How do you solve for a²+b²=289 when b+a=23

1 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

​a

​​ x

​b

​​ =x

​a+b

​​  

{a}^{2}+bb=289a

​2

​​ +bb=289

2 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

​a

​​ x

​b

​​ =x

​a+b

​​  

{a}^{2}+{b}^{2}=289a

​2

​​ +b

​2

​​ =289

3 Subtract {b}^{2}b

​2

​​  from both sides

{a}^{2}=289-{b}^{2}a

​2

​​ =289−b

​2

​​  

4 Take the square root of both sides

a=\pm \sqrt{289-{b}^{2}}a=±√

​289−b

​2

​​  

​

​​  

5 Rewrite 289-{b}^{2}289−b

​2

​​  in the form {a}^{2}-{b}^{2}a

​2

​​ −b

​2

​​ , where a=17a=17 and b=bb=b

a=\pm \sqrt{{17}^{2}-{b}^{2}}a=±√

​17

​2

​​ −b

​2

​​

6 Use Difference of Squares: {a}^{2}-{b}^{2}=(a+b)(a-b)a

​2

​​ −b

​2

​​ =(a+b)(a−b)

a=\pm \sqrt{(17+b)(17-b)}a=±√

​(17+b)(17−b)

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step-by-step explanation:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."