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3 years ago

2=x(7)-3 step by step

Mathematics

1 answer:

lubasha [3.4K]3 years ago

8 0

Regroup terms

2 = 7x - 3

Add 3 to both sides

2 + 3 = 7x

Simplify 2 + 3 to 5

5 = 7x

Divide both sides by 7

5/7 = x

Switch sides

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The sale price should be $45

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3 years ago

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If he rows with the current, Mishka rows 48 km in 3 hours. If he rows against the current, it takes 4 hours to travel the same d

Zarrin [17]

<h2>Mishka's speed = 14 km/hr</h2><h2>Current speed = 2 km/hr</h2>

Step-by-step explanation:

Let the speed of rowing be r and speed of current be c.

If he rows with the current, Mishka rows 48 km in 3 hours.

Relative speed = r + c

Distance = 48 km

Time = 3 hours

48 = 3 x ( r+c)

r + c = 16 ---------------------------eqn 1

If he rows against the current, it takes 4 hours to travel the same distance.

Relative speed = r - c

Distance = 48 km

Time = 4 hours

48 = 4 x ( r - c)

r - c = 12 ---------------------------eqn 2

eqn 1 + eqn 2

2r = 28

r = 14 km/hr

Substituting in eqn 1

14 + c = 16

c = 2 km/hr

Mishka's speed = 14 km/hr

Current speed = 2 km/hr

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3 years ago

Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}

Deffense [45]

Start with a line segment connecting two points, A and B. $\dfrac{DA}{BA}=3$ means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.

But with $\dfrac{DB}{BA}=2$ , or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.

Finally, $\dfrac{AB}{CB}=\dfrac12$ , so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.

Now let $d$ be the length of AB. Then the length of CD in terms of $d$ is $4d$ . We have the coordinates of C and D, and the distance between them is $\sqrt{(4-0)^2+(0-4)^2}=4\sqrt2$ . So

$4d=4\sqrt2\implies d=\sqrt2$

The slope of the line through C and D is

$\dfrac{0-4}{4-0}=-1$

and so the equation of the line through these points is

$y-4=-(x-0)\implies x+y=4$

So the coordinates of A are $(x,y)=(x,4-x)$ . The distance between C and A is $d=\sqrt2$ , so we have

$\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1$

Since A falls to the right of C (in the $x,y$ plane, not just in the sketch), we know to take the positive value $x=1$ . Then the $y$ coordinate is $y=4-1=3$ .