Please help!! Evaluate the expression for h = 14. -51 + 13(-4h + 57)^2

1 answer:

Anna [14]3 years ago

8 0

Answer:

-38

Step-by-step explanation:

-51 + 13(-4h + 57)^2

-51 + 13 [-4(14) + 57]²

-51 + 13 (-56 + 57)²

-51 + 13(1)²

-51 + 13(1)

-51 + 13

= -38

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pogonyaev

Steps to solve:

g(x) = 3 - 2x when g(-1)

~Substitute

g(-1) = 3 - 2(-1)

~Simplify

g(-1) = 3 - (-2)

~Simplify

g(-1) = 3 + 2

~Add

g(-1) = 5

Best of Luck!

3 0

4 years ago

Irina18 [472]

Answer:

$(y-7)^2+(x-2)^2=16$

and

$(x+2)^2+(y-15)^2 = 9$

Step-by-step explanation:

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where the coordinate (h,k) is the center of the circle.

Second Problem:

We can start with the second problem which uses this info very easily.
(h,k) in this problem is (-2,15) simply plug these into the equation. $(x--2)^2+(y-15)^2=r^2$ .
We can also add the radius 3 and square it so it becomes 9. The equation.
This simplifies to $(x+2)^2+(y-15)^2 = 9$ .

First Problem:

The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
$y^2-14y+x^2-4x+37=0$ first subtract 37 from both sides so the equation is now $y^2-14y+x^2-4x=-37$ .
$y^2-14y+x^2-4x+37=0$ by adding $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ which equals 49 and 4.
Add 49 and 4 to both sides and the equation is now: $y^2-14y+49+x^2-4x+4=-37+49+4$ You can simplify the y and x portions of the equations into the perfect squares or factored form $(y-7)^2$ and $(x-2)^2$ .
Finally put the whole thing together. $(y-7)^2+(x-2)^2=16$ .