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2 years ago

Compound inequalitiesSOLVE FOR X

Mathematics

2 answers:

snow_tiger [21]2 years ago

6 0

Answer:

X $\geq$ 2

Step-by-step explanation:

PolarNik [594]2 years ago

5 0

X is 2!!! Haha good luck :)

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Find the distance between the points (2, 4) and (8,-8) on a coordinate plane, to the nearest whole number.

Alisiya [41]

Answer:

2,5

Step-by-step explanation:

2 to the right and 5 up

8 0

2 years ago

Question 1 (1 point)<br> Which equation can be represented by a graph with a vertex at (1,3)?

madam [21]

Answer:

See explanation

Step-by-step explanation:

You have not provided the options to help us give you a specific answer.

If a quadratic function is written in the form

$f(x) = a {(x - h)}^{2} + k$

Then (h,k) is the vertex, and 'a' is a constant.

For instance

$f(x) = 5 {(x - 1)}^{2} + 3$

has vertex at (2,3).

Also, an absolute function in the form

$g(x) = a |x - h| + k$

has vertex at (h,k).

This means

$g(x) = |x - 1| + 3$

also has vertex at (1,3).

I hope this explanation is helpful.

4 0

3 years ago

Can someone pls help me

skelet666 [1.2K]

The midpoint is

(-4+0) / 2, (7 + (-3))/ 2 The first one is x coordinate and 2nd is the y cood)

= (-2, 2) answer

6 0

3 years ago

Solve the inequality 3x^2 + 10x - 8 < 0

Ksivusya [100]

Step-by-step explanation:

Given, 3x−2<2x+1

⇒3x−2x<1+2

⇒x<3orx∈(−∞,3)

The lines y=3x−2 and y=2x+1 both will intersect at x=3

Clearly, the dark line shows the solution of 3x−2<2x+1.

8 0

2 years ago

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(1-4)^2+(-5+3)^2}$

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

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

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

5 0

3 years ago

Compound inequalitiesSOLVE FOR X​

Compound inequalitiesSOLVE FOR X