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Stels [109]
3 years ago
7

What is 11 ≥ x > −4 written in interval notation?

Mathematics
2 answers:
Ghella [55]3 years ago
8 0

Answer:

( -4, 11 ]

Step-by-step explanation:

GaryK [48]3 years ago
5 0

Answer:

(-4,11]

Step-by-step explanation:

Round bracket for excluded boundaries,

Square bracket for included

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In square YOLA, RL=5x+28 and RO=8X-11
Alexeev081 [22]

Answer:

RL=5x+28 and

RO=8X-11

diagonal of square bisect equally the side

:.5x+28=8x-11

11+28=8x-5x

39=3x

x=39/3=13

<u>RY</u><u>=</u><u>RL</u><u>=</u><u>5</u><u>x</u><u>+</u><u>2</u><u>8</u><u>=</u><u>5</u><u>×</u><u>1</u><u>3</u><u>+</u><u>2</u><u>8</u><u>=</u><u>9</u><u>3</u><u>If the answer is 93, move to </u><u>answer</u><u>.</u>

6 0
2 years ago
a famer has 120 feet of fencing with whcih to enclose two adjacent rectangular pens as shown. what dimeensions should be used th
Nat2105 [25]

The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.

Let the area be y .

Area = (base) × (height)

Base = 2x

Height = h

Let the area of the rectangular pens be y .

∴ y = 2xh

Perimeter of all the fencing = 4x+3h

∴ 4x+3h = 120

now we solve for h

3h = 120-4x

h = 40 - 4/3 x

Now we will substitute this value in the above first equation:

y = 2xh

or, y = 2x (40 - 4/3 x)

or, y = 80x - 8/3 x²

Now for the maximum area we have to find the first order differentiation of y

now,

dy /dx = 80 - 16/3 x

At dy/dx = 0 we get the value of x for which y is maximum.

80 - 16/3 x = 0

or, - 16/3 x = -80

or, x = 15 feet

Hence height =  40 - 4/3 x = 40 - 20 = 20feet

Maximum area = 2xh = 2×15×40 = 1200 square feet

The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.

Disclaimer : The missing figure for the question is attached below.

To learn more about area visit:

brainly.com/question/27531272

#SPJ4

6 0
1 year ago
The solution of a+b&gt;−4 is a&gt;−9. What is the value of b?
evablogger [386]
A><span><span>−b</span>−<span>4
Hope this helps. c:</span></span>
5 0
3 years ago
How could you graph an inequality like this x&gt;3?
Firlakuza [10]

Answer:

  see below

Step-by-step explanation:

The graph of it on a number line is an open circle at x=3 with a line extending to the right through larger numbers.

When the inequality does not include the "or equal to" case, the boundary is graphed as a dashed line (on an x-y plane) or open circle (on a number line). The shaded area covers values of the variable that meet the condition of the inequality. Here, those are values of x that are more than 3.

6 0
3 years ago
Please help me now ! Thank you
jarptica [38.1K]
First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now

2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4

At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real. 

Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result. 

For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196

No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.

So x^2 = -4 has no real solutions. Taking the square root of both sides leads to

x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values


5 0
3 years ago
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