The cube of the difference of 1 and a number.

A triangle has a perimeter of 165 cm. The first side is 65 cm less than twice the second side. The third side is 10 cm less than

Dima020 [189]

Answer:

first side: 55 cm
second side: 60 cm
third side: 50 cm

Step-by-step explanation:

The problem statement gives all the side lengths in terms of that of the second side. Let s represent the length of the second side in cm. Then the length of the first side is (2s-65), and the length of the third side is (s-10). The perimeter is the sum of all the side lengths:

165 = (2s -65) +s +(s-10)

165 = 4s-75

240 = 4s

60 = s . . . . . . . . second side

2s-65 = 55 . . . . first side

s-10 = 50 . . . . . . third side

In order, the side lengths are 55 cm, 60 cm, 50 cm.

4 0

3 years ago

joel made some muffins. He gave 1/4 of the muffins to a neighbor. He took 3/8 of the muffins to school. What fraction of the muf

salantis [7]

Okay so 3/8+1/4 can be 3/8+2/8 and you get 5/8. This means that 3/8 of the muffins are left.

5 0

3 years ago

Rounding to the nearest tenth which of the following gives an overestimate

anastassius [24]

A is you like the answer

3 0

4 years ago

Read 2 more answers

The first person to get this right gets brainliest

Art [367]

Answer:

the slope of 2 has a greater slope

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A woman has n keys, of which one will open her door. If she tries the keys at random, discarding those that do not work, what is

enot [183]

Answer:

If she discard previously tried keys the probability is 1/n

If she doesn't discard previously tried keys the probability is:

$\frac{(n-1)^{k} }{n^{k} }$

Step-by-step explanation:

If she discard previously tried keys, the probability of open the door in the 1st, 2nd and 3rd try are:

1st try: $\frac{1}{n}$

2nd try: $(\frac{n-1}{n})*(\frac{1}{n-1})=\frac{1}{n}$

3rd try: ( $\frac{n-1}{n})*(\frac{n-2}{n-1})*(\frac{1}{n}) =\frac{1}{n}$

Where, for example, the probability of open the door in the 2nd try is the multiplication of the probability that the woman didn't open the door in the 1st try with the probability that she open the door in the 2nd try.

So, the probability that she open the door in the kth try is: 1/n

On the other hand, if she doesn't discard previously tried keys, the probability to open the door in the 1st, 2nd and 3rd try are:

1st try: $\frac{1}{n}$

2nd try: $(\frac{n-1}{n})*(\frac{1}{n})=\frac{n-1}{n^2}$

3rd try: ( $\frac{n-1}{n})*(\frac{n-1}{n})*(\frac{1}{n}) =\frac{(n-1)^2}{n^3}$

So, the probability that she open the door in the kth try is:

$\frac{(n-1)^{k-1}}{n^k}$

5 0

3 years ago