the function shown above

Mnenie [13.5K]

Answer:

where is the function? where are the equations?

7 0

3 years ago

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

7 0

3 years ago