Geometry homework help???

1 answer:

3 0

The explorer's running velocity will be 1/3 the velocity of the train, 45/3 = 15. From those issue statement, the train will a chance to be at the begin of the span The point when those voyagers have run 1/3 of the period of the span. In they run of the much end of the bridge, they will bring run 1/3 of the length of the span when those prepare achieves the close to wind. Beginning starting with those 1/3 point, they will make toward those 2/3 perspective At those prepare achieves those span. The prepare will travel the full period of the span in the the long haul it takes those voyagers to run those remaining 1/3. Accordingly the train must travel toward three times the running speed of the voyager. Train = 45 mph, voyagers = 15 mph. The speed of P&D = 15 mph

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10) g(n) = -21+ 3n; Find gl-6) A) 152 B) -20 C) 1 D) 90

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A) 152
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If this helped can I please get brainliest

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2 years ago

Solve for x.

Ivahew [28]

ANSWER

$x=\frac{2-\sqrt{10}} {3}$

or

$x=\frac{\sqrt{10}+2} {3}$

We have

$3x^2-4x-2=0$

Since we cannot factor easily, we complete the square.

Adding 2 to both sides give,

$3x^2-4x=2$

Dividing through by 3 gives

$x^2-\frac{4}{3}x= \frac{2}{3}$

Adding $(-\frac{2}{3})^2$ to both sides gives

$x^2-\frac{4}{3}x+(-\frac{2}{3})^2= \frac{2}{3}+(-\frac{2}{3})^2$

The expression on the Left Hand side is a perfect square.

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$\Rightarrow (x-\frac{2}{3})^2= \frac{10}{9}$

$\Rightarrow (x-\frac{2}{3})=\pm \sqrt{\frac{10}{9}}$

$\Rightarrow (x)=\frac{2}{3} \pm {\frac{\sqrt{10}}{3}$

Splitting the plus or minus sign gives

$x=\frac{2- \sqrt{10}} {3}$

or

$x=\frac{\sqrt{10}+2} {3}$

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2 years ago

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