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Mamont248 [21]
3 years ago
15

the amount of mil available per child in a day care centrr is given by the function m(x) =25/x, where x is the number of childre

n and m is the quantity of available milk in liters. if 50 children are present on a day how much milk is available per child
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

0.5 liters of milk are available per child.

Step-by-step explanation:

Amount of milk available per children:

The amount of milk, in liters, available for x children is given by:

m(x) = \frac{25}{x}

50 children are present on a day

This means that x = 50

How much milk is available per child?

This is m(50). So

m(50) = \frac{25}{50} = 0.5

0.5 liters of milk are available per child.

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Y = 3sin(2x - 5)
y = 3sin(2x - 5)

6 0
3 years ago
SIMPLIFY the expression! Please!
Levart [38]

1+3^2 = 1+9 = 10

10/5 = 2

2-6+2 = -4+2

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5 0
3 years ago
Read 2 more answers
Let D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36} Determine which of the following statements are true and which are false. a)
morpeh [17]

Answer:

\forall x\in D if x is odd then x> 0 is true statement.

\forall x\in D if x is odd then x> 0 is true statement.

\forall x\in D if x is even then x≤0 is false statement.

\forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

\forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

Step-by-step explanation:

Consider the provided information.

D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36}

Part (A) \forall x\in D if x is odd then x> 0

Here only even numbers are less than 0 that means the statement is true.

\forall x\in D if x is odd then x> 0 is true statement.

Part (B) \forall x\in D if x is less than 0 then x is even.

Here only even numbers are less than 0 that means the statement is true.

\forall x\in D if x is odd then x> 0 is true statement.

Part (C) \forall x\in D if x is even then x≤0

Here we can see that 16, 26, 32, 36 are even number and also greater than 0. Thus the statement is false.

\forall x\in D if x is even then x≤0 is false statement.

Part (D) \forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4.

There is only one number whose ones digit is 2. i.e. 32 also the tens digit of the number 32 is 3. Which makes the above statement true.

\forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

Part (E) \forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2.

Numbers having ones digit 6 are: 16, 26 and 36

Here, the tens digits are 1, 2 and 3 which is contradict to our statement. Hence the provided statement is false.

\forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

8 0
3 years ago
PLZZ HELLP ME !!!! WILL GIVE BRANLIEST !!!!
Zielflug [23.3K]

Answer:

24mm

Step-by-step explanation:

since it's a similar triangle, we solve;

EH/EG=DH/DG

EH=56mm;

EG=44.8mm;

DH=35mm;

DG=X+4.

Fix them,

56/44.8=35/x+4

cross multiply

56(x+4)=35×44.8

56x+224=1,568

collect the like term

56x=1,344

divide via by 56

56x/56=1344/56

x=24mm

Check/ verify

EH/EG=DH/DG

56/44.8=35/24+4

56/44.8=35/28

CROSS MULTIPLY OVER THE EQUAL SIGN.

56×28=35×44.8

1,568=1,568

THAT'S CORRECT.

6 0
3 years ago
Can someone tell me the answer and explanation plz I’ll give brainlist points
MAVERICK [17]

Answer:

5

Step-by-step explanation:

to cancel out the bottom part of the fraction, we multiply by 3. so now we have y+1= 6. we subtract 1, and get y=5

4 0
4 years ago
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