So, the best way to solve this problem is by creating a equation for which we can substitute any number of days. For this problem that equation would be W=5+3(t-1)
W=number of words
T=time or number of days
-1= This is here because we are already accounting for the first day of school so we need to have 1 less day than the number of days they have been in school.
So in 20 days the students had received
5+3(19) or 62words.
Answer: 62 words
Answer:
Its degree can be at least 1970
Step-by-step explanation:
for each root of the form √q, where q is not a square, we have a root -√q. Therefore, we need to find, among the numbers below to 1000, how many sqaures there are.
Since √1000 = 31.6, we have a total of 30 squares:
2², 3², 4², ...., 30², 31²
Each square gives one root and the non squares (there are 1000-30 = 970 of them) gives 2 roots (one for them and one for the opposite). Hence the smallest degree a rational polynomial can have is
970*2 + 30 = 1970
Angle 2 is half of 115 so it would be 57.5 and angle 1 would be 122.5 degrees
Answer:
17 inches
Step-by-step explanation:
The original square had sides of 4.5 (18/4)
To get a perimeter of 10, we know that two sides would already equal 9 (4.5+4.5). That means we only have 1 inch for the other two sides, so they are each 1/2 inch.
That leaves the second rectangle with a length and width of 4.5 and 4.
2(4.5) + 2(4) =
9+8=
17
Answer: 15 units .
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In this case, a square, the two sides of the square (forming a right triangle) are equal), and the "diagonal" forming is the hypotenuse of the right triangle.
In these cases, the measurements of the angles of the right triangle are "45, 45, 90" ; and the measurements of the sides are: "a, a, a√2" ; in which "a√2" is the hypotenuse.
We are given: "15√2" is the hypotenuse" ; and we are given that this is a right triangle of a square with a diagonal length (i.e. "hypotenuse" of "15√2" ; so the measure of the side of the "square" (and other two sides of the triangle formed) is: 15 units. (i.e., 15, 15, 15√2 ).
