Can anyone find the answers this for these question but make them fractions
1 answer:
We are given fractions and
.
In order to write them in simplest fractions, we need to find a common number that can divide both top and bottom.
Let us work on simplifying first fraction first.
We have 77672 on the top and 5548 in the bottom.
The greatest number 5548 can divide both 77672 and 5548 both.
Dividing 77672 by 5548, we get 14 and dividing 5548 by 5548 we get 1.
Putting quotent 14 on the top and 1 in the bottom, we get
. This is the answer for first part.
Let us work on simplifying second fraction .
Top number 288184 and bottom number 34, both can be divided by 34.
On dividing 288184 by 34, we get 8476 and dividing 34 by 34, we get 1.
Putting quotent 8476 on the top and 1 in the bottom, we get
. This is the answer for second part.
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Greetings!
We can factor this expression/ the GCF is:
Hope this helps.
-Benjamin
We can factor this expression/ the GCF is:
Hope this helps.
-Benjamin
• First way to solve:
We'll manipulate the expression of the equation:
If we have y=0:
Then, the function has one real zero (x=3) and two imaginary zeros (4i and -4i).
Answer: B
• Second way to solve:
The degree of the function is 3. So, the function has 3 complex zeros.
Since the coefficients of the function are reals, the imaginary roots are in a even number (a imaginary number and its conjugated)
The function "has only one non-repeated x-intercept", then there is only one real zero.
The number of zeros is 3 and there is 1 real zero. So, there are 2 imaginary zeros.
Answer: B.
We'll manipulate the expression of the equation:
If we have y=0:
Then, the function has one real zero (x=3) and two imaginary zeros (4i and -4i).
Answer: B
• Second way to solve:
The degree of the function is 3. So, the function has 3 complex zeros.
Since the coefficients of the function are reals, the imaginary roots are in a even number (a imaginary number and its conjugated)
The function "has only one non-repeated x-intercept", then there is only one real zero.
The number of zeros is 3 and there is 1 real zero. So, there are 2 imaginary zeros.
Answer: B.