Help me answer this thanks

Mathematics

2 answers:

zloy xaker [14]3 years ago

8 0

Answer:

C = 73

Step-by-step explanation:

Formula used: a²+b²=c²

A and B represent the numbers other than the hypotenuse, so C equals the hypotenuse. It doesn't matter which number you put as A or B. Here's the equation we're working with:

(55)² + (48)² = C²

That simplifies to:

3025 + 2305 =C²

5329 = C²

Now, C needs to be by itself in order for us to get the answer, so the next step is to do inverse operations. The inverse of sqauring a number (x²) is to square root it (√x). Here is the new equation we are working with:

√5329 = √C

That simplifies to:

73 = C

Hope this is correct!

zysi [14]3 years ago

5 0

It should be 77 because 180-55-48=77 and I took the test

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A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$