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larisa [96]
3 years ago
13

Which equations have a leading coefficient of 3 and a constant term of –2? Check all that apply.

Mathematics
2 answers:
Svetllana [295]3 years ago
8 0
Answers are: choice A, choice C, choice E

Explanation: The leading coefficient is the number that is to the left of the term with the largest exponent, which in this case is 2. The term 3x^2 is the leading term with the coefficient of 3. This is why the leading coefficient is 3 for choice A, choice C, and choice E. Choice B has a leading coefficient of -3 so we can rule that out. Choice D has a leading coefficient of 3, but the constant term is NOT -2. Instead the constant term is +2, so we can rule out choice D as well. The other choices that haven't been eliminated all have 3x^2 somewhere in them, as well as the constant term -2. The other x term isn't relevant to the restrictions placed in the instructions. 

Phoenix [80]3 years ago
5 0

Answer:

a,c,e

Step-by-step explanation:

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3 years ago
Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
Mnenie [13.5K]
One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
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the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
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D=\sqrt{100+1}
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6 0
3 years ago
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Fofino [41]

Answer:

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Step-by-step explanation:

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The slope is \frac{1}{2}

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3 years ago
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[tex]\begin{gathered} \text{The inequality is,} \\ 5
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