All categories

3 years ago

Can someone please help me with this problem???

Mathematics

1 answer:

hammer [34]3 years ago

8 0

Answer:

x + 4 = 5x - 6

x = 2.5

Step-by-step explanation:

x + 4 = 5x - 6

4 = 4x -6

10 = 4x

x = 2.5

You might be interested in

Kayden wants to ride his bicycle 29.5 miles this week. He has

geniusboy [140]

Answer: 5.5

Step-by-step explanation:

3m+13=29.5

3 0

3 years ago

Please help me answer this with the correct answer :)

BartSMP [9]

Pretty sure is 63 degrees since the one next to it is exactly the same and its 63 degrees

7 0

3 years ago

-2x^(3) +3=-1x^(3) -1x^(2) +1x+1

Natali5045456 [20]

-2x³ + 3 = -x³ - x² + x + 1

Add x³ to both sides, and subtract 3 from both sides.

-2x³ + x³ = -x² + x + 1 - 3

Add like terms.

-x³ = -x²+ x - 2

Move all of the (X's) to both sides, by adding x² to both sides, and subtracting x form both sides.

-x³ + x² - x = -2

~Hope I helped!~

7 0

3 years ago

Smart enough to answer this question? Gracie, Mary, and Nancy each have a small collection of seashells. Gracie has 5 more than

adelina 88 [10]

X = mary's shells
g = 1 1/4x + 5
n = 1 1/2x+ 1
g = n

1 1/4x + 5 = 1 1/2x + 1
5/4x + 5 = 3/2x + 1...multiply by common denominator of 4, this will get rid of fractions...however, this is optional...u can work with fractions if u wish.
5x + 20 = 6x + 4
5x - 6x = 4 - 20
-x = - 16
x = 16....so Mary has 16 shells

6 0

3 years ago

Read 2 more answers

Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r

erastova [34]

Answer:

The probability is $\frac{56!}{64!}$

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, $64 \choose 8 .$

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function $f : A \rightarrow A ,$ with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

$\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}$

7 0

3 years ago