Ask question

Ask question

All categories

2 years ago

7

Which of the following is not true for adding mixed numbers?

1 answer:

chubhunter [2.5K]2 years ago

5 0

1.which of the following is not true for adding mixed numbers?

A.the denominators of the fractions must be the same in order to add them.

B.you can add mixed numbers by changing them into improper fractions or by using a number line.

C.the sum of the mixed numbers written as an improper fraction is in simplest form.

D.the sum of the mixed numbers must be written in simplest form

The answer is C

You might be interested in

Whats the slope between (4,-2) and (12,10)

yKpoI14uk [10]

M=0 The slope is zero
Hope this helps

4 0

2 years ago

Help i think it’s the second one but i’m not sure

ozzi

It's the third one.

The second one is a parabola so that is not the answer. Hope I helped :)

3 0

3 years ago

Read 2 more answers

Plssssssssss help because i need to go to basketball practice

inna [77]

Answer:

The answer is a= 5

Step-by-step explanation:

hope this helps

4 0

2 years ago

Read 2 more answers

Zher and Gabbie are making cookies for the Hospitable Pantry. They only have up to 6 hours to

GrogVix [38]

Answer:

32x3=96. 20x3= 60. 60 + 96=156 if you make 3 hours of chocolate chip and 3 hours snickerdoodles you will have 156 cookies

4 0

2 years ago

Solve the initial value problem y′+y=f(t),y(0)=0 where f(t)={1,−1, if t<4 if t≥4 Use h(t−a) for the Heaviside function shifte

Anton [14]

Looks like the function on the right hand side is

$f(t)=\begin{cases}1&\text{for }t$

We can write it in terms of the Heaviside function,

$h(t-a)=\begin{cases}1&\text{for }t\ge a\\0&\text{for }t>a\end{cases}$

as

$f(t)=h(t)-2h(t-4)$

Now for the ODE: take the Laplace transform of both sides:

$y'(t)+y(t)=f(t)$

$\implies s Y(s)-y(0)+Y(s)=\dfrac{1-2e^{-4s}}s$

Solve for <em>Y</em>(<em>s</em>), then take the inverse transform to solve for <em>y</em>(<em>t</em>):

$(s+1)Y(s)=\dfrac{1-e^{-4s}}s$

$Y(s)=\dfrac{1-e^{-4s}}{s(s+1)}$

$Y(s)=(1-e^{-4s})\left(\dfrac1s-\dfrac1{s+1}\right)$

$Y(s)=\dfrac1s-\dfrac{e^{-4s}}s-\dfrac1{s+1}+\dfrac{e^{-4s}}{s+1}$

$\implies y(t)=1-h(t-4)-e^{-t}+e^{-(t-4)}h(t-4)$

$\boxed{y(t)=1-e^{-t}-h(t-4)(1-e^{-(t-4)})}$

4 0

2 years ago