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LUCKY_DIMON [66]
3 years ago
7

What is a valid prediction about the continuos function

Mathematics
1 answer:
Elodia [21]3 years ago
8 0

Answer:

EX: f(x) ≥ 0 over the interval [-1, 1].

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If 1 pound = 16 ounces, how many pounds are in 435 ounces?
docker41 [41]

Answer: 27.1875 pounds

Step-by-step explanation:

To find out how many pounds are in 435 ounces, we need to divide the number of ounces by the number of ounces per pound. Since 1 pound is equal to 16 ounces, we can divide 435 ounces by 16 ounces per pound to get the number of pounds. This gives us 435 ounces / 16 ounces/pound = 27.1875 pounds.

Therefore, there are approximately 27.1875 pounds in 435 ounces.

5 0
2 years ago
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Celia and Ryan are starting a nutrition program. Celia currently consumes 1200 calories a day and will increase that number by 1
GenaCL600 [577]
C is Celia's calorie count and R is Ryan's. d = days
C = 1200 + 100d
R = 3230 - 190d
Since you want to find when they will consume the <em>same</em> amount of calories, you set the two equations equal to each other: 1200 + 100d = 3230 - 190d.
Then, subtract 1200 from both sides: 100d = 2030 - 190d. Next, add 190d to both sides: 290d = 2030. Therefore, the number of days is eqal to 2030/290, which is 7.
4 0
3 years ago
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vladimir2022 [97]

Answer:

coordintes its lile coord

Step-by-step explanation:

ita like coordinates

6 0
4 years ago
How can you use fractions to place the decimal in the product?
Oxana [17]

Answer:

it is easier to multiply first and then to place the decimal point. Multiply the factors as if they were whole numbers. Disregard the decimal point until you place it in the product.

6 0
3 years ago
The roof of a new sports arena rests on round concrete pillars. The maximum load a cylindrical column of a circular cross sectio
KATRIN_1 [288]

Answer: There are 3.5 metric tons will be supported by a column 8 m high and two thirds m in diameter.

Step-by-step explanation:

Since we have given that

Height of columns of arena = 9 m

Diameter = 1 m

Load = 14 metric tons

Since Load is directly varies as the fourth power of the diameter 'd', and inversely as the square of the height.

so, Mathematically, it is expressed as

L=\dfrac{kd^4}{h^2}\\\\14=\dfrac{1.k}{9^2}\\\\14\times 81=k\\\\k=1134

If the height of columns = 8 m

Diameter = \dfrac{2}{3}\ m

So, Load would be

L=\dfrac{1134\times (\dfrac{2}{3})^4}{8^2}\\\\L=3.5\ tons

Hence, there are 3.5 metric tons will be supported by a column 8 m high and two thirds m in diameter.

7 0
3 years ago
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