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Margarita [4]
3 years ago
10

0+0???????????? ??????

Mathematics
1 answer:
Arisa [49]3 years ago
5 0
The answer for 0+0 is 0.
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jack has 14 cards. by the end of the week his amount of cards has doubled. how many cards does jack have now
slava [35]

Answer:

Jack has 28 cards.

Step-by-step explanation:

To double 14, multiply 14 times 2. 14x2=28

6 0
3 years ago
Read 2 more answers
Amy, Bodie, Claire, and Dejan are in the same math class. The locations of their desks in relation to each other are graphed on
Xelga [282]

Answer:10 ft

Step-by-step explanation:

5 0
3 years ago
Choose the constant term that completes the perfect square trinomial.<br><br> y2 + y
Oduvanchick [21]

Hello!

First off, please write y2 + y as y^2 + y. The " ^ " symbol denotes exponentiation, whereas y2 is meaningless.

To find the constant term in question, take half of the coefficient of y (that is, take 1/2) and square it. Then we have y2 + y + 1/4.

The constant term in question is 1/4.


3 0
3 years ago
Could someone help me please
Rina8888 [55]

Answer:

the answer should be 30

Step-by-step explanation:

since you can divided 15 by 5 to get 3, you can multiply 10 x 3 to get 30

8 0
3 years ago
PLEASE HELP!!!
pogonyaev

Step-by-step explanation:

Part A:

Let m be the number of mittens and s be the number of scarves. Then we have the inequalities:

s+m\leq 30. <em>This says Nivyana and Ana cannot make more than 30 scarves</em>

50s+25m\geq 1000. <em>This says that</em> <em>Nivyana and Ana have to earn at least $1000.</em>

Part B:

The graph is attached.

Notice that the graphs of the inequalities are solid lines, this just means that the points on these lines included to the solutions of each inequality.

The darker shaded region and the solid lines bounding it, are the solutions to the inequalities because that's where the values common to both inequalities are found.

Part C:

From the graph we get two possible solutions:

15 scarves & 10 mittens

25 scarves & 5 mittens.

These two points lie on the solid lines that bound the darker shaded region<em> (I picked those points to stress that the lines bounding the dark region are also solutions.)</em>

8 0
3 years ago
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