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Mazyrski [523]
3 years ago
14

An automated machine takes any cardboard rectangle and cuts off a square whose side length is equal to the shorter side length o

f the rectangle. Peter had a rectangle and, using only the machine, got out of it 2 large squares, 3 medium squares, and 5 small squares whose side length is 10 cm. What were the dimensions of Peter's rectangle?
Mathematics
2 answers:
oee [108]3 years ago
7 0

5 x 10 = 50

20+30=50

a=70 b=120

suter [353]3 years ago
3 0

Answer:

160 cm by 370 cm

Step-by-step explanation:

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Carlos bought 24 pounds of flour for $11 How many pounds of flour did he get per dollar?
Snezhnost [94]

Answer:

2.2 or 2.18

Step-by-step explanation:

3 0
3 years ago
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Can you please solve these equations by elimination? -10x -10y=20 , -7x -7y=14.
mash [69]

Answer:

-10x -10y=20

-10x - 20=10y

-10x - 20/10=y

-10x - 2=y

y=-10x-2

-7x -7y=14

-7x - 14 = 7y

-7x - 14/7=y

-7x - 2=y

y=-7x-2

6 0
3 years ago
Can yall help me with this please
puteri [66]

Answer:

1 x=4

y=-1

2 has no solution

3 x=5y -/8 + 27/4

y= 54-8x/5

4x=y=0

5 x=2 y=1

Step-by-step explanation:

8 0
2 years ago
True or false each triangular face has the same base and height​
Montano1993 [528]
True is going to be the correct answer.
5 0
2 years ago
In an arithmetic sequence, a17 = -40 and
Viktor [21]

Answer:

Tn = 2Tn-1 - Tn-2

Step-by-step explanation:

Before we can generate the recursive sequence, we need to find the nth term of the given sequence.

nth term of an AP is given as:

Tn = a+(n-1)d

If a17 = -40

T17 = a+(17-1)d = -40

a+16d = -40 ...(1)

If a28 = -73

T28 = a+(28-1)d = -73

a+27d = -73 ...(2)

Solving both equations simultaneously using elimination method.

Subtracting 1 from 2 we have:

27d - 16d = -73-(-40)

11d = -73+40

11d = -33

d = -3

Substituting d = -3 into 1

a+16(-3) = -40

a - 48 = -40

a = -40+48

a = 8

Given a = 8, d = -3, the nth term of the sequence will be

Tn = 8+(n-1) (-3)

Tn = 8+(-3n+3)

Tn = 8-3n+3

Tn = 11-3n

Given Tn = 11-3n and d = -3

Tn-1 = Tn - d... (3)

Tn-1 = 11-3n +3

Tn-1 = 14-3n

Tn-2 = Tn-2d...(4)

Tn-2 = 11-3n-2(-3)

Tn-2 = 11-3n+6

Tn-2 = 17-3n

From 3, d = Tn - Tn-1

From 4, d = (Tn - Tn-2)/2

Equating both common difference

(Tn - Tn-2)/2 = Tn - Tn-1

Tn - Tn-2 = 2(Tn - Tn-1)

Tn - Tn-2 = 2Tn-2Tn-1

2Tn-Tn = 2Tn-1 - Tn-2

Tn = 2Tn-1 - Tn-2

The recursive formula will be

Tn = 2Tn-1 - Tn-2

5 0
3 years ago
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