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diamong [38]
3 years ago
13

How many integer values of x are there so that x, 12,and 9 could be the lengths of the sides of a triangle?

Mathematics
1 answer:
OlgaM077 [116]3 years ago
8 0
Triangle properties.

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A medium-sized has 70 calories. this is 10 calories less than 1/4 of the calories in an Old Westie Chocolate Bar. We need to fin
OLEGan [10]

Answer:

320 calories

Step-by-step explanation:

1/4 would be 80 calories because 70 is ten less so it would be 80 = 1/4 so 80 times 4 would be 320 calories

5 0
4 years ago
Calculaye the perimeter and area of these composite shapes
lora16 [44]

For this case we have that the perimeter of the figure is given by the sum of the lengths of the sides, that is:

14 + 18 + 3 + 4 + 3 + 4 + 3 + 4 + (14-9) + (18-12) = 64

Thus, the perimeter of the figure is 64 centimeters.

Now, we find the area of the figure:

We have that by definition, the area of a rectangle is given by:

A = a * b

Where:

a and b are the sides of the rectangle

We have 4 vertical rectangles from left to right:

A_ {1} = 14 * (18-12) = 14 * 6 = 84 \ cm ^ 2\\A_ {2} = (3 + 3 + 3) * 4 = 9 * 4 = 36 \ cm ^ 2\\A_ {3} = (3 + 3) * 4 = 6 * 4 = 24 \ cm ^ 2\\A_ {4} = 3 * 4 = 12 \ cm ^ 2

Thus, the total area isA_ {t} = 156 \ cm ^ 2

Answer:

The perimeter of the figure is 64 centimeters.

A_ {t} = 156 \ cm ^ 2

4 0
4 years ago
17 1/4 over 100 as a decimal
Andrew [12]

Answer:

0.1725

Step-by-step explanation:

17 1/4 /100

17.25/100

0.1725

4 0
2 years ago
Read 2 more answers
Can help me with this?
Tanzania [10]

Answer:I’m sorry this is hard I don’t think I have the right answer chose C

Step-by-step explanation:

3 0
3 years ago
A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If
Blababa [14]

Answer: Our required probability is \dfrac{1}{2}

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = \dfrac{1}{3}

So, Probability of getting head from fair coin = \dfrac{1}{2}

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}

Hence, our required probability is \dfrac{1}{2}

No, the answer is not \dfrac{1}{3}

5 0
3 years ago
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