29 I think but iam not sure if it’s right
Answer:
i cant help with that cause i have no way of drawing
Step-by-step explanation:
Any time you have a fraction within an equation, multiply the entire equation by the denominator to clear the fraction. Since the lead term is negative, we can multiply that away as well
(-14) (0=-1/14x²+4x+5) [now distribute]
0=x²-56x+70 [try to factor into binomials first]
Since 70 only has prime factors of 2·5·7, there is no combination which equals (-56). Use the quadratic formula, or complete the square. I'll use quadratic:
x=<u>-b+/-√(b²-4ac)</u>
2a
a=1, b=(-56), c=70
x= <u>-(-56)+/- √((-56)²-4(1)(70)</u>
2(1)
x= <u>56+/- √(3136-280)
</u> 2
<u />x=<u>56+/-√(2856)</u>
2
x=<u>56+/-√(2³·3·119)</u>
2
x=<u>56+/-2√(714)</u>
2
x=28+√714; x=28-√714
Answer:
The towel bar should be placed at a distance of 
from each edge of the door.
Step-by-step explanation:
Given:
Length of the towel bar = 
Now given length is in mixed fraction we will convert in fraction.
To Convert mixed fraction into fraction Multiply the whole number part by the fraction's denominator, then Add that to the numerator, then write the result on top of the denominator.
can be Rewritten as 
Length of the towel bar =
Length of the door = 
can be Rewritten as 
Length of the door =
We need to find the distance bar should be place at from each edge of the door.
Solution:
Let the distance of bar from each edge of the door be 'x'.
So as we placed the towel bar in the center of the door it divides into two i.e. '2x'
Now we can say that;
Now we will take LCM to make the denominators common we get;
Now denominators are common so we will solve the numerators.
Or 
Hence The towel bar should be placed at a distance of from each edge of the door.
Answer:
No. See explanation below.
Step-by-step explanation:
Since the cards are being selected <u>without replacement,</u> every time we select a card, <u>the probability varies</u> (since there is one less card) and therefore, the probability doesn't remain the same for every trial and therefore, the probability of success changes for every trial.
It is because of this that this probability experiment doesn't represent a binomial experiment.
