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rusak2 [61]
3 years ago
11

Can you use the SAS postulate, the AAS theorem, or bothto prove the triangles are congruent

Mathematics
1 answer:
NikAS [45]3 years ago
3 0
Yes, most definitely
I hope that this helps!

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Write an equation in<br> y = mx + b form for<br> the line graphed.
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Answer:

y=2x-5

explanation

m is gradient so rise over run and b is the y intercept which is -5

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Given f(x) = -x - 2, find x when f(x) = 12.
Flauer [41]

Answer:

We substitute

12 = -x - 2

14 = -x or -x = 14

so the answer is -14!

I am in honors high school algebra and I am 11. Trust me bro.

Step-by-step explanation:

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Find expressions for the partial derivatives of the following functions:
nlexa [21]

Answer:

Step-by-step explanation: partial derivative is the differentiation of one variable e.g. X while leaving the values of the other variable e.g. Y

These four questions A, B, C and D have different functions separated by commas. I will not assume the commas to be something else like a plus sign.

A. f(x) = g'(x).k(y) , g'(x) + h(y)

f(y) = k'(y).g(x) , g(x) + h'(y)

B. f(x) = g'x (x+y)

f(y) = g'y (x+y) , h'y (y+z)

f(z) = h'z (y+z)

C. f(x) = f'x (xy) , f'x (zx)

f(y) = f'y (xy) , f'y (yz)

f(z) = f'z (yz) , f'z (zx)

D. f(x) = f'x (x) , g'(x) , h'x (x,y)

f(y) = h'x (x,y)

These are the partial derivative expressions for each variable in each function. You will need to pay a lot of attention to understand:

* while differentiating X alone, functions in Y which are separated by commas from the functions in X, are ignored totally because they are different questions

* In functions where X added to Y is in a bracket e.g. (x+y), to find the derivative of X, Y isn't thrown away because they are joined (by a plus sign) the derivative of X alone in this case would be f'x (x+y)

* f(x), just like g(x), simply means/represents a function in X hence f'(x) means the differentiation of all X-terms in that function

6 0
3 years ago
Help me in this one pls I give 15 pts
scoundrel [369]

Answer:

23.7555090109  

Step-by-step explanation:  

sin= opposite/hypotenuse  

sin(53) = BC/60cm  

BC= sin(53)*60 = 23.7555090109

6 0
3 years ago
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