Answer:
y = -5 + 6=1 is your answer
Draw the triangle and then use Pythagoras' theorem
distance =
= 19.70 (2dp)
<span>l ≤ 12
2l + 2w < 30
the second and fourth option are saying she can make the length longer than twelve and the third option forgets to double the length and width so it is accurate. The first option is the right answer.</span>
Answers:
10.) 
9.) 
8.) 
7.)
6.) 
Step-by-step explanations:
10.) 
9.) ![\displaystyle \frac{\sqrt[3]{135}}{\sqrt[3]{40}} \hookrightarrow \sqrt[3]{3\frac{3}{8}} \hookrightarrow \frac{3\sqrt[3]{5}}{2\sqrt[3]{5}} \\ \\ \boxed{1\frac{1}{2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csqrt%5B3%5D%7B135%7D%7D%7B%5Csqrt%5B3%5D%7B40%7D%7D%20%5Chookrightarrow%20%5Csqrt%5B3%5D%7B3%5Cfrac%7B3%7D%7B8%7D%7D%20%5Chookrightarrow%20%5Cfrac%7B3%5Csqrt%5B3%5D%7B5%7D%7D%7B2%5Csqrt%5B3%5D%7B5%7D%7D%20%5C%5C%20%5C%5C%20%5Cboxed%7B1%5Cfrac%7B1%7D%7B2%7D%7D)
8.) ![\displaystyle \frac{\sqrt[4]{162}}{\sqrt[4]{32}} \hookrightarrow \sqrt[4]{5\frac{1}{16}} \hookrightarrow \frac{\pm{3\sqrt[4]{2}}}{\pm{2\sqrt[4]{2}}} \\ \\ \boxed{\pm{1\frac{1}{2}}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csqrt%5B4%5D%7B162%7D%7D%7B%5Csqrt%5B4%5D%7B32%7D%7D%20%5Chookrightarrow%20%5Csqrt%5B4%5D%7B5%5Cfrac%7B1%7D%7B16%7D%7D%20%5Chookrightarrow%20%5Cfrac%7B%5Cpm%7B3%5Csqrt%5B4%5D%7B2%7D%7D%7D%7B%5Cpm%7B2%5Csqrt%5B4%5D%7B2%7D%7D%7D%20%5C%5C%20%5C%5C%20%5Cboxed%7B%5Cpm%7B1%5Cfrac%7B1%7D%7B2%7D%7D%7D)
7.) 
6.) 
I am joyous to assist you at any time.
Answer:
9
Step-by-step explanation:
To calculate c from the first right angle triangle,
C2 = 36 + 16
C = sqrt(52)
To calculate a from the bigger right angle triangle,
B2 + 52 = (a + 4)2
B2 - a2 = -36 + 8a
A2 + 36 = B2
Solving both by elimination,
A = 72/8
=9
