The first 5 figures in a pattern are shown. If the patterncontinues, which expression can be used to find the number of circles
to make up Figure n?
f.n^2+ 2n
h. 2n^2+ 1
g. n^2+ 2
j. 2n^2+n
f.n^2+ 2n
h. 2n^2+ 1
g. n^2+ 2
j. 2n^2+n
2 answers:
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9514 1404 393
Answer:
(i) x° = 70°, y° = 20°
(ii) ∠BAC ≈ 50.2°
(iii) 120
(iv) 300
Step-by-step explanation:
(i) Angle x° is congruent with the one marked 70°, as they are "alternate interior angles" with respect to the parallel north-south lines and transversal AB.
x = 70
The angle marked y° is the supplement to the one marked 160°.
y = 20
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(ii) The triangle interior angle at B is x° +y° = 70° +20° = 90°, so triangle ABC is a right triangle. With respect to angle BAC, side BA is adjacent, and side BC is opposite. Then ...
tan(∠BAC) = BC/BA = 120/100 = 1.2
∠BAC = arctan(1.2) ≈ 50.2°
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(iii) The bearing of C from A is the sum of the bearing of B from A and angle BAC.
bearing of C = 70° +50.2° = 120.2°
The three-digit bearing of C from A is 120.
__
(iv) The bearing of A from C is 180 added to the bearing of C from A:
120 +180 = 300
The three-digit bearing of A from C is 300.
To find the z-score for a weight of 196 oz., use
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.