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BartSMP [9]
2 years ago
5

On a class test, Lesley scored 4.5 points more than her friend Cathy. Lesley scored 81.5 points. Cathy's score, x, can be found

using the equation below.
The solution to the equation was calculated to be x = 36. Is this solution reasonable?

A.
Yes, it is reasonable because Lesley scored 4.5 points more than Cathy. Cathy scored more than 81.5 points.
B.
No, it is not reasonable because Lesley scored 4.5 points less than Cathy. Cathy scored less than 81.5 points, not more.
C.
Yes, it is reasonable because Lesley scored 4.5 points more than Cathy. Cathy scored 86 points.
D.
No, it is not reasonable because Lesley scored 4.5 points more than Cathy. Cathy scored less than 81.5 points, not more.
Mathematics
1 answer:
yaroslaw [1]2 years ago
3 0

Answer: B

Step-by-step explanation:

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In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t
Sergeeva-Olga [200]

Answer:

Perimeter  = (2 + √3)·a

Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

MB = \frac{a}{2}

Now in ΔAMB , By using Pythagoras theorem

AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}

Now, in ΔBMP :

sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}

Perimeter of ABM = AB + PB + PM + AM

\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a

Hence, Perimeter of ΔABP = (2 + √3)·a units

3 0
2 years ago
A side of a regular six - sided polygon is 8cm long. The perimeter of a similar polygon is 60cm. What is the length of a side of
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Answer:

  10 cm

Step-by-step explanation:

The larger polygon is also a regular 6-sided polygon, so each side is 1/6 of the perimeter length:

  (60 cm)/6 = 10 cm . . . length of a side

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inysia [295]
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Just add all of them and that's it :)
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For each system of equations, drag the true statement about its solution set to the box under the system?
natta225 [31]

Answer:

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

 of x and y are the same and the numerical terms are different

- The system of equation has infinity many solutions if the

   coefficients of x and y are the same and the numerical terms

   are the same

- The system of equation has one solution if at least one of the

  coefficient of x and y are different

* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

  numerical terms

∴ They have zero equation

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

- Substitute the value of x in equation (1) or (2) to find y

∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

3 0
2 years ago
The diameter of Circle Q terminates on the circumference of the circle at (0,3) and (0,-4). Write the equation of the circle in
Gnesinka [82]
First, determine the center of the circle by getting the midpoint of the points given for the circumference.
                    midpoint = ((0 + 0)/2, (3 + -4)/2)
                          midpoint (0, -0.5)
Then, we get the radius by determining the distance from either of the circumferential point to the center. 
                        radius = √(0 -  0)² + (3 +4)²  = 7
The equation for the circle would be,
                        x² + (y + 0.5)² = 7²
8 0
2 years ago
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