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Ivahew [28]
3 years ago
12

Find the perimeter and the area of a rectangle with the sides 3 3/10 meters and 6 2/3 meters.

Mathematics
2 answers:
Rus_ich [418]3 years ago
6 0

Answer:

37.5

Step-by-step explanation:

nika2105 [10]3 years ago
5 0

Answer:

Step-by-step explanation:

3 3/10=33/10

6 2/3=20/3

-------------------

33/10*20/3=660/30=22

The area is 22 m^2.

------------------------------------

2(33/10)+2(20/3)=66/10+40/3=299/15

The perimeter is 299/15 meters.

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1/2x-5=10-3/4x please helpppppp
GarryVolchara [31]

Answer:

x=12

Step-by-step explanation:

1/2x - 5 = 10 - 3/4x ---> move like terms to the same side

1/2x + 3/4x = 10 + 5 ---> calculate

5/4x = 15 ---> multiply both sides by 4/5

x = 12

8 0
3 years ago
The original price of a sweater was $40. When Mandy went shopping, it was on sale for 50% off. What was the cost of the sweater
goldenfox [79]

Answer: $20

Step-by-step explanation:

half of 40 is 20.....

7 0
3 years ago
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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.
In-s [12.5K]

Answer:

C

Step-by-step explanation:

We are given

The student council earned $1.25 for every cup of hot chocolate it sold

and $0.75 for every cup of apple cider it sold

total number of cup is 375

We can see that first term is 1.25x

and we have

The student council earned $1.25 for every cup of hot chocolate

so, to find total cost of hot chocolate , we multiplied 1.25 with x

that's why

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So the answer must be C

7 0
3 years ago
Determine the number of possible triangles, ABC, that can be formed given A = 30°, a = 4, and b = 10.
sweet [91]

Answer:

None

Step-by-step explanation:

We are given a triangle ABC with ∠A = 30°, sides a = 4 and b = 10.

According to the 'Law of Sines- Ambiguous Case', we have,

If a < b×sinA, then no triangle is possible.

If a = b×sinA, only one triangle is possible

If a > b×sinA, two triangles are possible.

So, we have,

b×sinA = 10 × sin30 = 10 × 0.5 = 5.

Now, as

4 = a < bsinA = 5.

We get, according to the rule, no triangle is possible.

4 0
3 years ago
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BigorU [14]
The simplest way to solve this is to notice that letting t = 8/3 will let the term (3t-8) = 0. Substituting this will make v(t) = 2. Then we check which graphs pass through (8/3, 2). Only the second and third graphs do.

Next, we look at the behavior of the graph as t increases. Based on the equation, as t increases, (3t-8)^3 increases as well, so v(t) will increase as well. This is shown by the second graph, in which v(t) increases as t increases.
5 0
3 years ago
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