All categories

3 years ago

Which is the larger amount 4/10 of a dollar or 8/10 of a dollar

Mathematics

2 answers:

3241004551 [841]3 years ago

7 0

8/10 of a dollar is the correct answer

allochka39001 [22]3 years ago

7 0

8/10 is larger, we can look at it two ways:
4/10 of a dollar is 40 cents, and 8/10 of a dollar is 80 cents, so we know 8/10 is larger; or:
8/10 would be the bigger fraction, so 8/10 of a dollar is larger.
---
Hope this helps!
==jding713==

You might be interested in

The height of a door would be about A: 8 inches B: 8 feet C: 8 yards D: 8Miles

liq [111]

8 inches :) (im pretty sure)

8 0

3 years ago

Read 2 more answers

1. Write a survey question for which you would expect to collect numerical

Yakvenalex [24]

Answer:

How many siblings do you have?

I hope this helps you out :)

5 0

3 years ago

A little help? A set of numbers is given below. √72, 9.25, √83, 2, 8.7 Which list correctly orders the numbers from least to gre

myrzilka [38]

Answer:

9.25, 83^1/2, 8.7, 72^1/2, 2 From greatest to least

2, 72^1/2, 8.7, 83^1/2, 9.25 from least to greatest

Step-by-step explanation:

6 0

3 years ago

What is the standard form of the equation of the circle to the right?

olga_2 [115]

Use y=Mx+B plug in correct spot

3 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago