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Ipatiy [6.2K]
3 years ago
6

Which polynomial, when added to the polynomial 5x^2–3x–9, is equivalent to: x^2–5x+6?

Mathematics
1 answer:
inn [45]3 years ago
8 0
Look at the attachment

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Select the correct ratios.
Goshia [24]
65:100 because there are 65 girls and then 65 girls + 35 boys = 100 total students
4 0
2 years ago
How many ways can nine skiers finish in first, second, and third place? "fill in the blank text field 1"?
Yuri [45]
The answer would be 9 Permutation 3 (or 9P3) which is equal to 504 possible ways.
3 0
3 years ago
Given m<12= 121 and m<6= 75, find the measure of each missing angle.
slava [35]

Answer/Step-by-step explanation:

Given:

m<12 = 121°

m<6 = 75°

a. m<1 = m<6 (vertical angles)

m<1 = 75° (substitution)

b. m<12 = m<1 + m2 (alternate exterior angles)

121° = 75° + m<2 (substitution)

121° - 75° = m<2 (subtraction property of equality)

46° = m<2

m<2 = 46°

c. m<1 + m<2 + m<3 = 180° (angles on a straight line)

75° + 46° + m<3 = 180° (substitution)

121° + m<3 = 180°

m<3 = 180° - 121° (subtraction property of equality)

m<3 = 59°

d. m<4 = m<3 (vertical angles)

m<4 = 59° (substitution)

e. m<5 + m<4 + m<6 = 180° (angles on a straight line)

m<5 + 59° + 75° = 180° (substitution)

m<5 + 134° = 180°

m<5 = 180° - 134° (Subtraction property of equality)

m<5 = 46°

f. m<7 = m<12 (vertical angles)

m<7 = 121° (substitution)

g. m<8 = m<4 (vertical angles)

m<8 = 59° (substitution)

h. m<9 = m<6 (Alternate Interior Angles)

m<9 = 75° (substitution)

i. m<10 + m<9 = 180° (Linear Pair)

m<10 + 75° = 180° (substitution)

m<10 = 180° - 75° (Subtraction property of equality)

m<10 = 105°

j. m<11 = m<8 (vertical angles)

m<11 = 59° (substitution)

k. m<13 = m<10 (vertical angles)

m<13 = 105° (substitution)

l. m<14 = m<9 (vertical angles)

m<14 = 75° (substitution)

5 0
3 years ago
Can I please get help on this one.
irina1246 [14]
mmmmmmmmmkkkkkkkkkkk
3 0
2 years ago
The class sizes of the introductory psychology courses at a college are shown below 121,134,106,93,149,130,119,128 the college a
frozen [14]
Center : Mean  Before the introduction of the new course, center = average(121,134,106,93,149,130,119,128) = 122.5  After the introduction of the new course, center =  average(121,134,106,93,149,130,119,128,45) = 113.9  The center has moved to the left (if plotted in a graph) because of the low intake for the new course.  Spread before introduction of the new course :  Arrange the numbers in ascending order:  (93, 106,119, 121), (128, 130,134, 149) Q1=median(93,106,119,121) = 112.5 Q3=median(128,130,134,149) = 132  Spread = Interquartile range = Q3-Q1 = 19.5  After addition of the new course,

 (45,93, 106,119,) 121, (128, 130,134, 149)
 Q1=median(45,93,106,119)=99.5
 Q3=median (128, 130,134, 149)= 132
 Spread = Interquartile range = 132-99.5 =32.5 
 We see that the spread has increased after the addition of the new course.
3 0
3 years ago
Read 2 more answers
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