Hey!
I hope you don't mind, but before I answer this question I'd like to do a quick review of some general angles and how to tell which is which.
<span><em><u /></em><span><em><u>QUICK REVIEW</u></em>
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So, let's first review what an acute angle is. An acute angle is an angle that is smaller than 90<span>°. The word acute basically means having a sharp or pointy end. So this is a helpful way to remember what an acute angle is.
Now, let's review what an obtuse angle is. An obtuse angle is an angle that is more than 90</span>°. If an angle measures over 90<span>° that it is more than likely that it is an obtuse angle.
Last but not least a right angle. A right angle is an angle that has to be exactly 90</span>°. If an angle is 90<span>° than it is most definitely a right 
<span><em><u /></em><span><em><u>END QUICK REVIEW</u></em>
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Let's start by finding out the angle in the top left hand corner. The angle is clearly no more that 90</span>° and is not 90° exactly. This angle must be an acute angle. We can also tell that it is an acute angle because the angle is sharp.
Now let's look at the angle on the bottom left hand corner. The angle is clearly no less than 90° and more than 90° exactly. This angle must be an obtuse angle.
Since the angles are basically the same on the other side, we won't be reviewing those. Now we'll count all the angles we have.
Acute Angles - 2
Obtuse Angles - 2
Right Angles - 0
<em>So, this means that in the figure shown above,</em>  there are 2 acute angles, 2 obtuse angles, and no right angles.
Hope this helps!
- Lindsey Frazier ♥
        
             
        
        
        
Answer:
1. 34.42 Toneladas
2. 28.640 toneladas
3. 38 toneladas
4. o
 o 
5. o
 o 
6. o
 o 
1.1 Solucioné el problema convirtiendo el .25 en fracción.
2.1 Conviertes el denominador en el mismo buscando el minimo comun multiplo
3.1 0.099



0.7
0.75
1.1

Step-by-step explanation:
Acuerdate que para solucionar las fracciones mixtas solo tienes que dividir el denominador al nominador, por ejemplo en si tienes  entonces el 3 cabe 3 veces en el 10, y sobra 1, entonces quedamos que en mixta la fracción sería
 entonces el 3 cabe 3 veces en el 10, y sobra 1, entonces quedamos que en mixta la fracción sería 
Entonces una vez que recordamos eso, podemos resolver los problemas de fracciones sin ningún problema vamos a resolver la número 4:

Como el denominador es igual, sólo sumamos el nominador:


Cómo el 10 cabe 1 vez en el 15, tenemos un entero y sobran 5:

 
        
             
        
        
        
Answer:
The answer to your question is A = 648 u²
Step-by-step explanation:
I attach the solution because it said that I was writing improper words.
 
        
             
        
        
        
First, lets start with the greatest 4-digit number. That would be 9,999.99999999999, with 9 to ∞, because the next number is a 5-digit number. The smallest 6-digit number is 100,000, because the number preceding 100,000 is a 4 digit number. To find difference, you have to subtract. 100,000 minus 9,999.9999 continued gives you 90,000.000000 to ∞ with a one with 1 at the end. You would write 90,000.01 with bar notation over only the 0 to indicate that it goes on to infinity.
        
             
        
        
        
Answer:
Step-by-step explanation:
Number of students 10
Problem 1. $625 for the bus hire per friday, So 625*4=$2500
Problem 2. 2500/25=$100 each for the whole 4 weeks
Problem 3.
10 students tickets 220= 2200 for all tickets. The bus, 625/10 = $62.5*4= $250 dollars for the whole 4 weeks for the bus so in all each student pays $470 each
20 students, tickets 220=4400 for all tickets. The bus, 625/20=$31.25*4=$125 for the whole 4 weeks for the bus, so in all each student must pay $345 each
30 students, tickets 220 = 6600 for all tickets. The bus, 625/30 =$20.83*4=$83.32 for the whole 4 weeks for the bus, so in all each student must pay $303.32 each
41 students, tickets $160=$6560 for all tickets. The bus, because you need 2 buses at 625 each so $1250 for both buses 1250/41= 30.49*4=$121.96 for the whole 4 weeks for the bus. So in al each student must pay $281.96 each
Hope this is correct