21 ≥ a/4 ( a/4 is a fraction)

PLEASE help with these 2 questions<br><br> thank you!

Sergio039 [100]

Download the app photo math help a whole lot

6 0

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 $y=4-1=3$ .

All this to say that A is the point (1, 3), so

$2x+y=2+3=5$

5 0

3 years ago

A cement walkway of uniform width has been built around an in-ground rectangular pool. the area of the walkway is 1344 square fe

Elza [17]

Answer:

6 feet.

Step-by-step explanation:

Dimensions of the Pool =80 feet long by 20 feet wide

Area of the walkway =1344 square feet.

If the width of the walkway=w

Length of the Larger Rectangle =80+2w
Width of the Larger Rectangle =20+2w

Area of the Walkway =Area of the Larger Rectangle - Area of the Pool

$1344=(80+2w)(20+2w)-(80*20)\\1344=80*20+160w+40w+4w^2-80*20\\4w^2+200w=1344\\4w^2+200w-1344=0\\We factorize\\4(w^2+50w-336)=0\\4(w^2-6w+56w-336)=0\\w(w-6)+56(w-6)=0\\(w-6)(w+56)=0\\w-6=0$ or $ w+56=0\\w=6$ ft or $ w=-56$ \\Therefore, the width of the walkway , w=6 feet.$

6 0

2 years ago

5\2,3 and 3\2,8 find the distance

pochemuha

Answer:

√26 ≈ 5.1 units

Step-by-step explanation:

√(x₂ - x₁)² + (y₂ - y₁)²

√(3/2 - 5/2)² + (8 - 3)²

√(-1)² + (5)²

√1 + 25

√26

4 0

3 years ago

10) Choose the solution for each linear system.

hodyreva [135]

Answer:

Step-by-step explanation:

10.

$\left \{ {{x-2y=4} \atop {4x+2y+6}} \right.$

The y values cancel out if you add the equations

5x = 10

x = 2

plug in x

2 - 2y = 4

-2y = 2

y=-1

Answer: C. (2, -1)

11.

$\left \{ {{6x-2y=-4} \atop {-4x+3y=-4}} \right.$

In order to get terms to cancel out, multiply the top by 3 and bottom by 2

$\left \{ {{18x-6y=-12} \atop {-8x+6y=-8}} \right.$

The y values cancel out if you add the equations

10x = -20

x = -2

plug in x

6(-2) - 2y = -4

-12 - 2y = -4

-2y = 8

y = -4

Answer: B. (-2, -4)

8 0

2 years ago