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3 years ago

Simplify the following expression. one-fourth(16 – 12x)

Mathematics

1 answer:

Musya8 [376]3 years ago

5 0

The answer to the question is 4-3x

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galina1969 [7]

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3 years ago

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2x2=<br><br> what is the answer to this question?

posledela

Answer:

2×2=4

Step-by-step explanation:

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3 years ago

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Jim's uncle is 4 times as old as his nephew jim. in 10 years, jim's uncle's age will be 20 years more than twice jim's age. how

xxTIMURxx [149]

Jim is currently 15 years old.

5 0

3 years ago

The endpoints of a diameter of a circle are (6,2) and (−2,5) . What is the standard form of the equation of this circle?

Paul [167]

Answer:

The standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$

Step-by-step explanation:

Given that the end points of a diameter of a circle are (6,2) and (-2,5);

Now to find the standard form of the equation of this circle:

The center is (h,k) of the circle is the midpoint of the given diameter

midpoint formula is $M=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

Let $(x_{1},y_{1})$ and $(x_{2},y_{2})$ be the given points (6,2) and (-2,5) respectively.

$M=(\frac{6-2}{2},\frac{2+5}{2})$

$M=(\frac{4}{2},\frac{7}{2})$

$M=(2,\frac{7}{2})$

Therefore the center (h,k) is $(2,\frac{7}{2})$

now to find the radius:

The diameter is the distance between the given points (6,2) and (-2,5)

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$=\sqrt{(-2-6)^2+(5-2)^2}$

$=\sqrt{(-8)^2+(3)^2}$

$=\sqrt{64+9}$

$=\sqrt{73}$

Therefore the radius is $\sqrt{73}$

i.e., $r=\sqrt{73}$

Therefore the standard form of the circle is

$(x-h)^2+(y-k)^2=r^2$

Now substituting the center and radiuswe get

$(x-2)^2+(y-\frac{7}{2})^2=(\sqrt{73})^2$

$(x-2)^2+(y-\frac{7}{2})^2=73$

Therefore the standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$

5 0

3 years ago

What is g(–3) in the function g(x) = 4x2 + 3?

EastWind [94]

Answer:

x is equal to -3. So it would be g(-3)=4(-3)+3 and the solve that.

Step-by-step explanation:

Hope this is helpful

4 0

3 years ago