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WITCHER [35]
3 years ago
14

Use benchmark fractions to compare 2/9 and 5/6. Part A Which of the following is a benchmark fraction Heather can use to compare

the two fractions?
Mathematics
2 answers:
emmasim [6.3K]3 years ago
6 0
The answer I’ve got was 30 hope it’s right
Dimas [21]3 years ago
5 0

Answer:

30

Step-by-step explanation:

2/9 and 5/6 have a lcm of 30.

You might be interested in
Whats 42/99 in simplest form.
const2013 [10]

Answer:

14/33

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
UPS charges $6.00 for the first pound, $1.00 for the second pound, and $0.15 for each additional pound. FedEx charges $4.00 for
Pavlova-9 [17]

Answer:

12 pounds

Step-by-step explanation:

We we need to write equations to represent the situation in this problem.

x=weight of package

UPS cost=6+1+0.15(x-2)

UPS cost=6+1+0.15x-0.3

UPS cost=6.7+0.15x

FedEx cost=4+2+.25(x-2)

FedEx cost=4+2+0.25x-0.5

FedEx cost=5.5+0.25x

Next, we can set the two equations equal to each other to solve for x.

6.7+0.15x=5.5+0.25x

6.7+0.15x-5.5=5.5+0.25x-5.5

1.2+0.15x-0.15x=0.25x-0.15x

1.2/0.1=0.1x/0.1

x=12

It takes 12 pounds for the UPS and FedEx packages to cost the same amount to ship.

8 0
4 years ago
Does anyone know how to do this?
Lena [83]

Answer:

Each guest would get 6 party favors.

Step-by-step explanation:

First, we find out how many party favors Amy bought. 3 bags of 12 party favors equals 36 party favors (3x12)

So, divide 36 party favors by 6 guests, and you get the final answer of 6 party favors per guest.

Hope this helps!!

6 0
3 years ago
The horsepower , H(s), required for a racecar to overcome wind resistance is given by the function : H(s) = 0.003s^2+0.07s-0.027
mote1985 [20]
Average rate of change = [H(100) - H(80)] / (100 - 80)
H(100) = 0.003(100)^2 + 0.07(100) - 0.027 = 0.003(10000) + 0.07(100) - 0.027 = 30 + 7 - 0.027 = 36.973

H(80) = 0.003(80)^2 + 0.07(80) - 0.027 = 0.003(6400) + 0.07(80) - 0.027 = 19.2 + 5.6 - 0.027 = 24.773

Average rate of change = (36.973 - 24.773)/(100 - 80) = 12.2/20 = 0.61

Answer: B
8 0
3 years ago
Read 2 more answers
1. There are 78 sophomores at a school. Each is required to take at least one year of either chemistry or physics, but they may
SVETLANKA909090 [29]

We are given that there are a total of 78 students. If we set the following variables:

\begin{gathered} C=\text{students only in chemestry} \\ P=\text{students only in physics} \\ PC=\text{students in physics and chemistry} \end{gathered}

Then, the sum of all of these must be 78, that is:

C+P+PC=78

Since there are 15 in chemistry and physics and 47 in chemistry, we may replace that into the equation and we get:

47+P+15=78

Simplifying:

62+P=78

Now we solve for P by subtracting 62 on both sides:

\begin{gathered} 62-62+P=78-62 \\ P=16 \end{gathered}

Therefore, there are 16 students in physics

3 0
1 year ago
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