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skelet666 [1.2K]
3 years ago
8

I need some help please

Mathematics
1 answer:
dimulka [17.4K]3 years ago
4 0
I think the answer is the third one, but idk.
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If A and B are the points (5,3) and (15,-7) respectively. Find the coordinates of the point which divides AB externally in the r
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Grace paid $1.70 for 5 ounces of candy. How much did she pay for 24 ounces?
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A square has a side length of 9 feet. What is the area of the square
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Step-by-step explanation:

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3 years ago
The graph below represents a population
Nataliya [291]

Answer:

Average rate of  change for the function for the interval (6, 12] is 500 people per year.

Option A is correct.

Step-by-step explanation:

We need to find the average rate of  change for the function for the interval

(6, 12]

The formula used to calculate Average rate of change is:

Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}

We are given a=6 and b=12

Looking at the graph we can see that when x=6 y= 3000 so, f(a)=3000

and when x=12, y=6000 so, f(b)=6000

Putting values in formula and finding Average rate of change:

Average \ rate \ of \ change=\frac{f(b)-f(a)}{b-a}\\Average \ rate \ of \ change=\frac{6000-3000}{12-6}\\Average \ rate \ of \ change=\frac{3000}{6}\\Average \ rate \ of \ change=500

So, average rate of  change for the function for the interval (6, 12] is 500 people per year.

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6 0
3 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

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Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

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D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

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−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

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For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
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