Answer:
Irreducible factorization
Step-by-step explanation:
<span><span>y = 2 + 2sec(2x)
The upper part of the range will be when the secant has the smallest
positive value up to infinity.
The smallest positive value of the secant is 1
So the minimum of the upper part of the range of
y = 2 + 2sec(2x) is 2 + 2(1) = 2 + 2 = 4
So the upper part of the range is [4, )
The lower part of the range will be from negative infinity
up to when the secant has the largest negative value.
The largest negative value of the secant is -1
So the maximum of the lower part of the range of
y = 2 + 2sec(2x) is 2 + 2(-1) = 2 - 2 = 0
So the lower part of the range is (, 0].
Therefore the range is (, 0] U [4, )
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Answer:
0, 1, 2
Step-by-step explanation:
Euclid's division Lemma states that for any two positive integers ‘a’ and ‘b’ there exist two unique whole numbers ‘q’ and ‘r’ such that , a = bq + r, where 0≤ r < b.
Here, a= Dividend, b= Divisor, q= quotient and r = Remainder.
According to Euclid's division lemma a 3q+r, where 0≤r≤3 and r is an integer.
Therefore, the values of r can be 0, 1 or 2.
Y÷6=29
y=29 x 6
y=174
Hope I can help u
Answer:
The two numbers are 10 and 6.
Step-by-step explanation:
Let's begin by calling these two numbers x and y, and setting up a system of equations.
x-y=4
x*y=60
Now, you can rearrange the first equation to find the value of x expressed through y.
x=y+4
Now, you can substitute this into the second equation.
(y+4)*y=60
y^2+4y=60
y^2+4y-60=0
(y-6)(y+10)=0
y=6
x=6+4=10
Hope this helps!
