PLEASE ANSWER URGENT WILL GIVE BRAINLIEST AND 20 POINTS.

Find the vertex of this parabola:<br> y = -2x2 + 4x - 10

amm1812

The parabola equation in its vertex form is y = a(x-h)² + k , where:
a is the same as the a coefficient in the standard form;
h is the x-coordinate of the parabola vertex; and.
k is the y-coordinate of the parabola vertex.
that’s how you find it

3 0

3 years ago

A bird traveled 4 miles north before stopping to rest. The bird then flew 2,640 yards south in search of food. Finally, it flew

ra1l [238]

7 and 1/2 mile because 2,640=1.5 miles
4-1.5=2.5+5=7.5

8 0

3 years ago

How to find the common denominator of two numbers?

Snowcat [4.5K]

Find the number that both can be divided by i.e. the common denominator of 56/8 is 8 because 56 is divisible by 8 and 8 is also divisible by 8.
so by finding the common denominator and dividing the fraction, the equation would look like this: 7/1

3 0

3 years ago

There are eight different jobs in a printer queue. Each job has a distinct tag which is a string of three upper case letters. Th

Vikentia [17]

Answer:

a. 40320 ways

b. 10080 ways

c. 25200 ways

d. 10080 ways

e. 10080 ways

Step-by-step explanation:

There are 8 different jobs in a printer queue.

a. They can be arranged in the queue in 8! ways.

No. of ways to arrange the 8 jobs = 8!

= 8*7*6*5*4*3*2*1

No. of ways to arrange the 8 jobs = 40320 ways

b. USU comes immediately before CDP. This means that these two jobs must be one after the other. They can be arranged in 2! ways. Consider both of them as one unit. The remaining 6 together with both these jobs can be arranged in 7! ways. So,

No. of ways to arrange the 8 jobs if USU comes immediately before CDP

= 2! * 7!

= 2*1 * 7*6*5*4*3*2*1

= 10080 ways

c. First consider a gap of 1 space between the two jobs USU and CDP. One case can be that USU comes at the first place and CDP at the third place. The remaining 6 jobs can be arranged in 6! ways. Another case can be when USU comes at the second place and CDP at the fourth. This will go on until CDP is at the last place. So, we will have 5 such cases.

The no. of ways USU and CDP can be arranged with a gap of one space is:

6! * 6 = 4320

Then, with a gap of two spaces, USU can come at the first place and CDP at the fourth. This will go on until CDP is at the last place and USU at the sixth. So there will be 5 cases. No. of ways the rest of the jobs can be arranged is 6! and the total no. of ways in which USU and CDP can be arranged with a space of two is: 5 * 6! = 3600

Then, with a gap of three spaces, USU will come at the first place and CDP at the fifth. We will have four such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 4 * 6!

Then, with a gap of four spaces, USU will come at the first place and CDP at the sixth. We will have three such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 3 * 6!

Then, with a gap of five spaces, USU will come at the first place and CDP at the seventh. We will have two such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 2 * 6!

Finally, with a gap of 6 spaces, USU at first place and CDP at the last, we can arrange the rest of the jobs in 6! ways.

So, total no. of different ways to arrange the jobs such that USU comes before CDP = 10080 + 6*6! + 5*6! + 4*6! + 3*6! + 2*6! + 1*6!

= 10080 + 4320 + 3600 + 2880 + 2160 + 1440 + 720

= 25200 ways

d. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways. Similarly, if LPW comes last, the remaining 7 jobs can be arranged in 7! ways. so, total no. of different ways in which the eight jobs can be arranged is 7! + 7! = 10080 ways

e. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways in the queue. Similarly, if QKJ comes second-to-last then also the jobs can be arranged in the queue in 7! ways. So, total no. of ways to arrange the jobs in the queue is 7! + 7! = 10080 ways

5 0

4 years ago

How do you solve this??? I need it fast please help with steps.<br> (x+3)(x+1)^2(x−4)>0

scZoUnD [109]

Answer:

x < -3 or x > 4

Step-by-step explanation:

The product of the factors is 4th-degree, and the leading coefficient is 1 (positive).

The factors tell you the following about the zeros:

x = -3 . . . sign change from + to -

x = -1 . . . graph touches, but no sign change. - on either side of x=-1

x = 4 . . . sign change from - to +

__

The function will be positive for x < -3 and for x > +4.

_____

<em>Additional comments</em>

The value of the function is zero when any of its factors is zero. The value of a factor changes sign when the value of x changes from one side of the 0 to the other. For example, here, we have x+1 as a factor, so x=-1 is a zero. When x = -0.9, the factor is positive (-0.9 +1 = 0.1). When x = -1.1, the factor is negative (-1.1 +1 = -0.1). If this factor were to the first power, it would cause the value of the function to change sign at x=-1.

However, the factor (x+1) has an even power: (x+1)^2. That means when the factor is negative, its square is positive, and when the factor is positive, its square is positive. In short, the factor (x+1)^2 causes the function to be zero at x=-1, but does not make the function change sign there.

The product of all of these factors will result in a polynomial with x^4 as the highest-degree term. That means the function is of even degree. The leading coefficient is 1, so is positive, and the function will generally have a U-shape. The left-most (odd-degree) zero will be where the function changes sign from positive to negative. The right-most (odd-degree) zero will be where the function changes sign from negative to positive. Those zeros are x=-3 and x=+4, respectively. There are no places between these where the function changes sign, so these zeros are the ends of the regions where the function is positive (>0).

3 0

2 years ago