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pentagon [3]
4 years ago
13

What is the solution to the inequality below ?

Mathematics
1 answer:
Vlad1618 [11]4 years ago
8 0
[tex]Domain:x\geq0\\\\\sqrt{x}\leq5\ \ \ \ |square\ both\ sides\\\\x\leq25[tex]

Answer: B. 0 ≤ x ≤ 25
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84 km / h in m / s plz help me<br>​
PilotLPTM [1.2K]

Answer:

23\frac{1}{3}m/s

Step-by-step explanation:

We can change the km to m first. 1km = 1000m.

84km/h = 84 000m/h

Then, we can change the h to s. 1h = 3600s.

84 000m/h = 23\frac{1}{3}m/s

7 0
3 years ago
The first day of a water polo tournament the total value of tickets sold was $17,610. One-day passes sold for $20 and tournament
Alenkasestr [34]

Answer:

gfggfgkfmf

Step-by-step explanation:

3 0
3 years ago
?
lilavasa [31]

The count of the equilateral triangle is an illustration of areas

There are 150 small equilateral triangles in the regular hexagon

<h3>How to determine the number of equilateral triangle </h3>

The side length of the hexagon is given as:

L = 5

The area of the hexagon is calculated as:

A = \frac{3\sqrt 3}{2}L^2

This gives

A = \frac{3\sqrt 3}{2}* 5^2

A = \frac{75\sqrt 3}{2}

The side length of the equilateral triangle is

l = 1

The area of the triangle is calculated as:

a = \frac{\sqrt 3}{4}l^2

So, we have:

a = \frac{\sqrt 3}{4}*1^2

a = \frac{\sqrt 3}{4}

The number of equilateral triangles in the regular hexagon is then calculated as:

n = \frac Aa

This gives

n = \frac{75\sqrt 3}{2} \div \frac{\sqrt 3}4

So, we have:

n = \frac{75}{2} \div \frac{1}4

Rewrite as:

n = \frac{75}{2} *\frac{4}1

n = 150

Hence, there are 150 small equilateral triangles in the regular hexagon

Read more about areas at:

brainly.com/question/24487155

4 0
3 years ago
How do you solve 5/x=x-4
alexandr402 [8]
\frac{5}{x}=x-4\\&#10;x\not=0\\&#10;5=x^2-4x\\&#10;x^2-4x-5=0\\&#10;x^2+x-5x-5=0\\&#10;x(x+1)-5(x+1)=0\\&#10;(x-5)(x+1)=0\\&#10;x=5 \vee x=-1
6 0
4 years ago
Solve for x<br> −4x+60&lt;72 OR 14x+11&lt;−31
Iteru [2.4K]

Answer:

14x+11<-31

14x<-42

x=-3

Hope This Helps!!!

6 0
3 years ago
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