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3 years ago

Can someone help me with this? Plz. Solve the quadratic equation 2x^2-4x-1=0

1 answer:

6 0

You have to use the quadratic formula to solve this. The zeros of this quadratic are 2 + sqrt2/2 and 2 - sqrt2/2, which in "real" numbers is 1.707 and .2928

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3. Use the diagram 2 and 3 are ____ angles 1. complementary 2. vertical 3. right 4. supplementary

Vladimir79 [104]

2 and three are supplementary angles .

#4.

The answer is opposite angles .

Hence there is no option

Bisected angles are equal

$\\ \sf\longmapsto 5x=3x+10$

$\\ \sf\longmapsto 5x-3x=10$

$\\ \sf\longmapsto 2x=10$

$\\ \sf\longmapsto x=\dfrac{10}{2}$

$\\ \sf\longmapsto x=5$

Now

m<ABC

$\\ \sf\longmapsto 5x+3x+10$

$\\ \sf\longmapsto 8x+10$

$\\ \sf\longmapsto 8(5)+10$

$\\ \sf\longmapsto 40+10$

$\\ \sf\longmapsto 50$

5 0

3 years ago

Read 2 more answers

How do you do this function (-3p - 4)(2p + 5)

kakasveta [241]

Answer: The answer is −6p^2-23p-20

6 0

3 years ago

Use the following matrices, A, B, C and D to perform each operation.

Vinvika [58]

Step-by-step explanation:

$A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]$

$B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]$

$C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]$

$D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]$

$1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]$

$2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

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

$4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]$