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Zanzabum
3 years ago
13

I need help with my ADD and subtraction fraction ​

Mathematics
1 answer:
kotegsom [21]3 years ago
4 0

Answer:

What do you need help with?

Step-by-step explanation:

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What are two other ways to name plane C?
svetoff [14.1K]
Answer: Other names for plane C can be plane EFG or plane EBF hope that helps (:
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3 years ago
Find f(-2) for the function f(x)= 3x2-2x+7
4vir4ik [10]

Answer:

f( - 2) = 23

Step-by-step explanation:

f(x) = 3 {x}^{2}  - 2x + 7 \\ plug \: x =  - 2 \\ f( - 2) = 3 {( - 2)}^{2}  - 2( - 2)+ 7 \\  = 3 \times 4 + 4 + 7 \\  = 12 + 11 \\  \huge \red{ \boxed{ f( - 2) = 23}}

4 0
3 years ago
PLEASE HELPPPP (with steps)
yan [13]

Answer:

the answer is x=-25/56

Step-by-step explanation:

2 4/5x=-1 1/4

14/5x=-5/4

x=-25/56

7 0
3 years ago
HI I NEED HELP IVE BEEN DOING HOMEWORK FOR 6 HOURS ummmmm anyways, how do i express this (y^3•y^6 - in picture) in exponential f
LUCKY_DIMON [66]

y^3 = y * y * y

y^6 = y * y * y * y * y * y

6 0
3 years ago
Read 2 more answers
What are the zeros of the quadratic function f(x) = 6x2 – 24x + 1?<br><br> fast please
xxTIMURxx [149]

Answer:      6(x²-4x+4-4)+1=0, 6(x-2)²-24+1=0, 6(x-2)²=23, x-2=±√(23/6), x=2±√(23/6)=2±1.95789, so x=3.95789 or 0.04211 approx. these are the zeros.

step by step explanation:

\boxed{\boxed{\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}}}

Solution-

The quadratic function is,

6x^2-24x + 1

a = 6, b = -24, c = 1

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

=\dfrac{-(-24)\pm \sqrt{-24^2-4\cdot 6\cdot 1}}{2\cdot 6}

=\dfrac{24\pm \sqrt{576-24}}{12}

=\dfrac{24\pm \sqrt{552}}{12}

=\dfrac{24\pm 2\sqrt{138}}{12}

=\dfrac{12\pm \sqrt{138}}{6}

=\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}

4 0
4 years ago
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