Letter C is correct.
When the function is even, (all powers are even) both ends will go in the same direction.
In this case since the leading coefficient is negative, both ends will go down.

3 0

3 years ago

If you remove the last digit (one’s place) from a 4-digit whole number, the resulting number is a factor of the 4-digit number.

shutvik [7]

Answer:

900

Step-by-step explanation:

We assume that your 4-digit number must be in the range 1000 to 9999. Clearly, any number ending in zero will meet your requirement:

1000/100 = 10

3890/389 = 10

However, the requirement cannot be met when the 1s digit is other than zero.

For some 3-digit number N and some 1s digit x, the 4-digit number will be

4-digit number: 10N+x

Dividing this by N will give ...

(10N+x)/N = 10 remainder x

N will only be a factor of 10N+x when x=0.

So, there are 900 4-digit numbers that meet your requirement. They range from 1000 to 9990.

6 0

3 years ago

Determine the zeroes of the polynomial

Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

$(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)} \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)} -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\$

$\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}$

8 0

3 years ago

What is the best choice for the equation of the line of best fit shown?

vitfil [10]

Answer:

What is the best choice for the equation of the line of best fit shown?

b) y = -1.5x + 23

What would the value of y be for a point at x = 8?

b) 11

Did this on EDGE!

8 0

2 years ago