Solve the inequality 4(k − 5) + 12k ≥ −4

2 answers:

8 0

Isolate the variable by dividing each side by factors that don't contain the variable.
Inequality Form:
k
≥
1
k
≥
1
Interval Notation:
[
1
,
∞
)
[
1
,
∞
)

IgorC [24]3 years ago

5 0

Answer:

k ≥ 1

Step-by-step explanation:

Step 1: Write inequality

4(k - 5) + 12k ≥ -4

Step 2: Solve for <em>k</em>

Distribute 4: 4k - 20 + 12k ≥ -4
Combine like terms: 16k - 20 ≥ -4
Add 20 to both sides: 16k ≥ 16
Divide 16 on both sides: k ≥ 1

You might be interested in

9. Consider the circles with the following equations:

kenny6666 [7]

Answer with Step-by-step explanation:

We are given that

$x^2+y^2=(\sqrt 2)^2$

$(x-3)^2+(y-3)^2=32=(4\sqrt 2)^2$

Compare with the equation of circle

$(x-h)^2+(y-k)^2=r^2$

Where center of circle=(h,k)

r=Radius of circle

a.Center of circle=(0,0)

Radius= $\sqrt 2$ units

Center of second circle=(3,3)

Radius of second circle= $4\sqrt 2$ units

b.Distance formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

Using the formula

The distance between the centers of two circle

= $\sqrt{(3-0)^2+(3-0)^2}=3\sqrt 2$

Hence, the distance between the centers of two circle = $3\sqrt 2$ units.

Substitute x=-1 and -1

$1+1=2=$

$(1-3)^2+(1-3)^2=32$

The circle must be tangent because there is just one point (-1,-1) is common in both circles and satisfied the equations of circle.

8 0

3 years ago

Quadrilateral ABCD ABCD is similar to quadrilateral EFGH EFGH. The lengths of the three longest sides in quadrilateral ABCD ABCD

Kay [80]

Sketch the two quadrilaterals and label them as shown in the figure below.

Let x = length of the 4th side of ABCD.

Because ABCD ~ EFGH, therefore
x/5 = 14/6

That is,
x = (5/6)*14 = 11.67 ft

Answer: D. 11.7 ft