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Gala2k [10]
4 years ago
14

The number 28 please somebody help me...

Mathematics
1 answer:
MArishka [77]4 years ago
8 0
Where is the persons info
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Follow the process of completing the square to solve 2x2 + 8x - 12 = 0.
Rom4ik [11]
Please:  Use "^" to denote exponentiation:  <span>2x^2 + 8x - 12 = 0

Reduce this by div. every term by 2:             </span><span>x^2 + 4x - 6 = 0

Here a=1, b=4 and c = -6.  Square half of b, obtaining (4/2)^2 = 4, and add, and then subtract, this 4 to x^2 + 4x - 6:

</span> x^2 + 4x +4  - 4 - 6 = 0.  Rewrite the square as (x+2)^2, obtaining new equation

(x+2)^2 = 10.  Take the sqrt of both sides:   x+2 = plus or minus sqrt(10).

Finally, solve for x:  x = -2 plus or minus sqrt(10).



8 0
4 years ago
I need to find sec(theta)= 2(sqrt3)/3
Sloan [31]

Answer: 30° and 330°

<u>Step-by-step explanation:</u>

sec θ = \frac{2\sqrt{3}}{3}

sec θ = \frac{1}{cos(\theta)}

\frac{1}{cos(\theta)} = \frac{2\sqrt{3}}{3}

cos θ = \frac{3}{2\sqrt{3}}

cos θ = \frac{3}{2\sqrt{3}}(\frac{\sqrt{3}}{\sqrt{3}})

cos θ = \frac{3\sqrt{3}}{3*2}

cos θ = \frac{\sqrt{3}}{2}

Look at the Unit Circle to see that cos = \frac{\sqrt{3}}{2} at 30° and 330°  ( which is equivalent to π/6 and 11π/6)


4 0
3 years ago
Read 2 more answers
Andrew is showing his work in simplifying −4.5 + 4.2 + 5.6 − 7.3. Identify any errors in his work or in his reasoning.
julia-pushkina [17]

Answer:

  • D. Step 5

Step-by-step explanation:

Andrew made a mistake at last step.

<u>It should be:</u>

  • -11.8 + 9.8 = 9.8 - 11.8 = -2 but not -21.6
6 0
3 years ago
PLEASE HELP ASAP WILL GIVE BRAINLEIST
OleMash [197]
Angle Q and angle S equal 61
6 0
3 years ago
Use the following matrices, A, B, C and D to perform each operation.
wolverine [178]

Step-by-step explanation:

\bold{40.}\\\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right] +\left[\begin{array}{ccc}4&1\\6&0\end{array}\right] =\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&1\\11&7\end{array}\right]

\bold{41.}\\\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]

\bold{42.}\\\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]

\bold{43.}\\\\CD=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\\\=\left[\begin{array}{ccc}4+0+1&-6-6+4&-8+3-1\\2+0+12&-3+0+16&-4+0-4\end{array}\right]\\\\=\left[\begin{array}{ccc}5&-8&-6\\14&13&-8\end{array}\right]

\bold{44.}\\\\2D+3C=2\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]+3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\\\\=\left[\begin{array}{ccc}(2)(-2)&(2)(3)&(2)(4)\\(2)(0)&(2)(-2)&(2)(1)\\(2)(3)&(2)(4)&(2)(-1)\end{array}\right]+\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]\\\\=\left[\begin{array}{ccc}-4&6&8\\0&-4&2\\6&8&-2\end{array}\right]+\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]

\large\bold{You\ can\ not\ add\ matrices\ of\ different\ dimensions!!!}

4 0
3 years ago
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