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olya-2409 [2.1K]
3 years ago
12

How do you add fractions

Mathematics
2 answers:
Nadusha1986 [10]3 years ago
5 0

Hello There!

First, you need to make sure that the denominators are equal. You must have equal denominators to add fractions.

Next, add the numerators together. The numerator is the top number of the fraction and this is added when the denominators are equal. Do not add the denominators together leave them alone!

Finally, add up the numerators and get your sum.

Gwar [14]3 years ago
3 0
In order to add fractions you need to change the denominators to make sure they are the same. Then, all you have to do is add the two numerators like normal and keep the denominator the same. 

Hope this helps!!!! :D
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Please keep it simple.
Vikentia [17]

Answer:

z^{2} +\frac{1}{2} +11z

Step-by-step explanation:

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2 years ago
Find the mode of 4, 6, 2, 5, 3, 4, 2, 6, 1, 7, 4, 3, and 8.<br>​
BartSMP [9]

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2 years ago
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solutio
AnnZ [28]

Answer:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

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If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Step-by-step explanation:

For this case we have the following differential equation given:

t y'' + 7y = t

With the conditions y(1)= 1 and y'(1) = 7

The frist step on this case is divide both sides of the differential equation by t and we got:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

7 0
3 years ago
Read 2 more answers
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kherson [118]

Answer:

Results are below.

Step-by-step explanation:

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Sara:

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Number of periods (n)= 365 days

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Jon:

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FV= $15,238.71

Sara:

FV= 16,250*(1.00011^365)

FV= $16,915.67

Sara will have $1,676.96 more than Jon.

4 0
2 years ago
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