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SCORPION-xisa [38]
3 years ago
10

Line w ll segment YZ

Mathematics
1 answer:
Oduvanchick [21]3 years ago
8 0
Angle one because they are alternate interior angles.
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Can you help me to solve this<br>​
RUDIKE [14]

Answer:

Is the question even complete ?

Step-by-step explanation:

5 0
2 years ago
The length of a rectangle is 1 unit more than the width. The area of the rectangle is 56 units. What is the width in units of th
Mandarinka [93]

Answer:

Width = 7units

Step-by-step explanation:

The area of a rectangle

A = h * w

A - area of a rectangle

h - height of a rectangle

w - width of a rectangle

Or

A = length * breadth

It's all the same

Given :

h = 1 + w

w = w

A = 56

56 = ( 1 + w) * w

56 = ( 1 + w)w

56 = w + w^2

56 = w^2 + w

w^2 + w - 56 = 0

Find a factor that can be multiplied to give -56

and added to give + 1

The factor is 8 and -7

Substitute 8w - 7w for w

w^2 + 8w - 7w - 56 = 0

( w^2 + 8w) - (7w - 56) = 0

w( w + 8) - 7( w + 8) = 0

( w - 7) = 0

( w + 8) =0

w - 7 = 0

w = 7

w + 8 = 0

w = -8

Since a side of any rectangle can not be negative

w = 7

The width of the rectangle is 7units

The length of the rectangle is 1 + w = 1 + 7 = 8units

4 0
3 years ago
PLS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Nadusha1986 [10]

Answer:

after school and college football season

6 0
3 years ago
Read 2 more answers
A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for the machine to work pro
olchik [2.2K]

Answer:

0.756

Step-by-step explanation:

It is given that a machine has four components, A, B, C, and D.

P(A)=P(B)=0.93, P(C)=0.95,P(D)=0.92

If these components set up in such a manner that all four parts must work for the machine to work properly.

We need to find the probability that the machine works properly. It means we have to find the value of P(A\cap B\cap C\cap D).

If two events X and Y are independent, then

P(X\cap Y)=P(X)\times P(Y)

Assume the probability of one part working does not depend on the functionality of any of the other parts.

P(A\cap B\cap C\cap D)=P(A)\times P(B)\times P(C)\times P(D)

Substitute the given values.

P(A\cap B\cap C\cap D)=0.93\times 0.93\times 0.95\times 0.92

P(A\cap B\cap C\cap D)=0.7559226

P(A\cap B\cap C\cap D)\approx 0.756

Therefore, the probability that the machine works properly is 0.756.

7 0
3 years ago
Cual es el resultado?
Korolek [52]
I hope this helps you


x^2/3 [x^2/3.x^-1/4)^6.1/3.2


x^2/3[x^8-3/12]^4


x^2/3 [x^5/12]^4


x^2/3.x^5/12.4


x^2/3.x^5/3


x^2+5/3


x^7/3
5 0
4 years ago
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