Answer:
length PQ = 8.9 m
Step-by-step explanation:
The picture below completes the question you asked. From the picture above RSTU is a rectangle . The two shaded square has a length of sides 8 m and 4 m respectively. The legs of the triangle is the sides of the 2 squares. Therefore, the triangle is a right angle triangle. The adjacent side is 4 m while the opposite side is 8 m.
Using Pythagoras theorem the hypotenuse of the triangle can be calculated below.
Pythagoras theorem
c² = a² + b² Therefore,
c² = 4² + 8²
c² = 16 + 64
c² = 80
square root both sides
c = √80
c = 8.94427191
c ≈ 8.9 m
length PQ = 8.9 m
Answer:
4n and 5p
Step-by-step explanation:
Because if it the cost of each your mulitplying and because you wanting to just define the variable p for pen and n for notebook all a variable is a letter than let you know what something is for using in math later
Answer:
36 inches
Step-by-step explanation:
Given data
Area of triangle =2 1/4 square inches
Area =9/4 square inches
Height of triangle =1/8 inches
We know that the expression for the area of triangle is
Area = 1/2Base *Height
9/4=1/2*Base *1/8
9/4=Base/16
Cross multiply
16*9=Base*4
144=Base*4
Base =144/4
Base=36 inches
415+86+12(2+5) first start by multiplying 12 and 2 and you will get 24 then multiply 12 and 5 and you get 60 so 415+86+24+60=585
<h2>
Answer:</h2>
<h2>
Step-by-step explanation:</h2>
As the question states,
John's brother has Galactosemia which states that his parents were both the carriers.
Therefore, the chances for the John to have the disease is = 2/3
Now,
Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.
Now, one of those children married with a person.
So,
Probability for the child to have disease will be = 1/2
Now, again the child's child (Martha) probability for having the disease is = 1/2.
Therefore,
<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>
(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)
<em><u>Hence, the probability for the first child to have Galactosemia is 
</u></em>
