The width of a rectangle is 61 centimeters more than the length. The perimeter is 406 centimeters. Find the length and the width.

Answers

Answer 1

Answer:

$Step \; 1: \; Assign \; Variables \; for \; the \; unknown \; that \; we \; need \; to \; find$

$Let \; x \; be \; length \; of \; the \; rectangle$

$Step \; 2: \; Set \; up \; equation \; based \; on \; information \;\n given \; about \; the \; rectangle$

$Statement \; 1: Width \; of \; a \; rectangle \; \nis \; 61cm \; more \; than \; the \; length\n\nWidth \; = \; 61+x\n\nStatement \; 2: \; The \; perimeter \; is \; 406cm\n\nPerimeter=2(Length+Width)\nPerimeter =2(x+61+x)\n\nSo \; the \; mathematical \; equation \; would \; be \n 2(x+61+x)=406$

$Step \; 3: \; Solve \; the \; equation \; by \n undoing \; whatever \; is \; done \; x.\n\n2(x+61+x)=406\nGroup \; and \; Combine \; like \; terms \; inside \; the \; parenthesis\n\n2(2x+61)=406\nDistribute \; 2 \; in \; the \; left \; side \; of \; the \; equation\n\n4x+122=406\nSubtract \; 122 \; on \; both \; sides\n\n4x+122-122=406-122\nSimplify \; on \; both \; sides\n\n4x=284\nDivide \; on \; both \; sides\n\n(4x)/(4)=(284)/(4)\nSimplify \; fractions \; on \; both \; sides\n\nx=71$

$Conclusion:\nLength=x=71cm\nSubstituting \; 71 \; for \; x \; and \; find \; Width \; value.\nWidth=61+x=71+61=132cm\n\nLength \; is \; 71 cm \; and \; Width \; is 132cm$

Answer 2

Answer:

Final answer:

The length of the rectangle is 71 centimeters and the width is 132 centimeters.

Explanation:

To find the length and width of the rectangle, we can set up a system of equations. Let's denote the length of the rectangle as L and the width as W. We know that W = L + 61. The formula for the perimeter of a rectangle is P = 2L + 2W. Plugging in the given values, we have 406 = 2L + 2(L + 61). Simplifying this equation, we get 406 = 4L + 122. Subtracting 122 from both sides, we obtain 284 = 4L. Dividing both sides by 4, we get L = 71. Finally, substituting the value of L into the equation W = L + 61, we find W = 71 + 61 = 132. Therefore, the length of the rectangle measures 71 centimeters and the width measures 132 centimeters.

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Whats times what equals negative 48 and adds up to negative 2

Answers

The two figures that multiply to-48 and add up to-2 are-8 and 6.

Let's assume the two figures are x and y.

According to the problem, we've two conditions

x y = -48

x y = -2

To break this, we can rewrite the alternate equation as

x = -2- y

Now substitute this value of x in the first equation

(-2 - y) y = -48

- 2y- y² = -48

Rearranging the terms

y² + 2y- 48 = 0

Now, solving the Quadratic Equation

(y+ 8)( y- 6) = 0

Setting each factor to zero

y + 8 = 0--> y = -8

y- 6 = 0--> y = 6

So we've two possible values for y is -8 and 6.

Substituting these values back into the equation x + y = -2, we can break for x

For y = -8

x(- 8) = -2

x- 8 = -2

x = 6

For y = 6

x 6 = -2

x = -2- 6

x = -8

Thus, the two figures that multiply to-48 and add up to-2 are-8 and 6.

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6 × -8 = -48

6 + -8 = -2

What is a 2c − 5b − c a − b simplified?

Answers

All you have to do is combine like terms. 2c and (-c) are like terms (-5b) and (-b) are like terms and a is a term. So, 2c combined with (-c) is C. (-5b) and (-b) combined is (-6b). Since a doesnt need to be combined with anything, you leave it as it is so your answer will then be c - 6b + a (this is the answer assuming that there is a + in front of the a. If there is a - sign, then just change your +a to -a at the end

The tuition costs, C, for a local community college are modeled by C(h) = 250 + 200h, where h represents the number of credit hours taken. The local state university has tuition costs, S, modeled by the function S(h) = 300 + 180h. How many credit hours will a student have to take for the two tuition costs to be equal? Round the answer to the nearest tenth of an hour. 250 + 200h = 300 + 180h 250 + 200h = 300 + 180h − 180h − 180h 250 + 20h = 300 h = credit hours

Answers

To solve how many credit hours will a student have to take for the two tuition costs to be equal, the two functions should be equated and solve for the number of hours
C (h) = S (h)
250 + 200h = 300 + 180h
200h – 180h = 300 – 250
20h = 50
H = 2.5 credit hours

The required equation is: $\mathbf{150 + 200h = 300 + 180h}$ and the number of credit hours is 7.5