There are 2.54 centimeters in an inch. when using this conversion factor, how many significant figures are you limited to?

(3 + 8i) + (7 -- 2i) - (6 - 5i) AO 4+1 B.O4 + 117 c. 16 +1 D.O4 + 1143

Lubov Fominskaja [6]

Answer: (3 + 8i) + (7 -- 2i) - (6 - 5i) = 4 + 15i

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Simplify the expression.

4 + 15i

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3 0

3 years ago

Explain why triangles in the figure are similar. then find the missing length x

Strike441 [17]

Answer:

∨∨∨∨see below∨∨∨∨∨∨

Step-by-step explanation: 6 26 18 13

The two outside angles are congruent. The two inside angles are supplemental thus they are equal. The last two angles the high one and the lower one must sum to 180° in their respective triangles so they are equal since their similar angles are equal.

find x

4 is to x as 5 is to 7.5

4/x = 5/7.5 solve for x

4 × 7.5 / 5 = x

30 / 5 = x

6 = x

4 0

3 years ago

The point located at (3, -1) is reflected across the y-axis. what are it's new coordinates?

Whitepunk [10]

I believe the answer is (-3,-1)

3 0

3 years ago

Graph the following piecewise functions for the specified domain.

Anni [7]

Answer:

Answer is given in the pictures

Step-by-step explanation:

Refer the figure given to see the plots

7 0

3 years ago

Read 2 more answers

Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised t

zimovet [89]

Answer:

(b) 1/792

Step-by-step explanation:

The complete question is;

Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.

(a) How many different groups of 5 problems can be chosen from the 12

problems?

(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?

In (a)

Apply the formula

$\frac{n!}{(n-r)!(r!)}$

where n=12 and r=5

substitute values

$=\frac{12!}{(12-5)!(5!)} \\\\\\=\frac{12!}{(7!)(5!)} \\\\\\=\frac{12*11*10*9*8}{5*4*3*2*1} \\\\\\=\frac{95040}{120} \\\\\\=792$

In (b)

If Jerry did only 5 problems of one assignment then the probability will be

$\frac{5}{12} *\frac{4}{11} *\frac{3}{10} *\frac{2}{9} *\frac{1}{8} =\frac{1}{792}$

4 0

4 years ago