What are the solutions to the equation 10=490t^2 ?

The solutions to the equation 10=490t^2 are t= 1/7 ,t=-1/7

Find the zeros of 10=490t^2

The zeros of 10=490t^2 are t= 1/7 ,t=-1/7

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10 = 490 t^{2}

t = \frac{1}{7}, t = - \frac{1}{7}

Decimal Notation

t = 0.14285 \dots, t = - 0.14285 \dots

Solution steps

10 = 490 t^{2}

Switch sides

490 t^{2} = 10

Divide both sides by

490

t^{2} = \frac{1}{49}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

t = \frac{1}{49}, t = - \frac{1}{49}

\frac{1}{49} = \frac{1}{7}

- \frac{1}{49} = - \frac{1}{7}

t = \frac{1}{7}, t = - \frac{1}{7}

1 + 8 w = 10 - w

x^{4} - 4 x^{3} + 3 x^{2} - 16 x^{2} + 64 x - 48

\frac{13 !}{( 13 - 4 ) !}

x - 2 \leq 5

x^{2} + 25

What are the solutions to the equation 10=490t^2 ?
The solutions to the equation 10=490t^2 are t= 1/7 ,t=-1/7
Find the zeros of 10=490t^2
The zeros of 10=490t^2 are t= 1/7 ,t=-1/7