Answer:
to solve
you would first need to convert them from mixed numbers to improper fractions. You can do this by multiplying the denominator with the whole number and then adding the numerator. Then you should make sure that the denominators are the same. You can do this by first finding a number that they will both multiply into. In this example it is 12. And when changing the denominator you must apply it to the numerator. So if you multiplied the denominator by 3 then you must multiply the numerator by 3. After you can just minus the numerators. However remember to keep the denominator the same. Then you can change it back into a mixed number by dividing the numerator by the denominator and having that whole number on the side of the fraction . lets say it is
then you would divide 13 by 6 which is 2 and a decimal and then minus the number that is multiplied by the numerator and it will give a number that is smaller than the denominator and then it would become
.
Step-by-step explanation:
1) Change the fractions to improper fractions.

2) Change the denominator to a common denominator.

3) Minus the Numerators.

4) Changing it to a mixed number

Answer:
B
Step-by-step explanation:
Answer:
4
Step-by-step explanation:
The slope is 4 because in the y=mx+b format m is always the slope while b is your initial value or starting point
Answer: 31
Step-by-step explanation:
The two disks that we know Gina has are: 2 and 8.
We want to find tha largest sum that Gina's disks can have:
This means that the other 3 disks must be the largest possible ones.
We know that Monique has the disks 7 and 9.
The already taken values are: 2, 7, 8, and 9.
The largest values remaining are:
10, 6 and 5.
Then the largest sum that Gyna's disks can have is given by the scenario where Gina draws these 3 disks, and the sum will be:
2 + 8 + 10 + 6 + 5 = 31
Answer
Correct option is
D
5 cm
△ABC is an equilateral △
∴Area=
4
3
a
2
=75
3
⟹ a
2
=300
a=10
3
cm
AM=MC=
2
10
3
=5
3
cm
Area =
2
1
×AC×BM
75
3
=
2
1
×10
3
×BM
∴BM=15cm
△AOC,△BOC and △AOB are of equal area
∴ar(△AOC)=
3
1
ar(△ABC)=
3
1
×75
3
=25
3
cm
2
In △AOC,
ar(△AOC)=
2
1
×AC×OM
⇒25
3
=
2
1
×10
3
×OM
∴OM=5cm Ans.