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swat32
3 years ago
7

42 is 8% of what number

Mathematics
2 answers:
Setler [38]3 years ago
7 0
8%=0.08
42÷0.08=525
So your answer is 525
Genrish500 [490]3 years ago
3 0
The answer to 42 of 8% is 3.36
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P’(-1, 1), I’ (1, 2), G’(1,0)

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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a^2+b^2=c^2

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Replace the values in and solve for c:

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If the perimeters of each shape are equal, which equation can be used to find the value of x? A triangle with base x + 2, height
Tju [1.3M]

Correct question :

If the perimeters of each shape are equal, which equation can be used to find the value of x? A triangle with base x + 2, height x, and side length x + 4. A rectangle with length of x + 3 and width of one-half x. (x + 4) + x + (x + 2) = one-half x + (x + 3) (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3) 2 (x) + 2 (x + 2) = 2 (one-half x) + 2 (x + 3) x + (x + 2) + (x + 4) = 2 (x + 3 and one-half)

Answer: (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3)

Step-by-step explanation:

Given the following :

A triangle with base x + 2, height x, and side length x + 4 - - - -

b = x + 2 ; a = x ; c = x + 4

Perimeter (P) of a triangle :

P = a + b + c

P =( x + 2) + x + (x + 4) - - - (1)

A rectangle with length of x + 3 and width of one-half x

l = x + 3 ; w = 1/2 x

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P = 2(x+3) + 2(1/2x)

If perimeter of each same are the same ; then;

(1) = (2)

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7 0
3 years ago
b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum
Mariana [72]

Answer:

{52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in {52 \choose 7} ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in {7 \choose 2} ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in {45 \choose 7}. Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in {7 \choose 2} ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in {38 \choose 7}. And again, we have to choose which ones are the hidden ones, which can be chosen in {7 \choose 2} ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in {31 \choose 7} ways. And choosing which ones are the hidden ones for this player can be done in {7 \choose 2} ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

{52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

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3 years ago
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Answer:

OOF

Step-by-step explanation:

6 0
3 years ago
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