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saveliy_v [14]
4 years ago
9

On a number line, the distance from zero to -7 is 7 units.

Mathematics
1 answer:
aivan3 [116]4 years ago
4 0

Answer: Option A. |-7| = 7


Solution:

The distance is a positive value, then we must not take in account the sign, and using:

|a| =-a if a<0

with a=-7

|-7| = -(-7)

|-7| = 7

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Dmitri needs 6 liters of a 16% solution of sulfuric acid for a research project in molecular biology. He has two supplies of sul
mylen [45]

Answer:

4 litres of 12% solution

2 litre of 24% solution

Step-by-step explanation:

Let the 12% solution = l

Let the 24% solution = h

Dimitri needs a total of 6 litres of both :

l + h = 6 - - - - - - - - (1)

He needs a mixture of both volumes of both to give a 16% volume

12%l + 24%h = 16% * 6

0.12l + 0.24h = 0.96 ----(2)

From (1)

l + h = 6;

l = 6 - h

Substitute l =6-h into (2)

0.12(6 - h) + 0.24h = 0.96

0.72 - 0.12h + 0.24h = 0.96

0.72 + 0.12h = 0.96

0.12h = 0.96 - 0.72

0.12h = 0.24

h = 0.24 / 0.12

h = 2

From ;

l = 6 - h

l = 6 - 2

l = 4

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
4 years ago
A heart shaped chocolate box is composed of one square and two half circles. The total number of chocolates in the box is calcul
Scrat [10]

Answer:

4²+3²+2=69

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Compute the probability that a hand of 13 cards contains all 4 of at least 1 of the 13 denominations.answer: 0.0343
gogolik [260]
The probability of a hand of 13 cards contains all 4 of one of the denominations, say Aces, is given by:
\frac{4C4\times48C9}{52C13}=0.00264106
The same probability applies to getting all 4 of each of the other denominations.
Therefore the required probability is:
0.00264106\times13=0.0343
The answer is 0.0343.
5 0
3 years ago
Can anyone help me out ?
KengaRu [80]

She made her first mistake in step 2

-2x+5 =-11

add 5 to each side

-2x-5+5 = -11+5

-2x = -6     (not -16)

divide by -2

-2x/-2 = -6/-2

x = 3

She added 5 to 11 and got 16 and made it negative instead of subtracting 5 from 11 and making it negative

6 0
4 years ago
Read 2 more answers
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