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Komok [63]
2 years ago
12

Guys pls help with all the steps ​

Mathematics
1 answer:
Gwar [14]2 years ago
5 0

Step-by-step explanation:

To solve both questions you will have to use the Pythagoras theorem

{c}^{2}  =  {a}^{2}  +  {b}^{2}

c being the hypotenuse

so to find AB we say

{ap}^{2}  =  {pb}^{2}  +  {ab}^{2}

To make AB the subject of the formula, we take PB over. Therefore:

{ab}^{2}  =  {ap}^{2}  -  {pb}^{2}

now we just substitute

{ab}^{2}  =  {13}^{2}  -  {5}^{2}

{ab}^{2}  = 169 - 25

{ab}^{2}  = 144

square root both sides of the equation

ab = 12

We then use the same formula to find AQ(which is a hypotenuse)

{aq}^{2}  =  {ab}^{2}  +   {bq}^{2}

{aq}^{2}  =  {12}^{2}  +  {9}^{2}

{aq}^{2}  = 144 + 81

{aq}^{2}  = 225

square root both sides

aq = 15

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What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



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