Answer:
71.6°
Step-by-step explanation:
=210s÷60s(1 minutes)
=3.5mins
=3.5mins×3%(=1min)
=10.5%
=100%-10.5%
=89.5%
=89.5%/100×80°
=71.6°
Hope this will help ya!
The right answer for the question that is being asked and shown above is that: "13 ft."An iron tank is constructed in the form of an isosceles trapezoid. Each base angle measures 105°, the length of the base is 10 feet, and each of the <span>nonparallel sides is 12 feet long. The length of a diagonal brace to the nearest whole number is 13 ft</span>
Answer:
The correct option is D. Discontinuity at (1, 7), zero at (negative four thirds, 0)
Step-by-step explanation:
To find the point of discontinuity :
Put the denominator equal to 0
⇒ x - 1 = 0
⇒ x = 1
Also, if the factor (x - 1) gets cancel, then it becomes a hole rather than a asymptote , ⇒ y = 3x + 4 at x = 1
⇒ y = 7
So, Point of discontinuity : (1, 7)
And the zero is : after cancelling the factor (x - 1) put the remaining factor = 0
⇒ 3x + 4 = 0
⇒ 3x = -4
⇒ x = negative four thirds ( zero of the function)
Therefore, The correct option is D. Discontinuity at (1, 7), zero at (negative four thirds, 0)
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
