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SVETLANKA909090 [29]
3 years ago
8

John visited his grandma's house over the weekend.

Mathematics
1 answer:
ddd [48]3 years ago
3 0

Answer:

cool, is this a question?

Step-by-step explanation:

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Can someone PLEASE help me with questions #5 and 6
Hatshy [7]

Domain={8,9}

Range = { -3,-4,-6,-9}

No relation is not function as elements in domain are related to more than one element of codomain

5 0
3 years ago
Tanya purchased 4 hammers for $11.79 each and 7 screwdrivers for $6.65 each. Write an expression for the total cost of the tools
igor_vitrenko [27]
<span>Write an expression for the total cost of the tools.
</span>
total cost = <span>$11.79 (4)   + </span>$6.65 (7) 

<span>Then find the total cost.

</span>total cost = $11.79 (4)   + $6.65 (7) 
total cost = $47.16 + $46.55 = $93.71
7 0
3 years ago
Read 2 more answers
*NEED ANSWER QUICK!!*
Rom4ik [11]

Answer:

You would need 9/4 or 2 1/4 gallons of paint

Step-by-step explanation:

You would need 2 1/4 gallon of paint because if 1/3 of the room takes 1/4 paint, then 3/4 would finish a room, multiply the three by three for three rooms and there you have 9/4 or 2 1/4 gallons

6 0
2 years ago
D=7.9cm. what is the circumference and area of the circle?
In-s [12.5K]
Circumfrence: pi times diameter
7.9*\pi
circumfrence is approximatley 24.806

area: \pi r^2
a=\pi*(1/2(d))^2
a= 48.99
8 0
3 years ago
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A 2X2 square is centered at the origin. It is dilated by a factor of 3. What are coordinates of the vertices of the square?
Vesna [10]

Answer:

The vertices are:

A' = (-3, -3)

B' = (3, -3)

C' = (3, 3)

D' = (-3, 3)

Step-by-step explanation:

Given:

A 2 x 2 square is centered at the origin.

So, the center of the square is (0, 0)

Since it is 2 x 2 square, the side of the square is 2 units.

So, the vertices of the 2 x 2 square are A (-1, -1),  B(1, -1), C(1. 1), D(-1, 1)

The above square is dilated by a factor of 3.

Let's name the dilated square A'B'C'D'

To find the coordinates of the vertices of dilated square, we need to multiply each vertices of ABCD by 3.

A(-1, -1) = 3(-1, -1) = A'(-3, -3)

B(1, -1) = 3(1, -1) = B'(3, -3)

C(1, 1) = 3(1, 1) = C'(3, 3)

D(-1, 1) = 3(-1, 1) = D'(-3, 3)

7 0
3 years ago
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