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sesenic [268]
3 years ago
5

25 What is the value of the digit 5

Mathematics
2 answers:
Goshia [24]3 years ago
6 0
The digit 5 is in the ones place. The 2 is in the tens place. 2 tens is equal to 20, and 5 ones is equal to 5. Together they make 25.
zlopas [31]3 years ago
5 0
In 25, the value of the 5 is ones or in the ones place. Hope it helps.
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Item 6 Your city is represented in a coordinate plane where each unit represents 1 kilometer. The library is at (−3, −3), the po
Snezhnost [94]

We know, shortest distance between two points (x_1,y_1) and (x_2,y_2) is :

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

So, distance between house and library is :

d_1=\sqrt{(-5-(-3))^2+(2-(-3))^2}\\\\d_1=5.39\ km

Distance between library and post office :

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\D=\sqrt{(-3-2)^2+(-3-2)^2}\\\\D=7.07\ km

Distance between post office and house :

D=\sqrt{(2-(-5)^2)+(2-2)^2}\\\\D=7\ km

Therefore, minimum distance that you can ride your bike is (5.39+7.07+7) km = 19.46  km.

Hence, this is the required solution.

8 0
4 years ago
Which type of conic section is described by the following equation?​
krok68 [10]

Answer:

  C.  Hyperbola opening up and down

Step-by-step explanation:

A graph tells the tale.

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The term with the positive coefficient identifies the axis along which the hyperbola opens. Here, that is the y-axis, so the figure opens up and down.

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7 0
3 years ago
Find the measure of the inscribed angle.
igor_vitrenko [27]

Answer:

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Step-by-step explanation:

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Fill in the blanks to complete the equation that describes the diagram.
melamori03 [73]

Answer:

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Hope this helps! Have a nice day!

Step-by-step explanation:

5 0
3 years ago
We are going to fence in a rectangular field that encloses 75 ft2. Determine the dimensions of the field that will require the l
pychu [463]

Answer:

For width W and length L, we have:

W = L = √(75ft^2) = 8.66ft

Step-by-step explanation:

For a rectangle of length L and width W, the area is:

A = L*W

and the perimeter is:

P = 2*L + 2*W

First, we know that the area of our rectangle is:

A = 75ft^2 = L*W

And we want to minimize the perimeter of our rectangle, then we need to minimize:

P = 2*L + 2*W

From the equation:

75ft^2 = L*W

We can isolate one of the variables, let's isolate L

L = (75ft^2)/W

We could replace this in the perimeter equation:

P(W) = 2*( (75ft^2)/W) + 2*W

P(W) = (150 ft^2)/W + 2*W

To find the minimum of the perimeter we need to look at the zero of the first derivative of P(W).

P'(W) = dP(W)/dW = -(150ft^2)/W^2 + 2

Now we need to find the value of W such that:

P'(W) = 0

0 =  -(150ft^2)/W^2 + 2

(150 ft^2)/W^2 = 2

(150ft^2) = 2*W^2

(150ft^2)/2 = W^2

75 ft^2 = W^2

√(75ft^2) = W = 8.66ft

And remember that:

L =  (75ft^2)/W

replacing with W = √(75ft^2)

L =  (75ft^2)/√(75ft^2) = √(75ft^2)

Then:

W = L = √(75ft^2) = 8.66ft

6 0
3 years ago
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