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3 years ago

Maggie needs to spend at least six hours each week practicing the piano. She has already practiced 3

Mathematics

1 answer:

grandymaker [24]3 years ago

8 0

Answer:

t ≥ 1.5

Step-by-step explanation:

Lets take time for practicing the piano to be t

The number of hours to practice per week should be at least 6 hours

This is written as t ≥ 6

She already practiced 3 hours this week, thus the remaining hours to practice should be

t ≥ 3

The minimum remaining hours to practice for the remaining 2 days should be

t=3

If these hours are evenly divided, then in a single day she should practice

for t ≥ 1.5

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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

Sample Work Help Needed! Please answer these questions and show the steps used to solve them. NO DECIMAL ANSWERS (except on thre

anygoal [31]

1. Using the exponent rule (a^b)·(a^c) = a^(b+c) ...

$\left(-2x^3y^4\right)\left(5x^9y^{-2}\right)=(-2)(5)\left(x^{(3+9)}y^{(4-2)}\right)\\\\=-10x^{12}y^{2}$

Simplify. Write in Scientific Notation

2. You know that 256 = 2.56·100 = 2.56·10². After that, we use the same rule for exponents as above.

$8 \left(32\times 10^{11}\right)=(8)(32)\times 10^{11}=256\times 10^{11}=2.56\times 10^{13}$

3. The distributive property is useful for this.

(3x – 1)(5x + 4) = (3x)(5x + 4) – 1(5x + 4)

... = 15x² +12x – 5x –4

... = 15x² +7x -4

4. Look for factors of 8·(-3) = -24 that add to give 2, the x-coefficient.

-24 = -1×24 = -2×12 = -3×8 = -4×6

The last pair of factors adds to give 2. Now we can write

... (8x -4)(8x +6)/8 . . . . . where each of the instances of 8 is an instance of the coefficient of x² in the original expression. Factoring 4 from the first factor and 2 from the second factor gives

... (2x -1)(4x +3) . . . . . the factorization you require

8 0

3 years ago

I. Tell whether each statement is TRUE or FALSE. Refer to the figure below. Write your answer on the space provided.

Vika [28.1K]

Answer:

1. TRUE

2. TRUE

3. TRUE

4. TRUE

5. FALSE

6. TRUE

Step-by-step explanation:

1. From the diagram, we have that m∠2 and m∠7 are alternate interior angles

If m∠2 = 70° and m∠7 = 70°, then we have;

m∠2 = 70° = m∠7

m∠2 = m∠7

Therefore, the alternate interior angles of the lines l₁, and l₂ are equal, and therefore, the lines l₁ and l₂ are parallel, l₁ ║ l₂

TRUE

2. If m∠3 = 90° and m∠7 = 90°, therefore, the angle formed by the intersection of l₃ and l₂ = 90°

Therefore l₃ ⊥ l₂

TRUE

3. m∠5 and m∠7 are corresponding angles

If m∠5 = 85° and m∠7 = 85°, then, m∠5 = m∠7

Therefore, the corresponding angles formed by the lines l₁ and l₂ are equal, therefore;

l₁ ║ l₂

TRUE

4. Whereby we have, m∠1 = m∠5, we get;

m∠1 + m∠5 = 180° by sum of angles on a straight line

∴ m∠1 + m∠5 = m∠1 + m∠1 = 2·m∠1 = 180°

m∠1 = 180°/2 = 90°

∴ m∠1 = 90° = m∠5, and l₃ ⊥ l₁

TRUE

5. m∠1 and m∠8 are alternate exterior angles

If m∠1 = 98° and m∠8 = 82°

∴ m∠1 ≠ m∠8 and l₁ ∦ l₈

FALSE

6. Given that l₁║ l₂, then the angle formed between l₁ and l₃ will be equal to th angle formed between l₂ and l₃

Therefore;

If l₁║ l₂, and l₃ ⊥ l₁, then l₃ ⊥ l₂

(If l₁║ l₂, and l₃ is perpendicular to l₁, then l₃ is also perpendicular to l₂)

TRUE

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mamaluj [8]

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Step-by-step explanation:

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