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xeze [42]
3 years ago
12

Jana finished 1/4 of her homework at school and 2/3 of it before dinner. Which of the following statements is true?

Mathematics
1 answer:
jonny [76]3 years ago
4 0
<span>c.She has a little homework left to finish.
</span>
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Please help me with this as soon as possible!
ladessa [460]

Answer:

B

Step-by-step explanation:

it is the B because that's 2/5

6 0
3 years ago
Read 2 more answers
Apply the distributive property to create an equivalent expression. 2(3-8y)
sdas [7]
Using the distributive property, an equivalent expression would be 6-16y


Comment the correct answer. 
4 0
4 years ago
Read 2 more answers
Find the formula for the nth term of the arithmetic sequence -39, 61, 161, 261, ...
maksim [4K]
<h3>Given that it is arithmetic</h3>

d = a_{2} - a_{1} = 61 - ( - 39) = 100

a_{n} = a_{1} + (n - 1)d \\ a_{n} = - 39+ 100 (n - 1) \\ a_{n} =  - 39 + 100n - 100 \\ a_{n} =  - 139 + 100n

6 0
2 years ago
Which ordered pairs are solutions to the equation?
soldier1979 [14.2K]


To solve this you want to plug in your x's and y's to see if they match.

A. 3(11)-4=33-4=29 and 29 is not equal to 5 so A is not a solution

B.3(5)-4=15-4=11=11  3(3)-4=9-4=5 and 5 is not equal to 2 so B is not a solution

C. 3(2)-4= 6-4=2 2 is not equal to 3 so C is not a solution

D is the answer. 3(5)-4=11 and 3(2)-4= 2

6 0
3 years ago
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
4 years ago
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