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Umnica [9.8K]
3 years ago
15

A square with side lengths of 3 cm is reflected vertically over a horizontal line of reflection that is 2 cm below the bottom ed

ge of the square. What is the distance between the points C and C’? cm What is the perpendicular distance between the point B and the line of reflection? cm What is the distance between the points A and A’? cm

Mathematics
1 answer:
oksian1 [2.3K]3 years ago
5 0

Answer:

a) 4 cm

b) 5 cm

c) 10 cm

Step-by-step explanation:

The side lengths of the reflected square are equal to the original, and the distance from the axis(2) also remains the same.  From there, it is just addition.

Hope it helps <3

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Ship A receives a distress signal from the north, And ship B receives a distress signal from the same vessel from the southeast.
11Alexandr11 [23.1K]

The coordinates of Ship A and B are missing, so i have attached it

Answer:

The vessel in distress is located at the coordinate (1, 2)

Step-by-step explanation:

The coordinates attached shows that;

Coordinates of A are (3, 4) while that of B are (1, 1).

We are told that Ship A receives a distress signal from the north, And ship B receives a distress signal from the same vessel from the southeast.

Now, from the image attached, If we imagine extending the line of ship A that is getting distress signal from southeast and also same thing for ship B that is getting signal from north,we'll discover that the lines intersect at a point with co-ordinates of approximately; (1, 2)

8 0
3 years ago
10. An expression is shown.
dolphi86 [110]

Answer: -0.25n-0.15

Step-by-step explanation:

simplify

0.5n+0.3-0.75n-0.45

Combine like terms

-0.25n-0.15

6 0
1 year ago
What is the solution for the equation 5/3b^3-2b-5=2/b3-2
irinina [24]

Answer:

Option C is correct.

Step-by-step explanation:

Given Equation:

\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

To find: Solution of the Equation.

Consider,

5(b^3-2)=2(3b^3-2b^2-5)

5b^3-10=6b^3-4b^2-10

6b^3-5b^3-4b^2+10-10=0

b^3-4b^2=0

b^2(b-4)=0

b² = 0   ⇒ b = 0

b - 4 = 0 ⇒ b = 4

Therefore, Option C is correct.

7 0
3 years ago
Read 2 more answers
Solve the system of equations. y = x − 5 and y = x2 − 5x + 3 A. (-4,-9) and (-2,-7) B. (4,-9) and (2,-7) C. (4,-3) and (2,-1) D.
larisa [96]

Answer:

D

Step-by-step explanation:

Given the 2 equations

y = x - 5 → (1)

y = x² - 5x + 3 → (2)

Substitute y = x² - 5x + 3 into (1)

x² - 5x + 3 = x - 5 ← subtract x - 5 from both sides

x² - 6x + 8 = 0 ← in standard form

(x - 2)(x - 4) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 2 = 0 ⇒ x = 2

x - 4 = 0 ⇒ x = 4

Substitute each of these values into (1) for corresponding values of y

x = 2 → y = 2 - 5 = - 3 ⇒ (2, - 3 )

x = 4 → y = 4 - 5 = - 1 ⇒ (4, - 1 )

4 0
3 years ago
Read 2 more answers
How many solutions does 5x-y=2 5x-y=-2 have?
alekssr [168]

Answer:

No Solution.

Step-by-step explanation:

5x-y=2

5x-y=-2

-------------

-(5x-y)=-2

5x-y=-2

----------------

-5x+y=-2

5x-y=-2

-------------

0=-4

no solution

7 0
3 years ago
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