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Usimov [2.4K]
2 years ago
5

HELP PLEASE with this question

Mathematics
1 answer:
grin007 [14]2 years ago
7 0
No it's not a function pretty sure
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Brad the pair of fractions as a pair of fractions with a common denominator 3/10 and 1/2​
LuckyWell [14K]

Answer:

<em><u>2/10 and 3/10</u></em>

Step-by-step explanation:

Hello!

The best way of finding common denominators is using their factors!

(of denominators)

10 : 1, 2, 5, 10

2 : 1, 2

We also can see that 10 is divisible by 2.

For that reason, 1/2 is equivalent to 2 * 10 ÷ 2.

That's 10.

So <em>5/10 and 3/10</em> have a common denominator 10.

Hope this helps!

4 0
3 years ago
Hugo wants to make a sandwich using wrap or baguette cheese or turkey mayonnaise or ketchup , list all the possible types of san
S_A_V [24]

Answer:

What subject is this

Step-by-step explanation:

6 0
3 years ago
Jason's piggy bank contained $5.70 in nickels and pennies. If there were 30 more pennies than nickels, how many of each did he h
ValentinkaMS [17]

Answer:

98 nickles and 128 pennies

Step-by-step explanation:

You probably can't read my hand writing but I spent like, 15 minutes on this. This is the correct answer, but I don't know how to explain how it is but I promise it is!

8 0
3 years ago
Emely has 25 onions and 14 peppers that she grew in her garden. homw many more oinons does she have that peppers
stiks02 [169]

Answer:

11

Step-by-step explanation:

<u><em>SUBTRACT</em></u>

4 0
3 years ago
Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0&lt;×&lt
brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

f'(x) = \dfrac{(-x + 2)^4}{dx}  = -4 \cdot (-x + 2)^3 = 0

∴ (-x + 2)³ = 0

x = 2

f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx}  = -12 \cdot (-x + 2)^2

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0
3 years ago
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