The guy who links files is pretty annoying amiright...
Divide 18/35 by 3/5 to get your answer. You can also multiply 3/5 by 35/18.
5 goes into 35 7 times, so cross out the 5 and 35 and put 1 and 7. 3 goes into 18 6 times, so cross out the 3 and 18 and put 1 and 6.
Now you have 1/1 times 7/6. Your answer is 7/6.
250=100%
25=10%
12.5=5%
2.5=1%
25*8=200(80%)
12.5+2.5=15(6%)
200+12=215(86%)
250+200=450(186%)
so 450ml is your answer
Hope this helped :)
Answer:
So its pretty simple
The table provided the Xs so you can just substitute the x in the eqaution becuase the eqaution gives you Y and your looking for Y.
For the first one
-3/4*-16+3=y
12+3=15
So Y is 15
DO the same for the others
Also I think your rate of change is wrong, I think it’s -3/4 but your choice really, I might be wrong
To find the length of the diagonal, we need to calculate the length of the base first and then use the value to calculate the diagonal since we are given the value of height of the triangle.
<h3>Pythagorean Theorem</h3>
This theorem is used to find the missing side of a right angle triangle given that we have the length of two sides.
To calculate the base of the triangle, let's use Pythagorean theorem here
![x^2 = y^2 + z^2\\x^2 = 2^2 + 3^2\\x^2 = 4 + 9 \\x^2 = 13\\x = \sqrt{13} \\x = 3.61](https://tex.z-dn.net/?f=x%5E2%20%3D%20y%5E2%20%2B%20z%5E2%5C%5Cx%5E2%20%3D%202%5E2%20%2B%203%5E2%5C%5Cx%5E2%20%3D%204%20%2B%209%20%5C%5Cx%5E2%20%3D%2013%5C%5Cx%20%3D%20%5Csqrt%7B13%7D%20%5C%5Cx%20%3D%203.61)
Having the length of the base, let's consider the height of the triangle which is 6 and substitute the values.
![d^2 = 6^2 + 3.61^2\\d^2 = 36 + 13.0321\\d^2 = 49.0321\\d = \sqrt{49.0321} \\d = 7.00](https://tex.z-dn.net/?f=d%5E2%20%3D%206%5E2%20%2B%203.61%5E2%5C%5Cd%5E2%20%3D%2036%20%2B%2013.0321%5C%5Cd%5E2%20%3D%2049.0321%5C%5Cd%20%3D%20%5Csqrt%7B49.0321%7D%20%5C%5Cd%20%3D%207.00)
From the calculations above, the values of the triangle are
- diagonal = 7
- base = 3.61
- height = 6
Learn more on pythagorean theorem here;
brainly.com/question/231802
#SPJ1