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Elina [12.6K]
3 years ago
12

Complete the square to transform the quadratic equation into the form

Mathematics
1 answer:
faltersainse [42]3 years ago
7 0

The square of a binomial is written like

(x\pm a)^2 = x^2\pm 2ax + a^2

In your case, you have 2ax=-8x, which implies a=-4

So, we want to write

(x-4)^2 = x^2-8x+16

But our left hand side is

x^2-8x-10

If we add 26 to both sides, we have

x^2 - 8x - 10 +26 = 18+26 \iff x^2-8x+16=44 \iff (x-4)^2=44

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PLEASE HELP ASAP!!!!! please help me!!
Maru [420]

Answer:

Step-by-step explanation:

A)

The parallelogram area formula is  b*h  just like a square but the height is the distance from the base to the parallel top .. not the side length.. that's the only difference.

the base is  x+4

total area is 5x+20

so use those in our formula

area = b*h

5x+20 = x+4 * h   now solve for h   (yes we'll have an x in the other side, but that's okay b/c their formulas do also)

(5x + 20) / (x+4) = h

B)

area = b*h

8x-24 = b*8

(8x-24)/8 = b

8x/8  - 24/8  = b

x -3 = b

C)

area = b * h

12x + 6y = 6 * h

(12x + 6y) / 6 = h

12x/6  + 6y/6  = h

2x + y = h

voila you're don :)

5 0
3 years ago
Make g the subject of the formula in w = 7 - sqrt g​
sukhopar [10]

Answer: g = w² - 14w + 49

w=7-\sqrt{g}\\\\=> \sqrt{g} =7-w \\\\g=(7-w)^{2}=49-14w+w^{2}

Step-by-step explanation:

3 0
3 years ago
List the angles in order from largest to smallest. A) ∠R, ∠P, ∠Q B) ∠R, ∠Q, ∠P C) ∠Q, ∠P, ∠R D) ∠Q, ∠R, ∠P
mars1129 [50]

Answer: A

Step-by-step explanation:

Angles opposite shorter sides of a triangle are smaller, and angles opposite longer sides in a triangle are larger.

8 0
2 years ago
Is 2/1/2 greater then 2/3/5
zavuch27 [327]

Answer: yes you are correct 2 /1/2 is greater

Step-by-step explanation:

7 0
2 years ago
PLS HELP ME!!! which matrix represents the system of equations below {8x-9y+13z=11
Finger [1]

Answer:

The answer is below

Step-by-step explanation:

The system of equations:

8x-9y+13z=11

-8x-5y+5z=15

3x+4y-8z=-10

The equations can be represented in matrix form as:

AX = B

X = A⁻¹B

Therefore:

\left[\begin{array}{ccc}8&-9&13\\-8&-5&5\\3&4&-8\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}11\\15\\-10\end{array}\right]\\\\\\\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}8&-9&13\\-8&-5&5\\3&4&-8\end{array}\right] ^{-1}\left[\begin{array}{c}11\\15\\-10\end{array}\right]\\

\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{19} &-\frac{1}{19}&\frac{1}{19}\\-\frac{49}{380}&-\frac{103}{380}&-\frac{36}{95} \\-\frac{17}{380}&-\frac{69}{380}&-\frac{28}{95} \end{array}\right] \left[\begin{array}{c}11\\15\\-10\end{array}\right]\\\\\\\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}-0.74\\-1.69\\0.13\end{array}\right] \\

4 0
2 years ago
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