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ivanzaharov [21]
3 years ago
7

At 5:00 on june morning , the temperature was 62.4.By 2:00 the afternoon , the temperature had risin 21.7 . what was the tempera

ture at 2:00 that afternoon ?
Mathematics
1 answer:
Ivahew [28]3 years ago
6 0
I think it's 84.1 (degrees). Hope this helps you
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PPP Loan

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2 years ago
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Which relation below represents a one to one function
Scilla [17]

Answer:

<h2>Second table</h2>

Step-by-step explanation:

<em>One-to-one function is a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain. Every element of the function's domain is the image of at most one element of its domain.</em>

First table:

No. Because for x = 7.25 and x = 8.5 we have the same value of y = 11

Second table:

Yes. Because for all values of x we have different values of y.

5 0
3 years ago
5+3r=5r-19(if there is no solution,type in ''no solution'')r= Answer
TEA [102]

Answer: 12 = r

Step-by-step explanation: When we have this kind of a setup, we want to put our variables together on one side of the equation and our numbers together on the other side of the equation.

First, let's put our variables on the right side by subtracting 3r from both sides of the equation. That gives us 5 = 2r - 19.

Now we can move our numbers to the left by adding 19 to both sides of the equation and we get 24 = 2r.

Divide both sides by 2 and 12 = r

Note:

Don't just do this problem in your head. It's extremely important to develop the habit of putting all your steps down on paper or digitally. It will really pay off for you down the line.

4 0
3 years ago
How to solve for shaded area
tamaranim1 [39]

Answer:

(a)  12.96 ft²

(b)  21.5 in²

Step-by-step explanation:

(a) For the first diagram

Area of the shaded region (A) = Area of Tripezium- area of circle

A = [1/2(a+b)h]-[πr²]............... Equation 1

Where a and b are the parallel side of the tripezium respectively, h = height of the tripezium, r = radius of the circle.

From the diagram,

Given: a = 15 ft, b = 6 ft, h = 12 ft, r = h/2 = 12/2 = 6 ft.

Constant: π = 3.14

Substitute these values into equation 1

A = [12(15+6)/2]-(3.14×6²)

A = 126-113.04

A = 12.96 ft²

(b) For the second diagram,

Area of the shaded region (A') = Area of square- area of circle

A' = (L²)-(πr²)............. Equation 2

Where L = lenght of one side of the square, r = radius of the circle

From the diagram,

Given: L = 2r = (2×5) = 10 in, r = 5 in

Substitute these values into equation 2

A' = (10²)-(3.14×5²)

A' = 100-78.5

A = 21.5 in²

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2 years ago
How much netherite do i need to make a full beacon again?
Hatshy [7]

Answer:

5,904 scraps

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