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Ulleksa [173]
3 years ago
9

A SQUARE CARPET IS LAID IN ONE CORNER OF A RECTANGULAR ROOM, LEAVING STRIPS OF UNCOVERED FLOOR 2M WIDE ALONG ONE SIDE AND 1M ALO

NG OTHER . THE AREA OF THE ROOM IS 56m SQUARED .FIND THE DIMENSIONS OF THE CARPET

Mathematics
1 answer:
Nat2105 [25]3 years ago
8 0

Answer:

Step-by-step explanation:

A square has equal sides. Let x represent the length of each side of the square carpet. The diagram representing the room and the carpet is shown in the attached photo. Therefore, the length of the room would be (x + 2)m while the width of the room would be (x + 1)m

Since the area of the room is 56m², it means that

(x + 2)(x + 1) = 56

x² + x + 2x + 2 = 56

x² + 3x + 2 - 56 = 0

x² + 3x - 54 = 0

x² + 9x - 6x - 54 = 0

x(x + 9) - 6(x + 9) = 0

x - 6 = 0 or x + 9 = 0

x = 6 or x = - 9

Since the dimension of the carpet cannot be negative, then x = 6

The dimension of the carpet is 6m × 6m

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For an angle θ with the point (−5, −12) on its terminating side, what is the value of cosine?
viva [34]

Answer:

option C. -\frac{5}{13}

Step-by-step explanation:

we have  that

The point (-5,-12) belong to the III quadrant

so

The value of the cosine is negative

Applying the Pythagoras Theorem

Find the value of the hypotenuse

h^{2}=5^{2}+12^{2}\\ h^{2} =25+144\\ h^{2}=169\\h=13\ units

The value of cosine of angle θ is the ratio between the side adjacent to angle θ and the hypotenuse

cos(\theta)=-\frac{5}{13}

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3 years ago
Find The difference 25(d−10)−23(d+6) is
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Answer:

2d-388 or 2(d-194)

Step-by-step explanation:

25(d-10)-23(d+6)

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25d-23d-250-138

2d-250-138

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3 years ago
Suppose x=c1e−t+c2e3tx=c1e−t+c2e3t. Verify that x=c1e−t+c2e3tx=c1e−t+c2e3t is a solution to x′′−2x′−3x=0x′′−2x′−3x=0 by substitu
Harrizon [31]

The correct question is:

Suppose x = c1e^(-t) + c2e^(3t) a solution to x''- 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)

Answer:

x = c1e^(-t) + c2e^(3t)

is a solution to the differential equation

x''- 2x' - 3x = 0

Step-by-step explanation:

We need to verify that

x = c1e^(-t) + c2e^(3t)

is a solution to the differential equation

x''- 2x' - 3x = 0

We differentiate

x = c1e^(-t) + c2e^(3t)

twice in succession, and substitute the values of x, x', and x'' into the differential equation

x''- 2x' - 3x = 0

and see if it is satisfied.

Let us do that.

x = c1e^(-t) + c2e^(3t)

x' = -c1e^(-t) + 3c2e^(3t)

x'' = c1e^(-t) + 9c2e^(3t)

Now,

x''- 2x' - 3x = [c1e^(-t) + 9c2e^(3t)] - 2[-c1e^(-t) + 3c2e^(3t)] - 3[c1e^(-t) + c2e^(3t)]

= (1 + 2 - 3)c1e^(-t) + (9 - 6 - 3)c2e^(3t)

= 0

Therefore, the differential equation is satisfied, and hence, x is a solution.

4 0
3 years ago
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