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3 years ago

Please help me, I’m struggling!

Mathematics

2 answers:

Lesechka [4]3 years ago

7 0

Answer:

a. 1/2

b. 5/6

c. 4/9

d. 12/5

e. 2/9

f. 2/5

Step-by-step explanation:

a. 1/3 < 1/2

1(2)= 2. 3(2)= 6. 1/6

1(3)= 3. 2(3)=6. 3/6

b. 4/5 < 5/6

4(6)= 24. 5(6)= 30. 24/30

5(5)= 25. 6(5)= 30. 25/30

c. 3/8 < 4/9

3(9)= 27. 8(9)= 72. 27/72

4(8)= 32. 9(8)= 72. 32/72

d. 12/5 > 9/4

12(4)= 48. 5(4)= 20. 48/20

9(5)= 45. 4(5)= 20. 45/20

e. 1/5 < 2/9

1(9)= 9. 5(9)= 45. 9/45

2(5)= 10. 9(5)=45. 10/45

f. 2/5 > 3/9

2(9)= 18. 5(9)= 45. 18/45

3(5)= 15. 9(5)= 45. 15/45

Leno4ka [110]3 years ago

3 0

Answer:

Multiply the numbers until you get the same denominator (Bottom number) for both fractions. I will do #1 to show you.

1/3 1/2

Both 3 and 2 go into 6, so I will use that.

When multiplying the bottom , you have to do the same to the top, so...

2/6 OR 3/6

3/6 is greater, so that is the answer.

Therefore, 1/2 is more

Another way you can do this is to draw a pizza, a section off parts of it according to the fractions.

Hope this helped.

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Together their speed will be

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If they drill together, they will spend

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Answer: it would have taken nearly 3 years.

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3 years ago

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2 years ago

If the probability of an events is 2/7 what must be the probability of its compliments

monitta

Answer:

The probability of its compliments = 0.7143 = 5/7

Step-by-step explanation:

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3 years ago

Factor the expression using the GCF. 50x−60

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3 years ago

Instructions:Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has

Akimi4 [234]

to compare the triangles, first we will determine the distances of each side

Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
Solving

∆ABC A(11, 6), B(5, 6), and C(5, 17)

AB = 6 units BC = 11 units AC = 12.53 units
∆XYZ X(-10, 5), Y(-12, -2), and Z(-4, 15)
XY = 7.14 units YZ = 18.79 units XZ = 11.66 units

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

MN = 6 units NO = 11 units MO = 12.53 units
∆JKL J(17, -2), K(12, -2), and L(12, 7).
JK = 5 units KL = 9 units JL = 10.30 units
∆PQR P(12, 3), Q(12, -2), and R(3, -2)
PQ = 5 units QR = 9 units PR = 10.30 units
Therefore
we have the ∆ABC and the ∆MNO 
with all three sides equal ---------> are congruent
we have the ∆JKL and the ∆PQR
with all three sides equal ---------> are congruent

let's check

Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

1) If ∆MNO ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC

then ∆MNO ≅ ∆ABC

a) Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

M(-9, -4)-----------------> M1(-9,4)

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

b) Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

∆M1N1O1 M1(-9, 4), N1(-3, 4), and O1(-3, 15).

M1(-9, -4)-----------------> M2(9,4)

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

c) Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

∆M2N2O2 M2(9,4), N2(3,4), and O2(3, 15).

M2(9, 4)-----------------> M3(11,6)=A

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

∆ABC A(11, 6), B(5, 6), and C(5, 17)

∆MNO reflection------- > ∆M1N1O1 reflection---- > ∆M2N2O2 translation -- --> ∆M3N3O3

The ∆M3N3O3=∆ABC

Therefore ∆MNO ≅ ∆ABC - > check list

2) If ∆JKL -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR

then ∆JKL ≅ ∆PQR

d) Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

∆JKL J(17, -2), K(12, -2), and L(12, 7).

J(17, -2)-----------------> J1(2,17)

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

e) translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

∆J1K1L1 J1(2, 17), K1(2, 12), and L1(-7, 12).

J1(2, 17)-----------------> J2(12,3)=P

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

∆PQR P(12, 3), Q(12, -2), and R(3, -2)

∆JKL rotation------- > ∆J1K1L1 translation -- --> ∆J2K2L2=∆PQR

Therefore ∆JKL ≅ ∆PQR - > check list

6 0

3 years ago