165 ounces for 11 boxes. I want to know the weight of 1 box. How do I get 11 boxes to be 1 box. Divide it by 11.
11/11 = 1 box. Since I chose 11 I have to also use 11 for the ounces
165/11 = 15 ounces per box
Answer: 4 1/2 - 1 11/12 = 2 1/3 but in decimal form 2.3
For numbers 15-17, we need to remember that two of a triangle's angles are always acute and the third angle will allow us to classify the triangle based on its angles. now that we know this, let's look at #15. the first two angles listed are acute, and the third is an obtuse angle, therefore it is an obtuse triangle. on #16 we have three acute angles, so it is an acute triangle. #17 has two acute angles and a right angle so it is a right triangle.
on numbers 21-23, we need to know that a triangle with all congruent sides is called equilateral, a triangle with two equal sides is isosceles, and a triangle with no equal sides is called scalene. #21 shows two equal sides so it is an isosceles triangle. #22 has three equal sides so it is an equilateral triangle. #23 has no equal sides so it is scalene. hope this helped! :)
<h2>
Greetings!</h2>
Answer:
y = 
and x = 
Step-by-step explanation:
To solve simultaneous equations, you need to have the number in front of both x's or y's the same. (signs doesn't matter)
To get -x to -10x we simply need to multiply the first equation by 10:
-x * 10 = -10x
-9y * 10 = -90y
16 * 10 = 160
-10x - 90y = 160
Now we can add the two equations:
-10x + 10x = 0
-90y + 20y = -70y
160 + 20 = 180
-70y = 180
70y = -180
7y = -18
y =
Now plug into the second equation:
10x + 20() = 20
10x - 
= 20
Move the over to the other side, making it a positive:
10x = 20 +
10x = 
Divide both sides by 10:
x =
So y = and x =
<h2>Hope this helps!</h2>
<span>The answer is that x = 21/9.
Start with the original: 4(2x - 1) - 7 = 4 - x + 6
Now distribute the 4:</span> <span> 8x - 4 - 7 = 4 - x + 6.
Now combine the like terms. 8x - 11 = -x + 10.
Now add x to both sides. 9x - 11 = 10.
Now add 11 to both sides. 9x = 21.
And divide by 9 on both side. x = 21/9. </span>
