Answer:
C, 20 units
Step-by-step explanation:
We see that both angles QRS and QTR are 90 degrees. In addition, angles SQR and RQT are equivalent (because they're both angle Q).
By AA Similarity, we know that triangle QTR is similar to triangle QRS.
With this similarity in mind, we can look at the ratios of corresponding lengths to set up a proportion. QR from triangle QTR is the hypotenuse, and it corresponds to hypotenuse QS from triangle QRS. So, we can write the ratio x/(9 + 16) = x/25.
Now, we see that long leg QT of triangle QTR corresponds to long leg QR of triangle QRS. So, another ratio we can write is: 16/x.
Finally, we set these two ratios equal to each other:
Cross-multiplying, we get: 
.
Thus, x = 
. The answer is C, 20 units.
Hope this helps!
Let us take the second equation first
2a + 2b = 6
Dividing both sides by 2 we get
a + b = 3
a = 3 - b
Putting the value of a in the first equation we get
3a + 4b = 9
3(3 - b) + 4b = 9
9 - 3b + 4b = 9
b = 9 - 9
= 0
Now putting the value of b in the second equation we get
a + b = 3
a + 0 = 3
a = 3
So the value of the unknown variable a is 3 and the value of the unknown variable b is 0.
Answer:
The probability that it will still be working after one week is 
Step-by-step explanation:
Given :
Total number of bulbs = 25
Number of bulbs which are good condition and will function for at least 30 days = 5
Number of bulbs which are partially defective and will fail in their second day of use = 10
Number of bulbs which are totally defective and will not light up = 10
To find : What is the probability that it will still be working after one week?
Solution :
First condition is a randomly chosen bulb initially lights,
i.e. Either it is in good condition and partially defective.
Second condition is it will still be working after one week,
i.e. Bulbs which are good condition and will function for at least 30 days
So, favorable outcome is 5
The probability that it will still be working after one week is given by,
Answer:
Approximately 6.4
Step-by-step explanation:
We can use the pythagorean thereom here, that tells us (a^2)+(b^2)=c^2. C is the hypotenuse, the side opposite from the right angle, while a and b are the other sides. We can insert 5 and 4 as a and b, and solve for c
:(5^2)+(4^2)=c^2
:25+16=c^2
:41=c^2
:sqrt(41)=6.4=c (We square rooted both sides. 6.4 is only rounded to the nearest hundredths place.) Hope this helps!
