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Anna11 [10]
2 years ago
10

ANSWER IT PLEASEEEEEEEEEE THANK YOU SO MUCH HAVE A NICE DAY !!!!

Mathematics
1 answer:
Sonbull [250]2 years ago
3 0

Answer:

56 feet, I think.

Step-by-step explanation:

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Which equation is the equation of a line that is perpendicular to x + 2y = 16?
gogolik [260]

Answer:

For the equation given:

x + 2y = 16 \\ 2y =  - x + 16 \\ y =   - \frac{1}{2} x + 8

for perpendicular lines, the relationship between their slopes is given by:

m _{1} \times m _{2} =  - 1

m1 is slope for one line, and m2 is slope for another line.

-  \frac{1}{2}  \times m_{2} =  - 1 \\  \\ m_{2}  =  - 1 \times  - 2 \\ { \underline{ \:  \: m_{2} = 2 \:  \: }}

Therefore, equation is;

{ \boxed{ \boxed{y = 2x - 2}}}

4 0
3 years ago
Select the correct answer.<br> Which graph shows a function and its inverse?
labwork [276]

Answer:

See below

Step-by-step explanation:

If a function is bijective and 1-to-1, then it will have an inverse function. Consequentially, they will be symmetrical about the line y=x, which is a diagonal line passing through the origin at a 45 degree angle.

None of the graphs look correct though, but it also seems that some options are cut out, so make sure to choose the correct graph given the characteristics I've previously described.

6 0
2 years ago
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Which is the product of 4x^5-6x^4+4x^3-6x^2
Elanso [62]

Answer:

2 x^2 (2 x - 3) (x^2 + 1)

Step-by-step explanation:

Factor the following:

4 x^5 - 6 x^4 + 4 x^3 - 6 x^2

Factor 2 x^2 out of 4 x^5 - 6 x^4 + 4 x^3 - 6 x^2:

2 x^2 (2 x^3 - 3 x^2 + 2 x - 3)

Factor terms by grouping. 2 x^3 - 3 x^2 + 2 x - 3 = (2 x^3 - 3 x^2) + (2 x - 3) = x^2 (2 x - 3) + (2 x - 3):

2 x^2 x^2 (2 x - 3) + (2 x - 3)

Factor 2 x - 3 from x^2 (2 x - 3) + (2 x - 3):

Answer:  2 x^2 (2 x - 3) (x^2 + 1)

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4 years ago
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Please answer the question please. The question is down below.
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Jaiosofopfpfpspappapp
4 0
3 years ago
If I have 297 dogs but 5 wolves came and toke 131. How many are left?
spin [16.1K]
166 I guess you just subtract the numbers 297-131= Or there is something am I missing on here
8 0
3 years ago
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