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Tcecarenko [31]
1 year ago
6

Factor this polynomial completely x2 - x - 20

Mathematics
2 answers:
dimulka [17.4K]1 year ago
5 0

Answer:

(x +5) (x -4)

Step-by-step explanation:

Factors of 20:

1,20

2,10

4,5

4 and 5 combine to create 1

+5 -4 = 1

(x +5) (x -4)

mafiozo [28]1 year ago
3 0

Answer:

(x+4)(x-5)

Step-by-step explanation:

We can look at it by thinking of two numbers, that add up to -1 and multiply to -20. We can first do this by looking at factors of -20, which are 1, -20, -1, 20, 2, -10, -2, 10, and so on. After looking through it, we can notice that -5 and 4 added equal -1, and when multiplied equal to -20, so they work. So, we factor to (x+4)(x-5)

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Use triangles ABC and A'B'C' for Exercise. Use the coordinates to find the lengths of the sides.
mafiozo [28]

Answer:

AB = 2

BC = 2

A'B' = 6

B'C' = 6

Step-by-step explanation:

5 0
3 years ago
Solve the following for the variable x.<br> 3x - 16 = 38
shusha [124]

Answer: x= 18

Step-by-step explanation:

3x - 16 = 38

38 + 16 = 54

3x/3 = 54/3

X= 18

3 0
3 years ago
Read 2 more answers
I’m timed !!! Please help
Doss [256]
1,450 because you would plug 75 in for x
f(x)= 20(75)-50
4 0
3 years ago
The value of a car depreciates by 40% each year. At the end of 2007 the value of the car was £3600 Work out the value of the car
natima [27]

Answer: $6,000

Step-by-step explanation: If set up correctly, the value of the car at the end of 2006 times 0.6 (because the car is losing 0.4 of its value, so it keeps 0.6 of its value) should get you the value of the car at the end of 2007.

The equation should look like this:

(value of the car at the end of 2006) x 0.6 = (value of the car at the end of 2007)

Then you substitute: (value at the end of 2006) x 0.6 = 3600

Then divide: (value at the end of 2006) = 0.6 ÷ 3600

Then substitute again: (value at the end of 2006) = 6,000

Hope this helped you! Feel free to ask me any questions!

3 0
3 years ago
How do you find absolute extrema for a function?<br> f(x)= (8+x)/(8-x); Interval of [4,6]
Mkey [24]
\bf f(x)=\cfrac{8+x}{8-x}\implies \cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{1(8-x)-(8+x)(-1)}{(8-x)^2}}\implies \cfrac{dy}{dx}=\cfrac{16}{(8-x)^2}

now, we get critical points from zeroing out the derivative, and also from zeroing out the denominator, but those at the denominator are critical points where the function is not differentiable, namely a sharp spike or cusp or an asymptote.

so, from zeroing out the derivative we get no critical points there, from the denominator we get x = 8, but can't use it because f(x) is undefined.

therefore, we settle for the endpoints, 4 and 6,

f(4) =3    and      f(6) = 7

doing a first-derivative test, we see the slope just goes up at both points and in between, but the highest is f(6), so the absolute maximum is there, while we can take say f(4) as the only minimum and therefore the absolute minumum as well.
3 0
3 years ago
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