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Orlov [11]
3 years ago
13

Help ASAP please and thank you

Mathematics
1 answer:
MArishka [77]3 years ago
4 0
The answer to the question is b
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postnew [5]
6.75 ghubvyubbbgyyvhguhb
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The range of the function f(k)=k^2+2k+1 is (25,65).what is the fuction domain
suter [353]

domain: [-9, -4] ∪ [6, 7]

 

5 0
3 years ago
The function, f(x) = –2x2 + x + 5, is in standard form. The quadratic equation is 0 = –2x2 + x + 5, where a = –2, b = 1, and c =
avanturin [10]
Given:

The function, f(x) = -2x^2 + x + 5

Quadratic equation: 0 = -2x^2 + x +5
where a = -2
            b = 1
            c = 5

The discriminate b^2 - 4ac = 41

To solve for the zeros of the quadratic function, use this formula:

x = ( -b +-√ (b^2 - 4ac) ) / 2a

x = ( 1 + √41 ) / 4  or 1.85
x = ( 1 - √41 ) / 4  or -1.35

Therefore, the zeros of the quadratic equation are 1.85 and -1.35. 
4 0
3 years ago
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Find the distance between the two points (-8,-4) (7,4)
Leya [2.2K]
You need to attach the picture
5 0
3 years ago
Find the value of x that makes:
Svetllana [295]

Answer:

correct value of x = 5

Step-by-step explanation:

\frac{ {( - 2)}^{4x + 2} }{ {( - 2)}^{2x} }  =  {( - 2)}^{12}  \\  =  \frac{ {( - 2)}^{4 \times 5 + 2} }{ { ( - 2)}^{2 \times 5 } }  \\  =  \frac{ {( - 2)}^{22} }{ { ( - 2)}^{10} }  \\  = \frac{4194304}{1024} \\  = 4096 = 4096(i.e \:  {( - 12)}^{12}   = 4096) \\

8 0
3 years ago
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