Answer:
Step-by-step explanation:
(X+4)1999=1
Open the bracket by multiplying 1999 by the values in the bracket
1999x + 7996 = 1
Collect like terms
1999x = 1-7996
1999x = -7995
Divide both sides by 1999
X = -3.99
What a delightful little problem ! (Partly because I could see
right away how to do it, and had the answer in a few minutes,
after a lot of impressive-looking algebra on my scratch-paper.)
Three consecutive integers are . . . x, x+1, and x+2
The smallest two are . . . x and x+1
Their product is . . . . . x(x+1)
5 times the largest one is . . . 5(x+2)
5 less than that is . . . . . . 5(x+2)-5
Now, the conditions of the problem say that <u>x (x + 1) = 5 (x+2) - 5</u>
THAT's the equation we have to solve, to find 'x' .
Eliminate parentheses: x² + x = 5x + 10 - 5
Combine like terms: x² + x = 5x + 5
Subtract 5x from each side: x² - 4x = 5
Subtract 5 from each side: <u>x² - 4x - 5 = 0</u>
You could solve that by factoring it, or use the quadratic equation.
Factored, it says that (x + 1) (x - 5) = 0
From which <em>x = -1</em>
and <em>x = +5</em>
We only want the positive results, so our three consecutive integers are
5, 6, and 7 .
To answer the question, the smallest one is <em><u>5 </u></em>.
<u>Check</u>:
5 x 6 ? = ? (7 x 5) - 5
30 ? = ? (35) - 5
30 = 30
yay !
Let f(x) = x² + 6x²-x+ 5 then ,
number to be added be P
then,
f(x) = x² + 6x²-x+ 5 +P
According to the qn,
(x+3) is exactly divisible by zero then,
R=0
comparing .. we get a= -3
now by remainder theorm
R=f(a)
0=f(-3)
0=(-3)² + 6(-3)²-(-3)+ 5 + P
0= 9 + 54 + 3 + 5 + P
-71=P
therefore, -71 should be added.
Hope you understand
X(4*1) because I’m just doing what I’m told.
Answer:
A. Less than zero
Step-by-step explanation:
So If you graph the function, you'll see that it never actually crosses the y axis into the third or fourth quadrant, the quadrants where all y values become negative.
The answer A is also correct on Edge
Have a great day dude
