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3 years ago

5

Two cars start from towns 630 mi apart and travel toward each other they meet after 6hr find the speed of each car one travels 2

5 mph than the other

1 answer:

soldier1979 [14.2K]3 years ago

7 0

Answer:

We must think that the cars are traveling 630 miles in six hours.

Their combined speeds are x and x+25.

distance = velocity * time

Since, velocity = distance / time

(x + x + 25) = 630 / 6

2x + 25 = 105

2x = 80

So x = 40 mph and the other car's velocity is 65 mph.

Step-by-step explanation:

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How do I write 684297 in word form?

Daniel [21]

Answer:

Six hundred eighty-four thousand two hundred ninety-seventh

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3 years ago

Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u –

Verdich [7]

Answer:

The solutions of the original equation are x=-5 and x=-2

Step-by-step explanation:

we have

$(x+3)^2+(x+3)-2=0$

Let

$u=(x+3)$

Rewrite the equation

$(u)^2+(u)-2=0$

Complete the square

$u^2+u=2$

$u^2+u+1/4=2+1/4$

$u^2+u+1/4=9/4$

rewrite as perfect squares

$(u+1/2)^2=9/4$

square root both sides

$(u+1/2)=\pm\frac{3}{2}$

$u=(-1/2)\pm\frac{3}{2}$

$u=(-1/2)+\frac{3}{2}=1$

$u=(-1/2)-\frac{3}{2}=-2$

the solutions are

u=-2,u=1

<em>Alternative Method</em>

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$(u)^2+(u)-2=0$

so

$a=1\\b=1\\c=-2$

substitute in the formula

$u=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}$

$u=\frac{-1\pm\sqrt{9}} {2}$

$u=\frac{-1\pm3} {2}$

$u=\frac{-1+3} {2}=1$

$u=\frac{-1-3} {2}=-2$

the solutions are

u=-2,u=1

<em>Find the solutions of the original equation</em>

For u=-2

$-2=(x+3)$ ----> $x=-2-3=-5$

For u=1

$1=(x+3)$ ----> $x=1-3=-2$

therefore

The solutions of the original equation are

x=-5 and x=-2

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Answer:

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Step-by-step explanation:

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Answer:

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