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nikklg [1K]
3 years ago
10

The expression 9x2 - 100 is equivalent to

Mathematics
1 answer:
ANEK [815]3 years ago
4 0

Answer:

I searched it up

Step-by-step explanation:

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An equal number of pieces of candy were given to children on Halloween. 30 pieces of candy were given out to boys. 48 pieces of
scoray [572]

Answer:

6 pieces of candy

Step-by-step explanation:

Since an equal number of pieces of candy were given to children on Halloween, and obviously, a different amount of boys and girls were present on Halloween. Since 30 pieces of candy were given out to boys and 48 were given out to girls, a certain amount of candy was given to some amount of boys to equal 30, and that same amount of candy was given to another amount of girls to equal 48. Since we are looking for the greatest number of pieces given to each child, we are looking for the greatest common divisor of 30 and 48, which is 6.

4 0
3 years ago
Solve x2 + 8x – 3 = 0 using the completing-the-square method.
Inessa [10]
The answer is the last one !
8 0
3 years ago
A chalkboard has an area of 35.8 ft a smaller chalkboard is 0.7 times that size what is the area of the smaller chalkboard
Elenna [48]

The smaller chalkboard would be 25.06 ft

3 0
4 years ago
What is the Gcf of 6x^3?
lina2011 [118]

Answer:  6x^3 is the greatest common factor of 12x^7 and 18x^3.

Step-by-step explanation:

5 0
3 years ago
How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.
Svetllana [295]
Naturally, any integer n larger than 127 will return 127\equiv127\mod n, and of course 127\equiv0\mod127, so we restrict the possible solutions to 1\le n.

Now,

127\equiv7\mod n

is the same as saying there exists some integer k such that

127=nk+7

We have

\implies 120=nk

which means that any n that satisfies the modular equivalence must be a divisor of 120, of which there are 16: \{1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\}.

In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

127=21\cdot6+1\iff127\equiv1\equiv7\mod6

(If we're allowing n=1, then I see no reason we shouldn't also allow 2, 3, 4, 5, 6.)
5 0
3 years ago
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