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3 years ago

Solve the inequality graph the solution |x-5| _< 3

Mathematics

1 answer:

xenn [34]3 years ago

6 0

The first one is the answer because

|6-5|<3 and 1<3
|4-5|<3 and -1<3
|2-5|<3 and -3<3
|0-5|<3 and -5<3
|-2-5|<3 and -7<3
|-4-5|<3 and -9<3
|-6-5|<3 and -11<3

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Using plane K, name a point that is not on the line.

yuradex [85]

Answer:

Step-by-step explanation:

Plane K is shown with two lines. Points on one of them are A, B, G. Points on the other one are G, M, N.

The point in plane K that is not shown on any line is point H.

3 0

1 year ago

Please help i’m finding this hard

pav-90 [236]

Answer:

Area for Shape p is 12, the name for the shape q is a rhombus.

Step-by-step explanation:

The shape p is a triangle for if you find the height of the shape which is 6 and the length of the base which is 4 multiply them together then divide by 2 to get the area.

8 0

3 years ago

Find the area of each shape, please help and I will give a lot points

Alex73 [517]

Answer:

15. Area of parallelogram = bh

b = 9m

h = 4m

a = 9*4 = 36m^2

You can also use the rectangle property here.

a = l*b

l = 9

b = 6

a = 9*6 = 54 m^2

Though I would prefer the first one better.

16. Area of triangle = 1/2*bh

b = 17m

h = 3

a = 17*3/2 = 51/2 = 25.5m^2

18. Area of a trapezium = a+b/2*h (a and b are two parallel sides and h the height)

a = 15

b = 7

h = 15

a = 15+7/2*15 = 22/2*15 = 11*15 = 165m^2

19. Area of triangle = 1/2*bh

b = 24

h = 7

a = 24*7/2 = 168/2 = 84

Hope u understood

Please mark as brainliest

Thank You

5 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 $y=4-1=3$ .