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andrew11 [14]
3 years ago
6

The point of intersection of the diagonals of a rectangle is 4 cm further away from the smaller side then from the larger side o

f the rectangle. The perimeter of the rectangle is equal to 56 cm. Find the lengths of the sides of the rectangle.
Mathematics
1 answer:
tino4ka555 [31]3 years ago
5 0

Answer:

  The rectangle is 10 cm by 18 cm.

Step-by-step explanation:

Let x represent the distance from the center of the rectangle to the longer side. Then the distance to the shorter side is x+4. The short side of the rectangle will have length 2x, and the long side will have length 2(x+4) = 2x+8.

The perimeter is twice the sum of the side lengths, so is ...

  56 = 2(2x +(2x+8)) = 8x +16

  40 = 8x . . . . . . . subtract 16

  40/8 = x = 5 . . . divide by the coefficient of x

The side lengths are 2x=10 and 2x+8 = 18 centimeters.

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Csc 0 = r/y
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2 years ago
Evaluate the Riemann sum for f(x) = 3 - 1/2 times x between 2 and 14 where the endpoints are included with six subintervals taki
Digiron [165]

Answer:

-6

Step-by-step explanation:

Given that :

we are to evaluate the Riemann sum for f(x) = 3 - \dfrac{1}{2}x from 2 ≤ x ≤ 14

where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.

The Riemann sum can be computed as follows:

L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x

where:

\Delta x = \dfrac{b-a}{a}

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n = 6

∴

\Delta x = \dfrac{14-2}{6}

\Delta x = \dfrac{12}{6}

\Delta x =2

Hence;

x_0 = 2 \\ \\  x_1 = 2+2 =4\\ \\  x_2 = 2 + 2(2) \\ \\  x_i = 2 + 2i

Here, we are  using left end-points, then:

x_i-1 = 2+ 2(i-1)

Replacing it into Riemann equation;

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix}  2+2 (i-1)  \end {pmatrix} \end {pmatrix}2

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2i-2)

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -   \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - 2  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} i

Estimating the integrals, we have :

= 6n - 2 ( \dfrac{n(n-1)}{2})

= 6n - n(n+1)

replacing thevalue of n = 6 (i.e the sub interval number), we have:

= 6(6) - 6(6+1)

= 36 - 36 -6

= -6

5 0
3 years ago
Please help me solve inequalities and show work. Will get brainiest!!
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The answer depends on whether you're looking for the x or the y. If y, the answers are the ones on the left, and the x on the right.

1) -4x + 3y ≤ 15        -4x + 3y <span>≤ 15
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</span><span>     ---     ---------          ----   ------------
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2)  -6x - 4y </span>< 12         -6x - 4y < 12<span>
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