The answer:
according to the image, the main theorem concerning right triangle similarity is as follow:
the altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the original triangle and each other,
in our case, KL is the altitude, and by applying theorem, we get three triangles that are similar:
therefore:
<span>△JKL ~ △JKM
</span><span>△JKM ~ △JKM
</span><span>△JMK ~ △KML</span>
What needs to be done first is to add up females and males that have passed.
42 + 14 = 56
so out of 56 students who passed 42 females passed 42/56 = 3/4 = 0.75
out of 56 students who passed, 14 males passed which turns into 14/56 = 1/4 = 0.25
check work; 0.75 + 0.25 = 1.00
NOW WE ARE DOING FAILS.
15 + 5 = 20
so out of 20 students who failed, 15 females failed so it turns into 15/20 = 3/4 = 0.75
out of 20 students who failed, 5 males failed. 5/20 = 1/4 = 0.25
check work; 0.75 + 0.25 = 1.00
i hope this helped! :)
Absolute value equations have more than one solution because when the answer is in abs. value, it is positive, so there could be a negative answer and it would still be positive. For example |-5|=5 and |5|=5 as well.
4+|2x|=-1 would have no solution because you cannot have the equation equal a negative since the 2x is in absolute value.
let's firstly convert the mixed fractions to improper fractions and then to do away with the denominators, let's multiply both sides by the LCD of all denominators.
![\stackrel{mixed}{1\frac{3}{4}}\implies \cfrac{1\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{7}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{7}{4}-\cfrac{4}{5}=\cfrac{35}{20}-\boxed{?}\implies \stackrel{\textit{multipling both sides by }\stackrel{LCD}{20}}{20\left( \cfrac{7}{4}-\cfrac{4}{5} \right)=20\left( \cfrac{35}{20}-\boxed{?} \right)} \\\\\\ 35-16=35-20\boxed{?}\implies 19=35-20\boxed{?}\implies -16=-20\boxed{?} \\\\\\ \cfrac{-16}{-20}=\boxed{?}\implies \cfrac{4}{5}=\boxed{?}](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B1%5Cfrac%7B3%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B1%5Ccdot%204%2B3%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B7%7D%7B4%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B7%7D%7B4%7D-%5Ccfrac%7B4%7D%7B5%7D%3D%5Ccfrac%7B35%7D%7B20%7D-%5Cboxed%7B%3F%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultipling%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B20%7D%7D%7B20%5Cleft%28%20%5Ccfrac%7B7%7D%7B4%7D-%5Ccfrac%7B4%7D%7B5%7D%20%5Cright%29%3D20%5Cleft%28%20%5Ccfrac%7B35%7D%7B20%7D-%5Cboxed%7B%3F%7D%20%5Cright%29%7D%20%5C%5C%5C%5C%5C%5C%2035-16%3D35-20%5Cboxed%7B%3F%7D%5Cimplies%2019%3D35-20%5Cboxed%7B%3F%7D%5Cimplies%20-16%3D-20%5Cboxed%7B%3F%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B-16%7D%7B-20%7D%3D%5Cboxed%7B%3F%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B5%7D%3D%5Cboxed%7B%3F%7D)