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vaieri [72.5K]
3 years ago
11

Which equation would you use to solve for the value of a?

Mathematics
2 answers:
PIT_PIT [208]3 years ago
8 0

Step-by-step explanation:

Solution,

In above figure,

a+a+76°=180° (being straight angles)

2a+76°=180°

2a=180°-76°

2a=104°

a=104°/2

a=52°

Hence, I will use option c. 2a+76°=180° for calculation.

Sveta_85 [38]3 years ago
5 0

Answer:

i use option c to solve for the value of a.

2a+76°=180°

2a=180°-76°

2a=104°

a=104°/2

a=52°

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8 0
3 years ago
Rewrite the function y= x^2 + 14x + 4 in vertex form
Step2247 [10]
Vertex form is basically commplete the square
y=a(x-h)^2+k


y=x^2+14x+4
take 1/2 of 14 and square it, (7^2=49)
add that and its negative to right side
y=x^2+14x+49-49+4
factor perfect squaer
y=(x+7)^2-49+4
y=(x+7)^2-45

answer is A
4 0
3 years ago
Please help Need help
Anettt [7]
This problem can be readily solved if we are familiar with the point-slope form of straight lines:
y-y0=m(x-x0) ...................................(1)
where 
m=slope of line
(x0,y0) is a point through which the line passes.

We know that the line passes through A(3,-6), B(1,2)

All options have a slope of -4, so that should not be a problem.  In fact, if we check the slope=(yb-ya)/(xb-xa), we do find that the slope m=-4.

So we can check which line passes through which point:

a. y+6=-4(x-3)
Rearrange to the form of equation (1) above,
y-(-6)=-4(x-3)  means that line passes through A(3,-6) => ok

b. y-1=-4(x-2) means line passes through (2,1), which is neither A nor B
   ****** this equation is not the line passing through A & B *****

c. y=-4x+6  subtract 2 from both sides (to make the y-coordinate 2)
   y-2 = -4x+4, rearrange
   y-2 = -4(x-1)  
   which means that it passes through B(1,2), so ok

d. y-2=-4(x-1)
   this is the same as the previous equation, so it passes through B(1,2), 
   this equation is ok.

Answer: the equation y-1=-4(x-2) does NOT pass through both A and B.
   
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3 years ago
Find the number of two-letter permutations of the letters.<br><br> Q, I, E, R, T, Y, U
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5 0
3 years ago
What is the coordinates of the center of an ellipse defined by the equation 16x^2 + 25y^2 + 160x - 200y + 400 = 0 ? Please give
pickupchik [31]

16x^2 + 25y^2 + 160x - 200y + 400 = 0     Rearrange and regroup.

(16x^2 + 160x) + (25y^2 - 200y ) = 0-400.     Group the xs together and the ys together.

16(X^2 + 10x) + 25(y^2-8y) = -400.     Factorising.

We are going to use completing the square method.

Coefficient of x in the first expression = 10.

Half of it = 1/2 * 10 = 5. (Note this value)

Square it = 5^2  = 25.     (Note this value)


Coefficient of y in the second expression = -8.

Half of it = 1/2 * -8 = -4. (Note this value)

Square it = (-4)^2  = 16. (Note this value)


We are going to carry out a manipulation of completing the square with the values

25 and 16.  By adding and substracting it.


16(X^2 + 10x) + 25(y^2-8y) = -400

16(X^2 + 10x + 25 -25) + 25(y^2-8y + 16 -16) = -400

Note that +25 - 25 = 0.    +16 -16 = 0. So the equation is not altered.

16(X^2 + 10x + 25) -16(25) + 25(y^2-8y + 16) -25(16) = -400


16(X^2 + 10x + 25) + 25(y^2-8y + 16)  = -400 +16(25) + 25(16)    Transferring the terms -16(25) and -25(16)

to other side of equation.  And 16*25 = 400


16(X^2 + 10x + 25) + 25(y^2-8y + 16)  = 25(16)


16(X^2 + 10x + 25) + 25(y^2-8y + 16)  = 400

We now complete the square by using the value when coefficient was halved.


16(x-5)^2 + 25(y-4)^2  = 400

Divide both sides of the equation by 400


(16(x-5)^2)/400 + (25(y-4)^2)/400  = 400/400              Note also that, 16*25 = 400.


((x-5)^2)/25 + ((y-4)^2)/16  = 1

((x-5)^2)/(5^2) + ((y-4)^2)/(4^2)  = 1


Comparing to the general format of an ellipse.

((x-h)^2)/(a^2) + ((y-k)^2)/(b^2)  = 1


Coordinates of the center = (h,k).

Comparing   with above   (x-5) = (x - h) , h = 5.

Comparing   with above   (y-k) = (y - k) , k = 4.

Therefore center = (h,k) = (5,4).

Sorry the answer came a little late.  Cheers.

3 0
3 years ago
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