Answer:
Find the sum of the series
∑(4x−5)
such that
1≤x≤7
.
answer is 77
Step-by-step explanation:
the second 3 (from the left) is in the hundreds place so we look at the 1st 7 (from the left) to determine how to round
since 7 is 5 or greater, we round the 3 to a 4
3400
Choice D
Answer:
the answer is 3
Step-by-step explanation:
(-5x-3)(-5x+3)=(5x)^2 - (3)^2
A difference of squares is the difference of two squared terms.
We have
a^2 - b^2
We can factorize the difference of squared terms like this:
a^2-b^2 = (a+b)(a-b)
We have (-5x-3)(-5x+3)
Lets prove it:
(-5x-3)(-5x+3) = (-5x*-5x)+(-5x*3)+(-3*-5x)+(-3*3)
(-5x-3)(-5x+3) = 25x^2+(-15x)+(15x)+(-9)
(-5x-3)(-5x+3) = 25x^2-15x+15x-9
(-5x-3)(-5x+3) = 25x^2-9
(-5x-3)(-5x+3) = (5x)^2 - (3)^2
Answer:
x = 6, x = - 6
Step-by-step explanation:
Given
y = x² - 36
To find the zeros let y = 0, that is
x² - 36 = 0 ← x² - 36 is a difference of squares and factors in general as
a² - b² = (a - b)(a + b), thus
x² - 36 = 0
x² - 6² = 0
(x - 6)(x + 6) = 0 ← in factored form
Equate each factor to zero and solve for x
x - 6 = 0 ⇒ x = 6
x + 6 = 0 ⇒ x = - 6
